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1l6dwfnc0 | maths | 3d-geometry | lines-and-plane | <p>Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the point $$(2,-5,11)$$, then $$\mathrm{d}^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{147}{2}$$"}, {"identifier": "B", "content": "96"}, {"identifier": "C", "content": "$$\\frac{32}{3}$$"}, {"identifier": "D", "content": "54"}] | ["C"] | null | Let $\langle a, b, c\rangle$ be direction ratios of plane containing
<br/><br/>
$$
\begin{aligned}
&\text { lines } \frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text { and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8} . \\\\
&\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text { (i) } \\\\
&\text { and } 3 a+7 b+8 c=0 \quad \ldots \text { (ii) } \\\\
&\text { from eq. (i) and (ii) }: \frac{a}{24-35}=\frac{b}{15-16}=\frac{c}{14-9} \\\\
&\therefore \text { D. R's} \text {. of plane are }<11,1,-5>
\end{aligned}
$$
<br/><br/>
Let $D . R$'s of plane $P$ be $< a_{1}, b_{1}, c_{1} >$ then.
<br/><br/>
$11 a_{1}+b_{1}-5 c_{1}=0$
<br/><br/>
and $9 a_{1}-b_{1}-5 c_{1}=0$
<br/><br/>
From eq. (iii) and (iv) :
<br/><br/>
$$
\begin{aligned}
&\frac{a_{1}}{-5-5}=\frac{b_{1}}{-45+55}=\frac{c_{1}}{-11-9} \\\\
&\therefore \text { D.R's } \text { of plane } P \text { are }< 1,-1,2 >
\end{aligned}
$$
<br/><br/>
Equation plane $P$ is : $1(x-3)-1(y+4)+2(z-7)=0$
<br/><br/>
$$
\Rightarrow x-y+2 z-21=0
$$
<br/><br/>
Distance from point $(2,-5,11)$ is $d=\frac{|2+5+22-2|}{\sqrt{6}}$<br/><br/> $\therefore d^{2}=\frac{32}{3}$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6dxjgo4 | maths | 3d-geometry | lines-and-plane | <p>The line of shortest distance between the lines $$\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$$ and $$\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$$ makes an angle of $$\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$$ with the plane $$\mathrm{P}: \mathrm{a} x-y-z=0$$, $$(a>0)$$. If the image of the point $$(1,1,-5)$$ in the plane $$P$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta-\gamma$$ is equal to _________________.</p> | [] | null | 3 | DR's of line of shortest distance<br/><br/>
$$
\left|\begin{array}{lll}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
0 & 1 & 1 \\
2 & 2 & 1
\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}
$$<br/><br/>
angle between line and plane is $\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha$<br/><br/>
$$
\cos \alpha=\sqrt{\frac{2}{27}}, \sin \alpha=\frac{5}{3 \sqrt{3}}
$$<br/><br/>
DR's normal to plane $(1,-1,-1)$<br/><br/>
$$
\sin \alpha=\left|\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^2+1+1}}\right|=\frac{5}{3 \sqrt{3}}
$$<br/><br/>
$\sqrt{3}|a|=5 \sqrt{a^2+2}$<br/><br/>
$$
3 a^2=25 a^2+50
$$<br/><br/>
No value of (a) | integer | jee-main-2022-online-25th-july-morning-shift |
1l6hzmtfd | maths | 3d-geometry | lines-and-plane | <p>The largest value of $$a$$, for which the perpendicular distance of the plane containing the lines $$
\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a \hat{j}-\hat{k})$$ and $$\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a \hat{k})$$ from the point $$(2,1,4)$$ is $$\sqrt{3}$$, is _________.</p> | [] | null | 2 | <p>Normal to plane $$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & a & { - 1} \cr
{ - 1} & 1 & { - a} \cr
} } \right|$$</p>
<p>$$ = \widehat i(1 - {a^2}) - \widehat j( - a - 1) + \widehat k(1 + a)$$</p>
<p>$$ = (1 - a)\widehat i + \widehat j + \widehat k$$</p>
<p>$$\therefore$$ Plane $$(1 - a)(x - 1) + (y - 1) + z = 0$$</p>
<p>Distance from (2, 1, 4) is $$\sqrt 3 $$ i.e.</p>
<p>$$ \Rightarrow \left| {{{(1 - a) + 0 + 4} \over {\sqrt {{{(1 - a)}^2} + 1 + 1} }}} \right| = \sqrt 3 $$</p>
<p>$$ \Rightarrow 25 + {a^2} - 10a = 3{a^2} - 6a + 9$$</p>
<p>$$ \Rightarrow 2{a^2} + 4a - 16 = 0$$</p>
<p>$$ \Rightarrow {a^2} + 2a - 8 = 0$$</p>
<p>$$a = 2$$ or $$ - 4$$</p>
<p>$$\therefore$$ $${a_{\max }} = 2$$</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6hzyvwf | maths | 3d-geometry | lines-and-plane | <p>The plane passing through the line $$L: l x-y+3(1-l) z=1, x+2 y-z=2$$ and perpendicular to the plane $$3 x+2 y+z=6$$ is $$3 x-8 y+7 z=4$$. If $$\theta$$ is the acute angle between the line $$L$$ and the $$y$$-axis, then $$415 \cos ^{2} \theta$$ is equal to _____________.</p> | [] | null | 125 | <p>$$L:lx - y + 3(1 - l)z = 1$$, $$x + 2y - z = 2$$ and plane containing the line $$p:3x - 8y + 7z = 4$$</p>
<p>Let $$\overrightarrow n $$ be the vector parallel to L.</p>
<p>then $$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
l & { - 1} & {3(1 - l)} \cr
1 & 2 & { - 1} \cr
} } \right|$$</p>
<p>$$ = (6l - 5)\widehat i + (3 - 2l)\widehat j + (2l + 1)\widehat k$$</p>
<p>$$\because$$ R containing L</p>
<p>$$3(6l - 5) - 8(3 - 2l) + 7(2l + 1) = 0$$</p>
<p>$$18l + 16l + 14l - 15 - 24 + 7 = 0$$</p>
<p>$$\therefore$$ $$l = {{32} \over {48}} = {2 \over 3}$$</p>
<p>Let $$\theta$$ be the acute angle between L and y-axis</p>
<p>$$\therefore$$ $$\cos \theta = {{{5 \over 3}} \over {\sqrt {1 + {{25} \over 9} + {{49} \over 9}} }} = {5 \over {\sqrt {83} }}$$</p>
<p>$$\therefore$$ $$415{\cos ^2}\theta = 125$$</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6jdxtys | maths | 3d-geometry | lines-and-plane | <p>Let the line $$\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$$ intersect the plane containing the lines $$\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$$ and $$4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a \in \mathbb{R}$$ at the point $$P(\alpha, \beta, \gamma)$$. Then the value of $$\alpha+\beta+\gamma$$ equals _____________.</p> | [] | null | 12 | <p>Equation of plane containing the line $$4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3$$ can be written as</p>
<p>$$4ax - y + 5z - 7a + \lambda (2x - 5y - z - 3) = 0$$</p>
<p>$$(4a + 2\lambda )x - (1 + 5\lambda )y + (5 - \lambda )z - (7z + 3\lambda ) = 0$$</p>
<p>Which is coplanar with the line</p>
<p>$${{x - 4} \over 1} = {{y + 1} \over { - 2}} = {z \over 1}$$</p>
<p>$$4(4a + 2\lambda ) + (1 + 5\lambda ) - (7a + 3\lambda ) = 0$$</p>
<p>$$9a + 10\lambda + 1 = 0$$ ..... (1)</p>
<p>$$(4a + 2\lambda )1 + (1 + 5\lambda )2 + 5 - \lambda = 0$$</p>
<p>$$4a + 11\lambda + 7 = 0$$ ...... (2)</p>
<p>$$a = 1,\,\lambda = - 1$$</p>
<p>Equation of plane is $$x + 2y + 3z - 2 = 0$$</p>
<p>Intersection with the line</p>
<p>$${{x - 3} \over 7} = {{y - 2} \over { - 1}} = {{z - 3} \over { - 4}}$$</p>
<p>$$(7t + 3) + 2( - t + 2) + 3( - 4t + 3) - 2 = 0$$</p>
<p>$$ - 7t + 14 = 0$$</p>
<p>$$t = 2$$</p>
<p>So, the required point is $$(17,0, - 5)$$</p>
<p>$$\alpha + \beta + \gamma = 12$$</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6kkntgj | maths | 3d-geometry | lines-and-plane | <p>If the line of intersection of the planes $$a x+b y=3$$ and $$a x+b y+c z=0$$, a $$>0$$ makes an angle $$30^{\circ}$$ with the plane $$y-z+2=0$$, then the direction cosines of the line are :</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0$$"}, {"identifier": "B", "content": "$$\\frac{1}{\\sqrt{2}}, \\pm \\,\\frac{1}{\\sqrt{2}}, 0$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{5}},-\\frac{2}{\\sqrt{5}}, 0$$"}, {"identifier": "D", "content": "$$\\frac{1}{2},-\\frac{\\sqrt{3}}{2}, 0$$"}] | ["B"] | null | <p>$${P_1}:ax + by + 0z = 3$$, normal vector : $${\overrightarrow n _1} = (a,b,0)$$</p>
<p>$${P_2}:ax + by + cz = 0$$, normal vector : $${\overrightarrow n _2} = (a,b,c)$$</p>
<p>Vector parallel to the line of intersection $$ = {\overrightarrow n _1} \times {\overrightarrow n _2}$$</p>
<p>$${\overrightarrow n _1} \times {\overrightarrow n _2} = (bc, - ac,0)$$</p>
<p>Vector normal to $$0\,.\,x + y - z + 2 = 0$$ is $${\overrightarrow n _3} = (0,1, - 1)$$</p>
<p>Angle between line and plane is 30$$^\circ$$</p>
<p>$$ \Rightarrow \left| {{{0 - ac + 0} \over {\sqrt {{b^2}{c^2} + {c^2}{a^2}} \sqrt 2 }}} \right| = {1 \over 2}$$</p>
<p>$$ \Rightarrow {a^2} = {b^2}$$</p>
<p>Hence, $${\overrightarrow n _1} \times {\overrightarrow n _2} = (ac, - ac,0)$$</p>
<p>Direction ratios $$ = \left( {{1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }},0} \right)$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nnkfa5 | maths | 3d-geometry | lines-and-plane | <p>Let the lines <br/><br/>$$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$$ and <br/><br/>$$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$$ be coplanar <br/><br/>and $$\mathrm{P}$$ be the plane containing these two lines. <br/><br/>Then which of the following points does <b>NOT</b> lie on P?</p> | [{"identifier": "A", "content": "$$(0,-2,-2)$$"}, {"identifier": "B", "content": "$$(-5,0,-1)$$"}, {"identifier": "C", "content": "$$(3,-1,0)$$"}, {"identifier": "D", "content": "$$(0,4,5)$$"}] | ["D"] | null | <p>$${L_1}:{{x - 1} \over \lambda } = {{y - 2} \over 1} = {{z - 3} \over 2}$$,</p>
<p>through a point $${\overrightarrow a _1} \equiv (1,2,3)$$</p>
<p>parallel to $${\overrightarrow b _1} \equiv (\lambda ,1,2)$$</p>
<p>$${L_2}:{{x + 26} \over { - 2}} = {{y + 18} \over 3} = {{z + 28} \over \lambda }$$</p>
<p>through a point $${\overrightarrow a _2} = ( - 26, - 18, - 28)$$</p>
<p>parallel to $${\overrightarrow b _2} = ( - 2,3,1)$$</p>
<p>If lines are coplanar then, $$({\overrightarrow a _2} - {\overrightarrow a _1})\,.\,{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$$</p>
<p>$$ \Rightarrow \left| {\matrix{
{27} & {20} & {31} \cr
\lambda & 1 & 2 \cr
{ - 2} & 3 & \lambda \cr
} } \right| = 0 \Rightarrow \lambda = 3$$</p>
<p>Vector normal to the required plane $$\overrightarrow n = {\overrightarrow b _1} \times {\overrightarrow b _2}$$</p>
<p>$$ \Rightarrow \overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 1 & 2 \cr
{ - 2} & 3 & 3 \cr
} } \right| = - 3\widehat i - 13\widehat j + 11\widehat k$$</p>
<p>Equation of plane</p>
<p>$$ \equiv ((x - 1),(y - 2),(z - 3))\,.\,( - 3, - 13,11) = 0$$</p>
<p>$$ \Rightarrow 3x + 13y - 11z + 4 = 0$$</p>
<p>Checking the option gives (0, 4, 5) does not lie on the plane.</p>
| mcq | jee-main-2022-online-28th-july-evening-shift |
1l6p20npe | maths | 3d-geometry | lines-and-plane | <p>If the foot of the perpendicular from the point $$\mathrm{A}(-1,4,3)$$ on the plane $$\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$$, is $$\left(-2, \frac{7}{2}, \frac{3}{2}\right)$$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $$3,-1,-4$$, is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\\sqrt{26}$$"}, {"identifier": "C", "content": "2$$\\sqrt{2}$$"}, {"identifier": "D", "content": "$$\\sqrt{14}$$"}] | ["B"] | null | <p>$$\left( { - 2,{7 \over 2},{3 \over 2}} \right)$$ satisfies the plane $$P:2x + my + nz = 4$$</p>
<p>$$ - 4 + {{7m} \over 2} + {{3n} \over 2} = 4 \Rightarrow 7m + 3n = 16$$ ...... (i)</p>
<p>Line joining $$A( - 1,\,4,\,3)$$ and $$\left( { - 2,{7 \over 2},{3 \over 2}} \right)$$ is perpendicular to $$P:2x + my + nz = 4$$</p>
<p>$${1 \over 2} = {{{1 \over 2}} \over m} = {{{3 \over 2}} \over n} \Rightarrow m = 1$$ & $$n = 3$$</p>
<p>Plane $$P:2x + y + 3z = 4$$</p>
<p>Distance of P from $$A( - 1,\,4,\,3)$$ parallel to the line</p>
<p>$${{x + 1} \over 3} = {{y - 4} \over { - 1}} = {{z - 3} \over { - 4}}:L$$</p>
<p>for point of intersection of P & L</p>
<p>$$2(3r - 1) + ( - r + 4) + 3( - 4r + 3) = 4 \Rightarrow r = 1$$</p>
<p>Point of intersection : $$(2,\,3,\, - 1)$$</p>
<p>Required distance $$ = \sqrt {{3^2} + {1^2} + {4^2}} = \sqrt {26} $$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1l6p39yy3 | maths | 3d-geometry | lines-and-plane | <p>Let a line with direction ratios $$a,-4 a,-7$$ be perpendicular to the lines with direction ratios $$3,-1,2 b$$ and $$b, a,-2$$. If the point of intersection of the line $$\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$$ and the plane $$x-y+z=0$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+\gamma$$ is equal to _________.</p> | [] | null | 10 | <p>Given $$a\,.\,3 + ( - 4a)( - 1) + ( - 7)2b = 0$$ ...... (1)</p>
<p>and $$ab - 4{a^2} + 14 = 0$$ ....... (2)</p>
<p>$$ \Rightarrow {a^2} = 4$$ and $${b^2} = 1$$</p>
<p>$$\therefore$$ $$L \equiv {{x + 1} \over 5} = {{y - 2} \over 3} = {z \over 1} = \lambda $$ (say)</p>
<p>$$\Rightarrow$$ General point on line is $$(5\lambda - 1,\,3\lambda + 2,\,\lambda )$$ for finding point of intersection with $$x - y + z = 0$$ we get $$(5\lambda - 1) - (3\lambda + 2) + (\lambda ) = 0$$</p>
<p>$$ \Rightarrow 3\lambda - 3 = 0 \Rightarrow \lambda = 1$$</p>
<p>$$\therefore$$ Point at intersection $$(4,\,5,\,1)$$</p>
<p>$$\therefore$$ $$\alpha + \beta + \gamma = 4 + 5 + 1 = 10$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
1ldo4up2x | maths | 3d-geometry | lines-and-plane | <p>Let the plane P pass through the intersection of the planes $$2x+3y-z=2$$ and $$x+2y+3z=6$$, and be perpendicular to the plane $$2x+y-z+1=0$$. If d is the distance of P from the point ($$-$$7, 1, 1), then $$\mathrm{d^{2}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{250}{83}$$"}, {"identifier": "B", "content": "$$\\frac{250}{82}$$"}, {"identifier": "C", "content": "$$\\frac{15}{53}$$"}, {"identifier": "D", "content": "$$\\frac{25}{83}$$"}] | ["A"] | null | Plane $P$, is passing through intersection of the two planes, so,
<br/><br/>$$
\begin{aligned}
& 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\\\
& x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0
\end{aligned}
$$
<br/><br/>It is perpendicular with plane, $2 x+y-2+1=0$
<br/><br/>So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \lambda-1)(-1)=0$
<br/><br/>$\lambda=-8$
<br/><br/>So, plane $p_1:-6 x-13 y-25 z+46=0$
<br/><br/>Distance of plane $p$ from the point $(-7,1,1)$
<br/><br/>$$
\begin{aligned}
& d=\frac{|+42-13-25+46|}{\sqrt{36+169+625}}=\frac{50}{\sqrt{830}} \\\\
& d^2=\frac{2500}{830}=\frac{250}{83}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-evening-shift |
1ldo7avkk | maths | 3d-geometry | lines-and-plane | <p>The point of intersection $$\mathrm{C}$$ of the plane $$8 x+y+2 z=0$$ and the line joining the points $$\mathrm{A}(-3,-6,1)$$ and $$\mathrm{B}(2,4,-3)$$ divides the line segment $$\mathrm{AB}$$ internally in the ratio $$\mathrm{k}: 1$$. If $$\mathrm{a}, \mathrm{b}, \mathrm{c}(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|$$ are coprime) are the direction ratios of the perpendicular from the point $$\mathrm{C}$$ on the line $$\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}$$, then $$|\mathrm{a}+\mathrm{b}+\mathrm{c}|$$ is equal to ___________.</p> | [] | null | 10 | Plane: $8 x+y+2 z=0$
<br><br>Given line $\mathrm{AB}: \frac{\mathrm{x}-2}{5}=\frac{\mathrm{y}-4}{10}=\frac{\mathrm{z}+3}{-4}=\lambda$
<br><br>Any point on line $(5 \lambda+2,10 \lambda+4,-4 \lambda-3)$
<br><br>Point of intersection of line and plane
<br><br>$$
\begin{aligned}
& 8(5 \lambda+2)+10 \lambda+4-8 \lambda-6=0 \\\\
& \lambda=-\frac{1}{3}
\end{aligned}
$$
<br><br>$\mathrm{C}\left(\frac{1}{3}, \frac{2}{3},-\frac{5}{3}\right)$
<br><br>$\mathrm{L}: \frac{\mathrm{x}-1}{-1}=\frac{\mathrm{y}+4}{2}=\frac{\mathrm{z}+2}{3}=\mu$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5mhm7e/af25a820-04bf-4a34-b4b6-12307077c8f7/834270a0-ad28-11ed-831c-57e1cd79ae86/file-1le5mhm7f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5mhm7e/af25a820-04bf-4a34-b4b6-12307077c8f7/834270a0-ad28-11ed-831c-57e1cd79ae86/file-1le5mhm7f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Evening Shift Mathematics - 3D Geometry Question 112 English Explanation">
<br><br>$\begin{aligned} & \overrightarrow{\mathrm{CD}}=\left(-\mu+\frac{2}{3}\right) \hat{\mathrm{i}}+\left(2 \mu-\frac{14}{3}\right) \hat{\mathrm{j}}+\left(3 \mu-\frac{1}{3}\right) \hat{\mathrm{k}} \\\\ & \left(-\mu+\frac{2}{3}\right)(-1)+\left(2 \mu-\frac{14}{3}\right) 2+\left(3 \mu-\frac{1}{3}\right) 3=0 \\\\ & \Rightarrow \mu=\frac{11}{14} \\\\ & \overrightarrow{\mathrm{CD}}=\frac{-5}{42}, \frac{-130}{42}, \frac{85}{42} \\\\ & \text { Direction ratios } \rightarrow(-1,-26,17) \\\\ & |\mathrm{a}+\mathrm{b}+\mathrm{c}|=10\end{aligned}$ | integer | jee-main-2023-online-1st-february-evening-shift |
1ldo7ffl8 | maths | 3d-geometry | lines-and-plane | <p>Let $$\alpha x+\beta y+\gamma z=1$$ be the equation of a plane passing through the point $$(3,-2,5)$$ and perpendicular to the line joining the points $$(1,2,3)$$ and $$(-2,3,5)$$. Then the value of $$\alpha \beta y$$ is equal to _____________.</p> | [] | null | 6 | Plane :
<br/><br/>$$
a(x-3)+b(y+2)+c(z-5)=0
$$
<br/><br/>Dr's of plane : $3 \hat{i}-\hat{j}-2 \hat{k}$
<br/><br/>$$
\begin{aligned}
& <3,-1,-2> \\\\
& P: 3(x-3)-1(y+2)-2(z-5)=0 \\\\
& 3 x-9-y-2-2 z+10=0 \\\\
& 3 x-y-2 z=1 \\\\
& \therefore \alpha=3, \beta=-1, \gamma=-2 \\\\
& \Rightarrow \alpha \beta \gamma=6
\end{aligned}
$$ | integer | jee-main-2023-online-1st-february-evening-shift |
ldo7laxw | maths | 3d-geometry | lines-and-plane | Let the plane $\mathrm{P}: 8 x+\alpha_{1} y+\alpha_{2} z+12=0$ be parallel to<br/><br/> the line $\mathrm{L}: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the
intercept of $\mathrm{P}$<br/><br/> on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is : | [{"identifier": "A", "content": "$\\frac{6}{\\sqrt{14}}$"}, {"identifier": "B", "content": "$\\sqrt{14}$"}, {"identifier": "C", "content": "$\\sqrt{\\frac{2}{7}}$"}, {"identifier": "D", "content": "$\\sqrt{\\frac{7}{2}}$"}] | ["B"] | null | P: $8 x+\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}+12=0$
<br/><br/>L: $\frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}$
<br/><br/>$\because \mathrm{P}$ is parallel to $\mathrm{L}$
<br/><br/>$\Rightarrow 8(2)+\alpha_{1}(3)+5\left(\alpha_{2}\right)=0$
<br/><br/>$\Rightarrow 3 \alpha_{1}+5\left(\alpha_{2}\right)=-16$
<br/><br/>Also $y$-intercept of plane $P$ is 1
<br/><br/>$\Rightarrow \alpha_{1}=-12$
<br/><br/>And $\alpha_{2}=4$
<br/><br/>$\Rightarrow$ Equation of plane $\mathrm{P}$ is $2 \mathrm{x}-3 \mathrm{y}+\mathrm{z}+3=0$
<br/><br/>$\Rightarrow$ Distance of line L from Plane $\mathrm{P}$ is
<br/><br/>$=\left|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|$
$=\sqrt{14}$ | mcq | jee-main-2023-online-31st-january-evening-shift |
ldo9cplr | maths | 3d-geometry | lines-and-plane | The foot of perpendicular from the origin $\mathrm{O}$ to a plane $\mathrm{P}$ which meets the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ is $(2, \mathrm{a}, 4), \mathrm{a} \in \mathrm{N}$. If the volume of the tetrahedron $\mathrm{OABC}$ is 144 unit$^{3}$, then which of the following points is <b>NOT</b> on P ? | [{"identifier": "A", "content": "$(3,0,4)$"}, {"identifier": "B", "content": "$(0,6,3)$"}, {"identifier": "C", "content": "$(0,4,4)$"}, {"identifier": "D", "content": "$(2,2,4)$"}] | ["A"] | null | Equation of Plane:
<br/><br/>$$
\begin{aligned}
& (2 \hat{i}+a \hat{j}+4 \hat{\mathrm{k}}) \cdot[(\mathrm{x}-2) \hat{\mathrm{i}}+(\mathrm{y}-\mathrm{a}) \hat{\mathrm{j}}+(\mathrm{z}-4) \hat{\mathrm{k}}]=0 \\\\
& \Rightarrow 2 \mathrm{x}+\mathrm{ay}+4 \mathrm{z}=20+\mathrm{a}^{2} \\\\
& \Rightarrow \mathrm{A} \equiv\left(\frac{20+\mathrm{a}^{2}}{2}, 0,0\right) \\\\
& \mathrm{B} \equiv\left(0, \frac{20+\mathrm{a}^{2}}{\mathrm{a}}, 0\right) \\\\
& \mathrm{C} \equiv\left(0,0, \frac{20+\mathrm{a}^{2}}{4}\right) \\\\
& \Rightarrow \text { Volume of tetrahedron } \\\\
& =\frac{1}{6}\left[ {\matrix{
{\overrightarrow a } & {\overrightarrow b } & {\overrightarrow c } \cr
} } \right] \\\\
& =\frac{1}{6} \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) \\\\
& \Rightarrow \frac{1}{6}\left(\frac{20+\mathrm{a}^{2}}{2}\right) \cdot\left(\frac{20+\mathrm{a}^{2}}{\mathrm{a}}\right) \cdot\left(\frac{20+\mathrm{a}^{2}}{4}\right)=144 \\\\
& \Rightarrow\left(20+\mathrm{a}^{2}\right)^{3}=144 \times 48 \times \mathrm{a} \\\\
& \Rightarrow \mathrm{a}=2 \\\\
& \Rightarrow \text { Equation of plane is } 2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=24 \\\\
& \text { Or } \mathrm{x}+\mathrm{y}+2 \mathrm{z}=12 \\\\
& \Rightarrow(3,0,4) \quad \text { Not } \text { lies on } \text { the } \\\\
& \mathrm{x}+\mathrm{y}+2 \mathrm{z}=12
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-evening-shift |
ldo9f5ge | maths | 3d-geometry | lines-and-plane | Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}] | ["A"] | null | Equation of Plane :
<br/><br/>$$
2(x-1)-3(y+1)-6(z+5)=0
$$
<br/><br/>or $2 x-3 y-6 z=35$
<br/><br/>$\Rightarrow$ Required distance $=$
$$
\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}
$$ = 5 | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldptgy8w | maths | 3d-geometry | lines-and-plane | <p>Let the line $$L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$$ intersect the plane $$2 x+y+3 z=16$$ at the point
$$P$$. Let the point $$Q$$ be the foot of perpendicular from the point $$R(1,-1,-3)$$ on the line $$L$$. If $$\alpha$$ is the area of triangle $$P Q R$$, then $$\alpha^{2}$$ is equal to __________.</p> | [] | null | 180 | $\quad L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}=r$ (say)
<br/><br/>Let $P \equiv\left(2 r_{1}+1,-r_{1}, r_{1}+3\right)$
<br/><br/>$P$ lies on $2 x+y+3 z=16$
<br/><br/>$\therefore 2\left(2 r_{1}+1\right)+\left(-r_{1}-1\right)+3\left(r_{1}+3\right)=16$
<br/><br/>$r_{1}=1$
<br/><br/>$P \equiv(3,-2,4)$
<br/><br/>$R \equiv(1,-1,-3)$
<br/><br/>Let $Q \equiv\left(2 r_{2}+1,-r_{2}-1, r_{2}+3\right)$
<br/><br/>$D R$ s of $Q R \equiv\left(2 r_{2}-r_{2} r_{2}+6\right)$
<br/><br/>DRs of $L \equiv(2,-1,1)$
<br/><br/>$Q R \perp L \Rightarrow 4 r_{2}+r_{2}+r_{2}+6=0$
<br/><br/>$r_{2}=-1$
<br/><br/>$Q \equiv(-1,0,2)$
<br/><br/>$\overrightarrow{Q P} \times \overrightarrow{R P}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 2 \\ 2 & -1 & 7\end{array}\right|=-12 \hat{i}-24 \hat{j}+0 \hat{k}$
<br/><br/>$\alpha=[P Q R]=\frac{1}{2}|\overrightarrow{Q P} \times \overrightarrow{R P}|=\frac{1}{2} \times 12 \sqrt{5}$
<br/><br/>$=6 \sqrt{5}$
<br/><br/>$\alpha^{2}=180$ | integer | jee-main-2023-online-31st-january-morning-shift |
1ldptu28a | maths | 3d-geometry | lines-and-plane | <p>Let $$\theta$$ be the angle between the planes $$P_{1}: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9$$ and $$P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15$$. Let $$\mathrm{L}$$ be the line that meets $$P_{2}$$ at the point $$(4,-2,5)$$ and makes an angle $$\theta$$ with the normal of $$P_{2}$$. If $$\alpha$$ is the angle between $$\mathrm{L}$$ and $$P_{2}$$, then $$\left(\tan ^{2} \theta\right)\left(\cot ^{2} \alpha\right)$$ is equal to ____________.</p> | [] | null | 9 | $P_{1}: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9$
<br/><br/>$$
\begin{aligned}
& P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15 \\\\
& \text { then } \cos \theta=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{1}{2} \\\\
& \begin{aligned}
& \therefore \theta=\frac{\pi}{3}, \text { Now } \alpha=\frac{\pi}{2}-\theta \\\\
& \therefore \tan ^{2} \theta \cdot \cot ^{2} \alpha=\tan ^{4} \theta \\\\
&=(\sqrt{3})^{4}=9
\end{aligned}
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-morning-shift |
ldqv1su4 | maths | 3d-geometry | lines-and-plane | A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then : | [{"identifier": "A", "content": "$a+b+\\sqrt{2} c=1$"}, {"identifier": "B", "content": "$\\sqrt{2} a+b+c=1$"}, {"identifier": "C", "content": "$\\sqrt{2} a-b+c=1$"}, {"identifier": "D", "content": "$a+\\sqrt{2} b+c=1$"}] | ["D"] | null | <p>$$l = {1 \over 2},m = {1 \over {\sqrt 2 }},n = \cos \theta $$</p>
<p>$${l^2} + {m^2} + {n^2} = 1$$</p>
<p>$$ \Rightarrow {1 \over 4} + {1 \over 2} + {n^2} = 1 \Rightarrow {n^2} = {1 \over 4} \Rightarrow n = \, + \,{1 \over 2}$$</p>
<p>$$\theta$$ is acute $$\therefore$$ $$n = {1 \over 2}$$</p>
<p>$$\therefore$$ $$\overrightarrow v = k\left( {{1 \over 2}\widehat i + {1 \over {\sqrt 2 }}\widehat j + {1 \over 2}\widehat k} \right),k \in R$$</p>
<p>$$\overrightarrow v .\,(\overrightarrow a - \overrightarrow b ) = 0$$</p>
<p>$$(\sqrt 2 - a){1 \over 2} + ( - 1 - b){1 \over {\sqrt 2 }} + (1 - c){1 \over 2} = 0$$</p>
<p>$$ \Rightarrow {a \over 2} + {b \over {\sqrt 2 }} + {c \over 2} = {1 \over 2}$$</p>
<p>$$ \Rightarrow a + \sqrt 2 b + c = 1$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
ldqvn3td | maths | 3d-geometry | lines-and-plane | If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is : | [{"identifier": "A", "content": "$\\frac{17}{5}$"}, {"identifier": "B", "content": "$\\frac{6}{13}$"}, {"identifier": "C", "content": "$\\frac{13}{6}$"}, {"identifier": "D", "content": "$\\frac{5}{17}$"}] | ["C"] | null | <p>Let $$P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)$$ & $$R(1,1,2)$$</p>
<p>$$\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k$$</p>
<p>& $$\overrightarrow Q R = - \widehat i + (1 - k)\widehat j + 3\widehat k$$</p>
<p>$$\therefore$$ Normal to plane will be</p>
<p>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & {(1 - k)} & 2 \cr
{ - 1} & {(1 - k)} & 3 \cr
} } \right| = \widehat i(1 - k) - \widehat j(8) + 3\widehat k(1 - k)$$</p>
<p>If line is parallel to this we have</p>
<p>$$1(1 - k) + 1( - 8) + ( - 3)(1 - k) = 0$$</p>
<p>$$ \Rightarrow 2(1 - k) = - 8$$</p>
<p>$$ \Rightarrow 1 - k = - 4 \Rightarrow k = 5$$</p>
<p>$$\therefore$$ $${{{k^2} + 1} \over {(k - 1)(k - 2)}} = {{26} \over {4.3}} = {{13} \over 6}$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldr7gixq | maths | 3d-geometry | lines-and-plane | <p>The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "$$\\frac{19}{3}$$"}, {"identifier": "D", "content": "$$\\frac{13}{3}$$"}] | ["A"] | null | <p>Equation of $${l_1} = {{x - 2} \over 2} = {{y - 6} \over 1} = {{z - 2} \over { - 2}}$$</p>
<p>Shortest distance with $${{x + 1} \over 2} = {{y + 4} \over { - 3}} = {z \over 2}$$ is</p>
<p>S.d $$ = \left| {{{\matrix{
3 & {10} & 2 \cr
2 & 1 & { - 2} \cr
2 & { - 3} & 2 \cr
} } \over {\left| { - 4\widehat i - 8\widehat j - 8\widehat k} \right|}}} \right| = \left| {{{( - 12) - 10(8) + 2( - 8)} \over {12}}} \right|$$</p>
<p>= 9 units</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldr7v822 | maths | 3d-geometry | lines-and-plane | <p>If the equation of the plane passing through the point $$(1,1,2)$$ and perpendicular to the line $$x-3 y+ 2 z-1=0=4 x-y+z$$ is $$\mathrm{A} x+\mathrm{B} y+\mathrm{C} z=1$$, then $$140(\mathrm{C}-\mathrm{B}+\mathrm{A})$$ is equal to ___________.</p> | [] | null | 15 | <p>Line of intersection of the planes $$x - 3y + 2z - 1 = 0$$ and $$4x - y + z = 0$$ is normal $$(\overrightarrow n )$$ to the required plane.</p>
<p>$$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 3} & 2 \cr
4 & { - 1} & 1 \cr
} } \right| = - \widehat i + 7\widehat j + 11\widehat k$$</p>
<p>Equation of plane is</p>
<p>$$ - x + 7y + 11z = \lambda $$</p>
<p>It passes through (1, 1, 2)</p>
<p>$$\therefore$$ $$\lambda = 28$$</p>
<p>So, the plane is</p>
<p>$$ - x + 7y + 11z = 28$$</p>
<p>$$ \Rightarrow {{ - 1} \over {28}}x + {7 \over {28}}y + {{11} \over {28}}z = 1$$</p>
<p>$$A = {{ - 1} \over {28}},B = {7 \over {28}},C = {{11} \over {28}}$$</p>
<p>$$140(C - B + A) = 15$$</p> | integer | jee-main-2023-online-30th-january-morning-shift |
1ldsesbom | maths | 3d-geometry | lines-and-plane | <p>The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\sqrt5$$. If $$ab < 0$$ and P is the point $$(a-b,b,2b-a)$$ then CP$$^2$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{17}{3}$$"}, {"identifier": "B", "content": "$$\\frac{97}{3}$$"}, {"identifier": "C", "content": "$$\\frac{16}{3}$$"}, {"identifier": "D", "content": "$$\\frac{73}{3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1let6xptn/17e2a896-c42e-490f-a47a-4034b21a373c/808505a0-ba1e-11ed-b1c7-2f0d4a78b053/file-1let6xpto.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1let6xptn/17e2a896-c42e-490f-a47a-4034b21a373c/808505a0-ba1e-11ed-b1c7-2f0d4a78b053/file-1let6xpto.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Evening Shift Mathematics - 3D Geometry Question 95 English Explanation"></p>
<p>$$C:\left( {{{a + 4} \over 3},{{ - 2 + 2b} \over 3},{{4 - 6} \over 3}} \right)$$</p>
<p>$$ = \left( {{{a + 4} \over 3},{{ - 2 + 2b} \over 3},{{ - 2} \over 3}} \right)$$</p>
<p>C lies on plane $$2x - y + z = 4$$</p>
<p>$$2\left( {{{a + 4} \over 3}} \right) - \left( {{{2b - 2} \over 3}} \right) - {2 \over 3} = 4$$</p>
<p>$$ \Rightarrow 2a + 8 - 2b + 2 - 2 = 12$$</p>
<p>$$ \Rightarrow a - b = 2$$ ..... (i)</p>
<p>Now, $$OP = \sqrt 5 $$</p>
<p>$${\left( {{{a + 4} \over 3}} \right)^2} + {\left( {{{2b - 2} \over 3}} \right)^2} + {4 \over 9} = 5$$ and using (i)</p>
<p>$$a = {{11} \over 5},1$$</p>
<p>$$ \Rightarrow b = {1 \over 5}, - 1$$</p>
<p>as also $$ \Rightarrow a = 1,b = - 1$$</p>
<p>$$P(2, - 1, - 3),c\left( {{5 \over 3},{{ - 4} \over 3},{{ - 2} \over 3}} \right)$$</p>
<p>$$C{P^2} = {1 \over 9} + {1 \over 9} + {{49} \over 9} = {{17} \over 3}$$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsffwjf | maths | 3d-geometry | lines-and-plane | <p>If the lines $${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$$ and $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$$ intersect at the point P, then the distance of the point P from the plane $$z = a$$ is :</p> | [{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "16"}] | ["A"] | null | <p>$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} = \lambda $$ (say)</p>
<p>& $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} = \mu $$ (say)</p>
<p>$$\therefore$$ $$\lambda + 1 = 2\mu + a$$ ...... (i)</p>
<p>$$2\lambda + 2 = 3\mu - 2$$ ..... (ii)</p>
<p>$$\lambda - 3 = \mu + 3$$ .... (iii)</p>
<p>By (i) & (ii)</p>
<p>$$ \Rightarrow 3\mu - 2 = 4\mu + 2a + 2$$</p>
<p>$$\mu=-2(1+a)$$ & $$\lambda=5-3a$$</p>
<p>Put $$\lambda$$ & $$\mu$$ in (iii) we get</p>
<p>$$a=-9$$</p>
<p>$$\mu=16$$</p>
<p>$$\lambda=22$$</p>
<p>$$\therefore$$ Point of intersection $$\equiv(23,46,19)$$</p>
<p>Distance from $$z=-9$$ is 28</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldswkl3l | maths | 3d-geometry | lines-and-plane | <p>Let the equation of the plane P containing the line $$x+10=\frac{8-y}{2}=z$$ be $$ax+by+3z=2(a+b)$$ and the distance of the plane $$P$$ from the point (1, 27, 7) be $$c$$. Then $$a^2+b^2+c^2$$ is equal to __________.</p> | [] | null | 355 | The line $\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}$ have a point $(-10,8,0)$ with d. r. $(1,-2,1)$
<br/><br/>
$\because$ the plane $a x+b y+3 z=2(a+b)$
<br/><br/>
$\Rightarrow \mathrm{b}=2 \mathrm{a}$
<br/><br/>
$\&$ dot product of d.r.'s is zero
<br/><br/>
$\therefore \mathrm{a}-2 \mathrm{~b}+3=0$ $\therefore \mathrm{a}=1 \& \mathrm{~b}=2$
<br/><br/>
Distance from $(1,27,7)$ is
<br/><br/>
$c=\frac{1+54+21-6}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5 \sqrt{14}$
<br/><br/>
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1+4+350$
<br/><br/>
$=355$ | integer | jee-main-2023-online-29th-january-morning-shift |
1ldv2zcrz | maths | 3d-geometry | lines-and-plane | <p>Let the equation of the plane passing through the line $$x - 2y - z - 5 = 0 = x + y + 3z - 5$$ and parallel to the line $$x + y + 2z - 7 = 0 = 2x + 3y + z - 2$$ be $$ax + by + cz = 65$$. Then the distance of the point (a, b, c) from the plane $$2x + 2y - z + 16 = 0$$ is ____________.</p> | [] | null | 9 | Let the equation of the plane is
<br/><br/>
$(x-2 y-z-5)+\lambda(x+y+3 z-5)=0$
<br/><br/>
$\because \quad$ it's parallel to the line
<br/><br/>
$x+y+2 z-7=0=2 x+3 y+z-2$ <br/><br/>So, vector along the line $\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|$
<br/><br/>
$$
=-5 \hat{i}+3 \hat{j}+\hat{k}
$$
<br/><br/>
$\because $ Plane is parallel to line
<br/><br/>
$\therefore -5(1+\lambda)+3(-2+\lambda)+1(-1+3 \lambda)=0$
<br/><br/>
$$
\lambda=12
$$
<br/><br/>
So, by (i)
<br/><br/>
$13 x+10 y+35 z=65$
<br/><br/>
$\therefore a=13, b=10, c=35$
and $d=\frac{26+20-35+16}{\sqrt{9}}=9$ | integer | jee-main-2023-online-25th-january-morning-shift |
1ldww23tv | maths | 3d-geometry | lines-and-plane | <p>If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $$x+2y+z=0$$ and $$3y-z=3$$ is ($$\alpha,\beta,\gamma$$), then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "5"}] | ["D"] | null | Direction of line
<br/><br/>
$$
\begin{aligned}
\vec{b} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
0 & 3 & -1
\end{array}\right| \\\\
& =\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3) \\\\
& =-5 \hat{i}+\hat{j}+3 \hat{k}
\end{aligned}
$$
<br/><br/>
Equation of line
<br/><br/>
$$
\frac{x-3}{-5}=\frac{y-2}{1}=\frac{z-1}{3}
$$
<br/><br/>
Let foot of perpendicular be $=(-5 k+3, k+2,3 k+1)$
<br/><br/>
$\Rightarrow(-5 k+2)(-5)+(k-7)(1)+(3 k-6)(3)=0$
<br/><br/>
Or $25 k-10+k-7+9 k-18=0$
<br/><br/>
Or $k=1$
<br/><br/>
$\alpha+\beta+\gamma=-k+6=5$ | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldwwhcmx | maths | 3d-geometry | lines-and-plane | <p>Let the plane containing the line of intersection of the planes <br/><br/>P<sub>1</sub> : $$x+(\lambda+4)y+z=1$$ and <br/><br/>P<sub>2</sub> : $$2x+y+z=2$$ <br/><br/>pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of <br/><br/>the point (2$$\lambda,\lambda,-\lambda$$) from the plane P<sub>2</sub> is :</p> | [{"identifier": "A", "content": "$$2\\sqrt6$$"}, {"identifier": "B", "content": "$$3\\sqrt6$$"}, {"identifier": "C", "content": "$$4\\sqrt6$$"}, {"identifier": "D", "content": "$$5\\sqrt6$$"}] | ["B"] | null | Equation of plane passing through point of intersection of $\mathrm{P} 1$ and $\mathrm{P} 2$<br/><br/>
$$
\begin{aligned}
& \mathrm{P}_1+\mathrm{kP}_2 = 0 \\\\
& (\mathrm{x}+(\lambda+4) \mathrm{y}+\mathrm{z}-1)+\mathrm{k}(2 \mathrm{x}+\mathrm{y}+\mathrm{z}-2)=0
\end{aligned}
$$<br/><br/>
Passing through $(0,1,0)$ and $(1,0,1)$<br/><br/>
$$
\begin{aligned}
& (\lambda+4-1)+\mathrm{k}(1-2)=0 \\\\
& (\lambda+3)-\mathrm{k}=0\quad...(1)
\end{aligned}
$$<br/><br/>
Also passing $(1,0,1)$<br/><br/>
$$
\begin{aligned}
& (1+1-1)+\mathrm{k}(2+1-2)=0 \\\\
& 1+\mathrm{k}=0 \\\\
& \mathrm{k}=-1
\end{aligned}
$$<br/><br/>
put in (1)<br/><br/>
$$
\begin{aligned}
& \lambda+3+1=0 \\\\
& \lambda=-4
\end{aligned}
$$<br/><br/>
Then point $(2 \lambda, \lambda,-\lambda)$<br/><br/>
$$
\begin{aligned}
& (-8,-4,4) \\\\
& d=\left|\frac{-16-4+4-2}{\sqrt{6}}\right| \\\\
& d=\frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=3 \sqrt{6}
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldybfknv | maths | 3d-geometry | lines-and-plane | <p>The distance of the point ($$-1,9,-16$$) from the plane <br/><br/>$$2x+3y-z=5$$ measured parallel to the line <br/><br/>$${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$ is :</p> | [{"identifier": "A", "content": "13$$\\sqrt2$$"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "20$$\\sqrt2$$"}, {"identifier": "D", "content": "31"}] | ["B"] | null | Given, $${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$
<br><br>$$ \Rightarrow $$ $${{x + 4} \over 3} = {{y - 2} \over -4} = {{z - 3} \over {12}}$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leyorqpr/6043abe7-b1a8-4575-ada8-a5ec43b191a3/29ee5170-bd24-11ed-8df0-25e8b2ace386/file-1leyorqpz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leyorqpr/6043abe7-b1a8-4575-ada8-a5ec43b191a3/29ee5170-bd24-11ed-8df0-25e8b2ace386/file-1leyorqpz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Morning Shift Mathematics - 3D Geometry Question 83 English Explanation">
<br>Equation of line passing through the point P$(-1,9,-16)$ and parallel to line $\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}$ is $\frac{x+1}{3}=\frac{y-9}{-4}=\frac{z+16}{12}=\lambda$
<br><br>Any point on this line (A) $=(3 \lambda-1,-4 \lambda+9,12 \lambda-16)$
<br><br>Point of intersection line and plane
<br><br>$$
\begin{aligned}
& 2(3 \lambda-1)+3(-4 \lambda+9)-1(12 \lambda-16)=5 \\\\
\Rightarrow & 6 \lambda-2-12 \lambda+27-12 \lambda+16=5 \\\\
\Rightarrow & \lambda=2 \quad \\\\
\therefore & \text { Point A }=(5,1,8) \\\\
\therefore & \text { Distance (AP) } \\\\
= & \sqrt{(5+1)^2+(9-1)^2+(-16-8)^2}\\\\
= & \sqrt{36+64+576}\\\\
= & 26 \text { units }
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnyni4l | maths | 3d-geometry | lines-and-plane | Let the plane $P$ contain the line $2 x+y-z-3=0=5 x-3 y+4 z+9$ and be<br/><br/> parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point<br/><br/> $\mathrm{A}(8,-1,-19)$ from the plane $\mathrm{P}$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$<br/><br/> is equal to ______________. | [] | null | 26 | Plane $\mathrm{P} \equiv \mathrm{P}_1+\lambda \mathrm{P}_2=0$<br><br>
$$
\begin{aligned}
& (2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0 \\\\
& (5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0 \\\\
& \overrightarrow{\mathrm{n}} \cdot \overrightarrow{\mathrm{b}}=0 \text { where } \overrightarrow{\mathrm{b}}(2,4,5) \\\\
& 2(5 \lambda+2)+4(1-3 \lambda)+5(4 \lambda-1)=0 \\\\
& \lambda=-\frac{1}{6}
\end{aligned}
$$<br><br>
Plane $7 x+9 y-10 z-27=0$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrmtdff/7b21977a-d096-41a4-aa81-a71dc07e4eb1/773bf0b0-e0db-11ed-900a-094cf67a472f/file-1lgrmtdfg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrmtdff/7b21977a-d096-41a4-aa81-a71dc07e4eb1/773bf0b0-e0db-11ed-900a-094cf67a472f/file-1lgrmtdfg.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - 3D Geometry Question 77 English Explanation"><br>
Equation of line $\mathrm{AB}$ is<br><br>
$$
\frac{\mathrm{x}-8}{-3}=\frac{\mathrm{y}+1}{4}=\frac{z+19}{12}=\lambda
$$<br><br>
Let $B=(8-3 \lambda,-1+4 \lambda,-19+12 \lambda)$ lies on plane $P$<br><br>
$$
\begin{aligned}
& \therefore 7(8-3 \lambda)+9(4 \lambda-1)-10(12 \lambda-19)=27 \\\\
& \lambda=2 \\\\
& \therefore \text { Point } B=(2,7,5) \\\\
& A B=\sqrt{6^2+8^2+24^2}=26
\end{aligned}
$$ | integer | jee-main-2023-online-15th-april-morning-shift |
1lgowhqoz | maths | 3d-geometry | lines-and-plane | <p>The plane, passing through the points $$(0,-1,2)$$ and $$(-1,2,1)$$ and parallel to the line passing through $$(5,1,-7)$$ and $$(1,-1,-1)$$, also passes through the point :</p> | [{"identifier": "A", "content": "$$(0,5,-2)$$"}, {"identifier": "B", "content": "$$(2,0,1)$$"}, {"identifier": "C", "content": "$$(1,-2,1)$$"}, {"identifier": "D", "content": "$$(-2,5,0)$$"}] | ["D"] | null | <p>The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is (5-1, 1-(-1), -7-(-1)) = (4, 2, -6).</p>
<p>Next, let's find another vector that is parallel to the plane. This vector can be obtained by taking the difference between the coordinates of the points (0, -1, 2) and (-1, 2, 1), which the plane passes through. This gives us a vector of (0-(-1), -1-2, 2-1) = (1, -3, 1).</p>
<p>The normal to the plane is perpendicular to both these vectors. It can be found by taking the cross product of the two vectors. </p>
<p>The cross product of vectors (4, 2, -6) and (1, -3, 1) is :</p>
$$
\begin{aligned}
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & 2 & -6 \\
1 & -3 & 1
\end{array}\right|=\hat{i}(-16)-\hat{j}(+10)+\hat{k}(-14) \\\\
& =-16 \hat{i}-10 \hat{j}-14 \hat{k}
\end{aligned}
$$
<p>The equation of the plane can now be written in the form :</p>
<p>-16x - 10y - 14z = d</p>
<p>We can find the constant 'd' by substituting one of the points through which the plane passes, say (0, -1, 2) :</p>
<p>d = -16.0 -10.(-1) -14.2 = 10 - 28 = -18</p>
<p>So, the equation of the plane is -16x - 10y - 14z = -18.</p>
<p>Now, we substitute the given options into the equation to check which one satisfies it :</p>
<p>Option A: (-16.0 -10.5 -14.(-2) = -18) => -50 + 28 = -22 ≠ -18, so A is not correct.
<br/><br/>Option B: (-16.2 -10.0 -14.1 = -18) => -32 - 14 = -46 ≠ -18, so B is not correct.
<br/><br/>Option C: (-16.1 -10.(-2) -14.1 = -18) => -16 + 20 -14 = -10 ≠ -18, so C is not correct.
<br/><br/>Option D: (-16.(-2) -10.5 -14.0 = -18) => 32 - 50 = -18, so D is correct.</p>
<p>So, the plane also passes through the point given in Option D, which is (-2, 5, 0).</p>
| mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpyb2eu | maths | 3d-geometry | lines-and-plane | <p>The distance of the point $$(-1,2,3)$$ from the plane $$\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$$ and $$\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$$ is :</p> | [{"identifier": "A", "content": "$$3 \\sqrt{6}$$"}, {"identifier": "B", "content": "$$3 \\sqrt{5}$$"}, {"identifier": "C", "content": "$$2 \\sqrt{6}$$"}, {"identifier": "D", "content": "$$2 \\sqrt{5}$$"}] | ["C"] | null | 1. Determine the line of shortest distance between the given two lines:
<br/><br/>Direction vector of line 1: $$\vec{d_1} = 2\hat{i} + \hat{k}$$
<br/><br/>Direction vector of line 2: $$\vec{d_2} = \hat{i} - \hat{j} + \hat{k}$$
<br/><br/>Now, let's find the cross product $$\vec{N} = \vec{d_1} \times \vec{d_2}$$
<br/><br/>$$
\vec{N} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 1 \\
1 & -1 & 1 \\
\end{vmatrix}
= (\hat{i}(0+1) - \hat{j}(2-1) + \hat{k}(-2-0))
= \hat{i} - \hat{j} - 2\hat{k}
$$
<br/><br/>So, the direction vector of the line of shortest distance is $$\vec{N} = \hat{i} - \hat{j} - 2\hat{k}$$.
<br/><br/>2. Find the equation of a line passing through point $$A(-1, 2, 3)$$ and having the direction vector $$\vec{N}$$:
<br/><br/>$$
\frac{x + 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{-2} = \lambda
$$
<br/><br/>3. Find the point $$P$$ where the line intersects the given plane:
<br/><br/>Let the coordinates of point $$P$$ be in terms of $$\lambda$$:
<br/><br/>$$
P(\lambda - 1, -\lambda + 2, -2\lambda + 3)
$$
<br/><br/>Since $$P$$ lies on the plane $$x - 2y + 3z = 10$$, we can substitute the coordinates of $$P$$ in terms of $$\lambda$$ into the equation of the plane:
<br/><br/>$$
(\lambda - 1) - 2(-\lambda + 2) + 3(-2\lambda + 3) = 10 $$
<br/><br/>$$
\lambda = -2 $$
<br/><br/>$$
P(-3, 4, 7)
$$
<br/><br/>4. Calculate the distance between points $$A$$ and $$P$$:
<br/><br/>$$
AP = \sqrt{(-3 - (-1))^2 + (4 - 2)^2 + (7 - 3)^2} = 2\sqrt{6}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgrgb0dt | maths | 3d-geometry | lines-and-plane | <p>Let the plane P: $$4 x-y+z=10$$ be rotated by an angle $$\frac{\pi}{2}$$ about its line of intersection with the plane $$x+y-z=4$$. If $$\alpha$$ is the distance of the point $$(2,3,-4)$$ from the new position of the plane $$\mathrm{P}$$, then $$35 \alpha$$ is equal to :</p> | [{"identifier": "A", "content": "126"}, {"identifier": "B", "content": "105"}, {"identifier": "C", "content": "85"}, {"identifier": "D", "content": "90"}] | ["A"] | null | Equation of plane after rotation :
<br/><br/>$$
\begin{aligned}
& (4 x-y+z-10)+\lambda(x+y-z-y)=0 \\\\
\Rightarrow & (4+\lambda) x+y(\lambda-1)+z(1-\lambda)-4 \lambda-10=0 \\\\
& \overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 \\\\
\Rightarrow & (4+\lambda) 4+(\lambda-1)(-1)+(1-\lambda) 1=0 \\\\
\Rightarrow & 16+4 \lambda-\lambda+1+1-\lambda=0 \\\\
\Rightarrow & 2 \lambda=-18 \\\\
\Rightarrow & \lambda=-9
\end{aligned}
$$
<br/><br/>$\therefore$ equation of plane : $-5 x-10 y+10 z+26=0$
<br/><br/>Distance of plane from $(2,3,-4)$
<br/><br/>$$
\begin{aligned}
& =\left|\frac{-10-30-40+26}{\sqrt{100+100+26}}\right|=\frac{54}{15}=\alpha \\\\
\therefore 35 \alpha & =35 \cdot \frac{54}{15}=7 \times \frac{54}{3}=7 \times 18=126
\end{aligned}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgsw3mg4 | maths | 3d-geometry | lines-and-plane | <p>Let the line $$l: x=\frac{1-y}{-2}=\frac{z-3}{\lambda}, \lambda \in \mathbb{R}$$ meet the plane $$P: x+2 y+3 z=4$$ at the point $$(\alpha, \beta, \gamma)$$. If the angle between the line $$l$$ and the plane $$P$$ is $$\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$$, then $$\alpha+2 \beta+6 \gamma$$ is equal to ___________.</p> | [] | null | 11 | $L: \frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda} $
<br/><br/>$ P: x+2 y+3 z=4$
<br/><br/>Vector parallel to line : $\langle 1,2, \lambda\rangle=\bar{b}$
<br/><br/>Normal vector to plane $P:<1,2,3\rangle=\bar{n}$
<br/><br/>Angle between plane and line is $\theta$
<br/><br/>Then, $\sin \theta=\frac{<1,2, \lambda>\cdot<1,2,3>}{\sqrt{1^2+2^2+\lambda^2} \cdot \sqrt{1^2+2^2+3^2}}$
<br/><br/>$$
\Rightarrow \frac{3}{\sqrt{14}}=\frac{1+4+3 \lambda}{\sqrt{\lambda^2+5} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3}
$$
<br/><br/>$$
L_1 \equiv \frac{x-0}{3}=\frac{y-1}{6}=\frac{z-3}{2}=\mu
$$
<br/><br/>Any point on line : $(3 \mu, 6 \mu+1,2 \mu+3)$
<br/><br/>It lies on P
<br/><br/>$$
\begin{aligned}
& \therefore 3 \mu+12 \mu+2+6 \mu+9=4 \\\\
& \Rightarrow \mu=\frac{-1}{3}
\end{aligned}
$$
<br/><br/>Hence, $\alpha=3 \mu=-1, \beta=6 \mu+1=-1, \gamma=2 \mu+3=\frac{7}{3}$
<br/><br/>Now, $\alpha+2 \beta+6 \gamma=11$ | integer | jee-main-2023-online-11th-april-evening-shift |
1lgvqjg9j | maths | 3d-geometry | lines-and-plane | <p>Let the foot of perpendicular from the point $$\mathrm{A}(4,3,1)$$ on the plane $$\mathrm{P}: x-y+2 z+3=0$$ be N. If B$$(5, \alpha, \beta), \alpha, \beta \in \mathbb{Z}$$ is a point on plane P such that the area of the triangle ABN is $$3 \sqrt{2}$$, then $$\alpha^{2}+\beta^{2}+\alpha \beta$$ is equal to ___________.</p> | [] | null | 7 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkifvys/3c8e033b-fa69-4b24-96b7-4df950fb761d/62d0e540-6786-11ee-8adf-57893cbbad41/file-6y3zli1lnkifvyt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnkifvys/3c8e033b-fa69-4b24-96b7-4df950fb761d/62d0e540-6786-11ee-8adf-57893cbbad41/file-6y3zli1lnkifvyt.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Mathematics - 3D Geometry Question 60 English Explanation">
<br><br>We have, equation of plane $P$ is $x-y+2 z+3=0$
<br><br>Perpendicular distance from $A$ to a plane $P$ i.e. N
<br><br>$=\frac{|4-3+2(1)+3|}{\sqrt{1^2+(-1)^2+2^2}}$
<br><br>$B(5, \alpha, \beta)$ lies on the plane $P$, so
<br><br>$5-\alpha+2 \beta+3=0 \Rightarrow \alpha=2 \beta+8$ .........(i)
<br><br>Direction ratio of $A N$ is $<1,-1,2>$
<br><br>Equation of $A N$ is $\frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}$
<br><br>Co-ordinate of $N: \frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4}$ $=-1$
<br><br>$\Rightarrow x=3, y=4, z=-1$
<br><br>$\therefore$ Co-ordinate of $N$ is $(3,4,-1)$
<br><br>$$
\begin{aligned}
& \text { Also, } \operatorname{Area of}(\triangle \mathrm{ABN})=3 \sqrt{2} \\\\
& \Rightarrow \frac{1}{2} \times \mathrm{AN} \times \mathrm{BN}=3 \sqrt{2} \\\\
& \Rightarrow \frac{1}{2} \times \sqrt{6} \times \mathrm{BN}=3 \sqrt{2} \Rightarrow \mathrm{BN}=2 \sqrt{3} \\\\
& \text { or } \mathrm{BN}^2=12
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \Rightarrow(5-3)^2+(\alpha-4)^2+(\beta+1)^2=12 \\\\
& 4+(2 \beta+4)^2+(\beta+1)^2=12(\text { using eqn (i)) } \\\\
& \Rightarrow 4 \beta^2+16+16 \beta+\beta^2+1+2 \beta=8 \\\\
& \Rightarrow 5 \beta^2+18 \beta+9=0
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \Rightarrow(5 \beta+3)(\beta+3)=0 \\\\
& \Rightarrow \beta=-3 \left[\because \beta \in z \text {, so rejecting } \beta=\frac{-3}{5}\right]\\\\
&\text { and }
\alpha=2 \beta+8=2(-3)+8=2\\\\
& \text { also } \alpha^2+\beta^2+\alpha \beta=9+4-6=7
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-evening-shift |
1lgxh5015 | maths | 3d-geometry | lines-and-plane | <p>Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\alpha,\beta,\gamma)$$, then $$\alpha\beta+\beta\gamma+\gamma\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "74"}, {"identifier": "C", "content": "76"}, {"identifier": "D", "content": "70"}] | ["B"] | null | Given that two vertex of a $\triangle A B C$ be $A(2,4,6)$ and <br><br>$B(0,-2,-5)$ and $G=$ centroid $=(2,1,-1)$
[Given]
<br><br>Let, the other vertex is $C(x, y, z)$
<br><br>According to the question,
<br><br>$$
\begin{aligned}
& \frac{2+0+x}{3}=2 \\\\
&\Rightarrow x =4
\end{aligned}
$$
<br><br>$$
\begin{aligned}
&\frac{4-2+y}{3}=1 \\\\
&\Rightarrow y=1
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \frac{6-5+z}{3}=-1 \\\\
&\Rightarrow z=1
\end{aligned}
$$
<br><br>Hence, third vertex is $C(4,1,-4)$
<br><br>Now, if image of $C(4,1,-4)$ in the plane $x+2 y+4 z=11$ is $D(\alpha, \beta, \gamma)$.
<br><br>So, $\frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=k$
(say)
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbbqc7r/6c542301-d903-4fd1-a7ae-79fb8e9918bd/17b8dca0-6279-11ee-9a72-3138802cff89/file-6y3zli1lnbbqc8b.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnbbqc7r/6c542301-d903-4fd1-a7ae-79fb8e9918bd/17b8dca0-6279-11ee-9a72-3138802cff89/file-6y3zli1lnbbqc8b.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Morning Shift Mathematics - 3D Geometry Question 58 English Explanation 1">
<br><br>$$
\Rightarrow \alpha=k+4, \beta=2 k+1, \gamma=4 k-4
$$
<br><br>Then, co-ordinates of $D$ is $(k+4,2 k+1,4 k-4)$
<br><br>Let, $E$ be the mid-point of $C D$, which lies on the plane $x+2 y+4 z=11$.
<br><br>Then, co-ordinates of $E$ is
<br><br>$$
\left(\frac{k+8}{2}, \frac{2 k+2}{2}, \frac{4 k-8}{2}\right)=\left(\frac{k+8}{2}, k+1,2 k-4\right)
$$
<br><br>Since, $E$ lies on the plane, $x+2 y+4 z-11=0$
<br><br>$$
\begin{aligned}
& \text { So, } \frac{k+8}{2}+2 k+2+8 k-16-11=0 \\\\
& \Rightarrow \frac{k+8+4 k+4+16 k-54}{2}=0
\end{aligned}
$$
<br><br>$$
\begin{array}{rlrl}
& \Rightarrow 21 k =42 \\\\
& \Rightarrow k =2
\end{array}
$$
<br><br>Hence, $D \equiv(6,5,4) \equiv(\alpha, \beta, \gamma)$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbbyfva/975f2e66-a12d-4dc9-903b-55c6281e4359/f9019760-6279-11ee-9a72-3138802cff89/file-6y3zli1lnbbyfvb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnbbyfva/975f2e66-a12d-4dc9-903b-55c6281e4359/f9019760-6279-11ee-9a72-3138802cff89/file-6y3zli1lnbbyfvb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Morning Shift Mathematics - 3D Geometry Question 58 English Explanation 2">
<br><br>So, $\alpha=6, \beta=5, \gamma=4$
<br><br>Now, $\alpha \beta+\beta \gamma+\gamma \alpha=6 \times 5+5 \times 4+4 \times 6$
<br><br>$=30+20+24=74$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgxsud7o | maths | 3d-geometry | lines-and-plane | <p>Let P be the point of intersection of the line $${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$$ and the plane $$x+y+z=2$$. If the distance of the point P from the plane $$3x - 4y + 12z = 32$$ is q, then q and 2q are the roots of the equation :</p> | [{"identifier": "A", "content": "$${x^2} + 18x - 72 = 0$$"}, {"identifier": "B", "content": "$${x^2} - 18x - 72 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 18x + 72 = 0$$"}, {"identifier": "D", "content": "$${x^2} - 18x + 72 = 0$$"}] | ["D"] | null | Given, equation of line is
<br/><br/>$$
\begin{aligned}
& \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=k \\\\
& \therefore x=3 k-3, y=k-2, z=1-2 k
\end{aligned}
$$
<br/><br/>Since, given that $P \equiv(3 k-3, k-2,1-2 k)$ be the point of intersection of the given line and the plane $x+y+z=2$
<br/><br/>$$
\begin{aligned}
&\text { So, }(3 k-3)+(k-2)+(1-2 k) =2 \\\\
&\Rightarrow 2 k-4=2 \Rightarrow k =3
\end{aligned}
$$
<br/><br/>Thus, $P=(6,1,-5)$
<br/><br/>Now, distance of point $P$ from the plane $3 x-4 y+12 z=32$ is
<br/><br/>$$
\begin{array}{rlrl}
&q =\left|\frac{18-4-60-32}{\sqrt{9+16+144}}\right|=\left|\frac{-78}{\sqrt{169}}\right| \\\\
&\Rightarrow q =\left|\frac{-78}{13}\right|=\frac{78}{13}=6 \\\\
&\therefore q =6 \Rightarrow 2 q=12
\end{array}
$$
<br/><br/>Thus, $q$ and $2 q$ are roots of the equation $x^2-18 x+72=0$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgylgam9 | maths | 3d-geometry | lines-and-plane | <p>For $$\mathrm{a}, \mathrm{b} \in \mathbb{Z}$$ and $$|\mathrm{a}-\mathrm{b}| \leq 10$$, let the angle between the plane $$\mathrm{P}: \mathrm{ax}+y-\mathrm{z}=\mathrm{b}$$ and the line $$l: x-1=\mathrm{a}-y=z+1$$ be $$\cos ^{-1}\left(\frac{1}{3}\right)$$. If the distance of the point $$(6,-6,4)$$ from the plane P is $$3 \sqrt{6}$$, then $$a^{4}+b^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "85"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "25"}] | ["C"] | null | We have, $\theta=\cos ^{-1} \frac{1}{3}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \cos \theta=\frac{1}{3} \\\\
& \therefore \sin \theta=\sqrt{1-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}
$$
<br/><br/>The given plane line and are
<br/><br/>$$
a x+y-z=b $$
<br/><br/>$$ x-1=a-y=z+1
$$
<br/><br/>$$
\begin{aligned}
& \therefore \sin \theta=\frac{a \cdot 1+(1)(-1)+(-1)(1)}{\sqrt{a^2+1^2+1^2} \sqrt{1^2+1^2+1^2}} \\\\
& \Rightarrow \frac{a-1-1}{\sqrt{a^2+2} \sqrt{3}}=\frac{2 \sqrt{2}}{3} \\\\
& \Rightarrow 3(a-2)=2 \sqrt{6} \sqrt{a^2+2} \\\\
& \Rightarrow 9\left(a^2+4-4 a\right)=24\left(a^2+2\right) \\\\
& \Rightarrow 9 a^2+36-36 a=24 a^2+48 \\\\
& \Rightarrow 15 a^2+36 a+12=0 \\\\
& \Rightarrow 5 a^2+12 a+4=0 \\\\
& \Rightarrow 5 a^2+10 a+2 a+4=0 \\\\
& \Rightarrow 5 a(a+2)+2(a+2)=0 \\\\
& \Rightarrow a=\frac{-2}{5},-2
\end{aligned}
$$
<br/><br/>So, $a=-2$ [$$
\because a \in Z
$$]
<br/><br/>Hence, the eqn. of plane is $-2 x+y-z-b=0$
<br/><br/>$$
\begin{aligned}
& \text { Now, } d=\left|\frac{-12-6-4-b}{\sqrt{4+1+1}}\right|=3 \sqrt{6} \\\\
& \Rightarrow|-(b+22)|=18 \\\\
& \Rightarrow b=18-22=-4 \\\\
& \therefore a^4+b^2=(-2)^4+(-4)^2 \\\\
& =16+16=32
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lgylmp2h | maths | 3d-geometry | lines-and-plane | <p>Let $$\mathrm{P}$$ be the plane passing through the line <br/><br/>$$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$$ and the point $$(2,4,-3)$$. <br/><br/>If the image of the point $$(-1,3,4)$$ in the plane P <br/><br/>is $$(\alpha, \beta, \gamma)$$ then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "11"}] | ["A"] | null | Equation of line : $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$
<br><br>Let $B \equiv(2,4,-3)$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyvtkfw/de189598-fad7-463c-a96b-e9cf1720e557/77a21fc0-5ba1-11ee-b31c-37f6bf9b942e/file-6y3zli1lmyvtkfx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lmyvtkfw/de189598-fad7-463c-a96b-e9cf1720e557/77a21fc0-5ba1-11ee-b31c-37f6bf9b942e/file-6y3zli1lmyvtkfx.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 8th April Evening Shift Mathematics - 3D Geometry Question 55 English Explanation">
<br>So, $\overrightarrow{\mathrm{AB}}=(2-1) \hat{i}+(4-2) \hat{j}+(-3+5) \hat{k}$ <br><br>$=\hat{i}+2 \hat{j}+2 \hat{k}$
<br><br>$$
\begin{aligned}
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 7 \\
1 & 2 & 2
\end{array}\right|=(-6-14) \hat{i}-(2-7) \hat{j}+(2+3) \hat{k} \\\\
& =-20 \hat{i}+5 \hat{j}+5 \hat{k} \\\\
& =-5(4 \hat{i}-\hat{j}-\hat{k})
\end{aligned}
$$
<br><br>$\therefore$ Equation of plane is :
<br><br>$$
4(x-1)+(-1)(y-2)-1(z+5)=0
$$
<br><br>$$
\begin{aligned}
& \Rightarrow 4 x-4-y+2-z-5=0 \\\\
& \Rightarrow 4 x-y-z-7=0
\end{aligned}
$$
<br><br>$\because$ Image of point $(-1,3,4)$ is $(\alpha, \beta, \gamma)$
<br><br>$$
\begin{aligned}
& \text { So, } \frac{\alpha+1}{4}=\frac{\beta-3}{-1}=\frac{\gamma-4}{-1}=\frac{-2(-4-3-4-7)}{16+1+1}=2 \\\\
& \Rightarrow \alpha=7, \beta=1, \gamma=2 \\\\
& \text { So, } \alpha+\beta+\gamma=10
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh00mq7x | maths | 3d-geometry | lines-and-plane | <p>Let $$\lambda_{1}, \lambda_{2}$$ be the values of $$\lambda$$ for which the points $$\left(\frac{5}{2}, 1, \lambda\right)$$ and $$(-2,0,1)$$ are at equal distance from the plane $$2 x+3 y-6 z+7=0$$. If $$\lambda_{1} > \lambda_{2}$$, then the distance of the point $$\left(\lambda_{1}-\lambda_{2}, \lambda_{2}, \lambda_{1}\right)$$ from the line $$\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$$ is ____________.</p> | [] | null | 9 | Since $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2,0,1)$ are equidistant
<br/><br/>from plane $2 x+3 y-6 z+7=0$
<br/><br/>$$
\begin{aligned}
& \therefore\left|\frac{2\left(\frac{5}{2}\right)+3(1)-6(\lambda)+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{2^2+3^2+6^2}}\right| \\\\
& \Rightarrow|5+3-6 \lambda+7|=|-4-6+7| \\\\
& \Rightarrow|15-6 \lambda|=|-3| \\\\
& \Rightarrow 15-6 \lambda= \pm 3 \\\\
& \Rightarrow 15-6 \lambda=3 \text { or } 15-6 \lambda=-3 \\\\
& \Rightarrow 6 \lambda=12 \text { or } 6 \lambda=18 \\\\
& \Rightarrow \lambda=2 \text { or } \lambda=3
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \because \lambda_1>\lambda_2 \\\\
& \therefore \lambda_1=3 \text { and } \lambda_2=2
\end{aligned}
$$
<br/><br/>So, point will be $(1,2,3)$
<br/><br/>Let $\mathrm{M}_0=(1,2,3)$
<br/><br/>$M_1$ is the point through which line passes i.e, $(5,1,-7)$
<br/><br/>and $\vec{s}=\hat{i}+2 \hat{j}+2 \hat{k}$
<br/><br/>$$
\therefore \overrightarrow{\mathrm{M}_0 \mathrm{M}_1}=4 \hat{i}-\hat{j}-10 \hat{k}
$$
<br/><br/>Now, required distance $=\left|\frac{\overrightarrow{\mathrm{M}_0 \mathrm{M}_1} \times \vec{s}}{|\vec{s}|}\right|$
<br/><br/>$$
\begin{aligned}
& =\frac{|(4 \hat{i}-\hat{j}-10 \hat{k}) \times(\hat{i}+2 \hat{j}+2 \hat{k})|}{\sqrt{1+4+4}} \\\\
& =\frac{|18 \hat{i}-18 \hat{j}+9 \hat{k}|}{3}=9
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-morning-shift |
1lh216te7 | maths | 3d-geometry | lines-and-plane | <p>If the equation of the plane passing through the line of intersection of the planes $$2 x-y+z=3,4 x-3 y+5 z+9=0$$ and parallel to the line $$\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$$ is $$a x+b y+c z+6=0$$, then $$a+b+c$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "14"}, {"identifier": "D", "content": "12"}] | ["C"] | null | Equation of plane intersection of two plane
<br/><br/>$$
\mathrm{P}_1+\lambda \mathrm{P}_2=0
$$
<br/><br/>$$
\mathrm{P}_1: 2 x-y+z=3, \text { and } \mathrm{P}_2: 4 x-3 y+5 z+9=0
$$
<br/><br/>Equation of any plane passing through the intersection of given planes is
<br/><br/>$(2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0$
<br/><br/>$$
\Rightarrow(2+4 \lambda) x+(-1-3 \lambda) y+(1+5 \lambda) z+(-3+9 \lambda)=0
$$ ..........(i)
<br/><br/>Plane (i) is parallel to given line
<br/><br/>$$
\begin{aligned}
& \therefore (-2)(2+4 \lambda)+4(-1-3 \lambda)+5(1+5 \lambda) =0 \\\\
& \Rightarrow -4-8 \lambda-4-12 \lambda+5+25 \lambda =0 \\\\
& \Rightarrow 5 \lambda-3 =0 \\\\
& \Rightarrow \lambda =\frac{3}{5}
\end{aligned}
$$
<br/><br/>Putting the value of $\lambda$ in Eq. (i), we get
<br/><br/>$$
\begin{aligned}
& (2 x-y+z-3)+\frac{3}{5}(4 x-3 y+5 z+9)=0 \\\\
& \Rightarrow 11 x-7 y+10 z+6=0 \\\\
& \therefore a=11, b=-7, c=10 \text { and } d=6 \\\\
& \therefore a+b+c=11-7+10=14
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2ybxdl | maths | 3d-geometry | lines-and-plane | <p>Let the line $$\mathrm{L}$$ pass through the point $$(0,1,2)$$, intersect the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ and be parallel to the plane $$2 x+y-3 z=4$$. Then the distance of the point $$\mathrm{P}(1,-9,2)$$ from the line $$\mathrm{L}$$ is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "$$\\sqrt{74}$$"}, {"identifier": "C", "content": "$$\\sqrt{69}$$"}, {"identifier": "D", "content": "$$\\sqrt{54}$$"}] | ["B"] | null | Given line, $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{1}=\lambda$ .........(i)
<br><br>$\therefore$ Any point on the line $A(2 \lambda+1,3 \lambda+2, \lambda+3)$
<br><br>Since, line $L$ pass through the point $B(0,1,2)$ and intersects line (i)
<br><br>$\therefore$ Direction ratio of $A B$ are
<br><br>$$
\begin{array}{ll}
& 0-2 \lambda-1,1-3 \lambda-2,2-\lambda-3 \\\\
&\text { i.e., } -2 \lambda-1,-3 \lambda-1,-\lambda-1
\end{array}
$$
<br><br>Line $L$ parallel to the plane $2 x+y-3 z=4$
<br><br>$$
\begin{aligned}
& \therefore 2(-2 \lambda-1)+1(-3 \lambda-1)-3(-\lambda-1)=0 \\\\
& \Rightarrow-4 \lambda-2-3 \lambda-1+3 \lambda+3=0 \\\\
& \Rightarrow -4 \lambda=0 \\\\
& \Rightarrow \lambda=0 \\\\
& \therefore \text { Direction ratio of } A B=-1,-1,-1 \\\\
& \text { i.e., } 1,1,1
\end{aligned}
$$
<br><br>$\therefore$ Equation of line $L$ is $\frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu$
<br><br>$$
\Rightarrow x=\mu, y=\mu+1, z=\mu+2
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1locplbnw/62b80666-6a84-4264-9637-5115e1fdf309/4c1321b0-7708-11ee-a563-572884a00dbd/file-6y3zli1locplbnx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1locplbnw/62b80666-6a84-4264-9637-5115e1fdf309/4c1321b0-7708-11ee-a563-572884a00dbd/file-6y3zli1locplbnx.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Evening Shift Mathematics - 3D Geometry Question 45 English Explanation">
<br><br>$\therefore$ Direction ratios of $P Q=\mu-1, \mu+10, \mu$
<br><br>Since, $P Q \perp L$
<br><br>$$
\begin{aligned}
& \therefore 1(\mu-1)+1(\mu+10)+\mu(1)=0 \\\\
& \Rightarrow \mu-1+\mu+10+\mu=0 \\\\
& \Rightarrow 3 \mu+9=0 \\\\
& \Rightarrow \mu=-3 \\\\
& \therefore Q \text { is }(-3,-2,-1)
\end{aligned}
$$
<br><br>$\therefore$ Distance of the point $P(1,-9,2)$ from the line $L$ is
<br><br>$$
\begin{array}{r}
= \sqrt{(-3-1)^2+(-2+9)^2+(-1-2)^2} \\\\
= \sqrt{16+49+9}=\sqrt{74}
\end{array}
$$ | mcq | jee-main-2023-online-6th-april-evening-shift |
1lsgafa13 | maths | 3d-geometry | lines-and-plane | <p>Let $$A(2,3,5)$$ and $$C(-3,4,-2)$$ be opposite vertices of a parallelogram $$A B C D$$. If the diagonal $$\overrightarrow{\mathrm{BD}}=\hat{i}+2 \hat{j}+3 \hat{k}$$, then the area of the parallelogram is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2} \\sqrt{410}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{2} \\sqrt{306}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2} \\sqrt{586}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{2} \\sqrt{474}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \text { Area }=|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}| \\
& =\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
5 & -1 & 7 \\
1 & 2 & 3
\end{array}\right| \\
& =\frac{1}{2}|-17 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{474}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
lvc57bf8 | maths | 3d-geometry | lines-and-plane | <p>If $$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is</p> | [{"identifier": "A", "content": "$$\\frac{4 \\sqrt{2}}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{2 \\sqrt{2}}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{5 \\sqrt{2}}{3}$$\n"}, {"identifier": "D", "content": "$$2 \\sqrt{2}$$"}] | ["A"] | null | <p>$$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are vertices of a quadrilateral</p>
<p>$$\begin{aligned}
& \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\
& =-\hat{i}+\hat{j}+2 \hat{k} \\
& \overrightarrow{B D}=\left(\frac{10}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)-\left(\frac{5}{3} \hat{i}+\frac{7}{3} \hat{j}+\frac{1}{3} \hat{k}\right) \\
& \overrightarrow{B D}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{2}{3} \hat{k} \\
& \text { Area }=\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}| \\
& =\frac{1}{2}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & 2 \\
\frac{5}{3} & \frac{-5}{3} & \frac{-2}{3}
\end{array}\right| \\
& =\frac{1}{2} \sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^2}\left[\because \overrightarrow{A C} \times \overrightarrow{B D}=\frac{8}{3} \hat{i}+\frac{8}{3} \hat{j}\right] \\
& =\frac{1}{2} \times \frac{8}{3} \times \sqrt{2}=\frac{4 \sqrt{2}}{3} \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
lCoOxMgGdxQ2IcRT | maths | 3d-geometry | lines-in-space | The lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar if : | [{"identifier": "A", "content": "$$k=3$$ or $$-2$$"}, {"identifier": "B", "content": "$$k=0$$ or $$-1$$"}, {"identifier": "C", "content": "$$k=1$$ or $$-1$$"}, {"identifier": "D", "content": "$$k=0$$ or $$-3$$"}] | ["D"] | null | Coplanar if
<br/><br/>$$\left| {\matrix{
{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr
{{l_1}} & {{m_1}} & {{n_1}} \cr
{{l_2}} & {{m_2}} & {{n_2}} \cr
} } \right| = 0$$
<br><br>$$\therefore$$ $$\left| {\matrix{
1 & { - 1} & { - 1} \cr
1 & 1 & { - k} \cr
k & 2 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
0 & 0 & { - 1} \cr
2 & {1 + k} & { - k} \cr
{k + 2} & 1 & 1 \cr
} } \right| = 0$$
<br><br>$${k^2} + 3k = 0 \Rightarrow k\left( {k + 3} \right) = 0$$
<br><br>or $$k = 0$$ or $$-3$$ | mcq | aieee-2003 |
QB1hUjgch8hPRWNt | maths | 3d-geometry | lines-in-space | A line with direction cosines proportional to $$2,1,2$$ meets each of the lines $$x=y+a=z$$ and $$x+a=2y=2z$$ . The co-ordinates of each of the points of intersection are given by : | [{"identifier": "A", "content": "$$\\left( {2a,3a,3a} \\right),\\left( {2a,a,a} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,a} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,2a} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {3a,3a,3a} \\right),\\left( {a,a,a} \\right)$$ "}] | ["B"] | null | Let a point on the line
<br><br>$$x = y + a = z$$ is $$\left( {\lambda ,\lambda - a,\lambda } \right)$$
<br><br>and a point on the line
<br><br>$$x + a = 2y = 2z$$ is $$\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),$$
<br><br>then direction ratio of the line joining these points are
<br><br>$$\lambda - \mu + a,\,\,\lambda - a - {\mu \over 2},\,\,\lambda - {\mu \over 2}$$
<br><br>If it represents the required line, then
<br><br>$${{\lambda - \mu + a} \over 2}$$
<br><br>$$ = {{\lambda - a - {\mu \over 2}} \over 1}$$
<br><br>$$ = {{\lambda - {\mu \over 2}} \over 2}$$
<br><br>on solving we get $$\lambda = 3a,\,\mu = 2a$$
<br><br>$$\therefore$$ The required points of intersection are
<br><br>$$\left( {3a,3a - a,3a} \right)$$ and $$\left( {2a - a,{{2a} \over 2},{{2a} \over 2}} \right)$$
<br><br>or $$\left( {3a,2a,3a} \right)$$ and $$\left( {a,a,a} \right)$$ | mcq | aieee-2004 |
4OZMVKi7SP8wxWVL | maths | 3d-geometry | lines-in-space | If the straight lines
<br/>$$x=1+s,y=-3$$$$ - \lambda s,$$ $$z = 1 + \lambda s$$ and $$x = {t \over 2},y = 1 + t,z = 2 - t,$$ with parameters $$s$$ and $$t$$ respectively, are co-planar, then $$\lambda $$ equals : | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "D", "content": "$$-2$$ "}] | ["D"] | null | The given lines are
<br><br>$$x - 1 = {{y + 3} \over { - \lambda }} = {{z - 1} \over \lambda } = s.........\left( 1 \right)$$
<br><br>and $$2x = y - 1 = {{z - 2} \over { - 1}} = t..........\left( 2 \right)$$
<br><br>The lines are coplanar, if
<br><br>$$\left| {\matrix{
{0 - \left( { - 1} \right)} & { - 1 - 3} & { - 2 - \left( { - 1} \right)} \cr
1 & { - \lambda } & \lambda \cr
{{1 \over 2}} & 1 & { - 1} \cr
} } \right| = 0$$
<br><br>$${c_2} \to {c_2} + {c_3};\left| {\matrix{
1 & { - 5} & 1 \cr
1 & 0 & \lambda \cr
{{1 \over 2}} & 0 & { - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow 5\left( { - 1 - {\lambda \over 2}} \right) = 0 \Rightarrow \lambda = - 2$$ | mcq | aieee-2004 |
u6SMbm7CaVhVyKjm | maths | 3d-geometry | lines-in-space | The angle between the lines $$2x=3y=-z$$ and $$6x=-y=-4z$$ is : | [{"identifier": "A", "content": "$${0^ \\circ }$$ "}, {"identifier": "B", "content": "$${90^ \\circ }$$"}, {"identifier": "C", "content": "$${45^ \\circ }$$"}, {"identifier": "D", "content": "$${30^ \\circ }$$"}] | ["B"] | null | The given lines are $$2x = 3y = - z$$
<br><br>or $$\,\,\,\,\,\,\,\,\,\,{x \over 3} = {y \over 2} = {z \over { - 6}}$$ $$\,\,\,\left[ {} \right.$$ Dividing by $$6$$ $$\left. {} \right]$$
<br><br>and $$6x = - y = - 4z$$
<br><br>or $$\,\,\,\,\,\,\,\,\,{x \over 2} = {y \over { - 12}} = {z \over { - 3}}$$ $$\,\,\,\,\left[ {} \right.$$ Dividing by $$12$$ $$\left. {} \right]$$
<br><br>$$\therefore$$ Angle between two lines is
<br><br>$$\cos \theta = {{3.2 + 2.\left( { - 12} \right) + \left( { - 6} \right).\left( { - 3} \right)} \over {\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }}$$
<br><br>$$ = {{6 - 24 + 18} \over {\sqrt {49} \sqrt {157} }} = 0 \Rightarrow \theta = {90^ \circ }$$ | mcq | aieee-2005 |
Z4uIG8y06yV4jPYd | maths | 3d-geometry | lines-in-space | The two lines $$x=ay+b, z=cy+d;$$ and $$x=a'y+b' ,$$ $$z=c'y+d'$$ are perpendicular to each other if : | [{"identifier": "A", "content": "$$aa'+cc'=-1$$"}, {"identifier": "B", "content": "$$aa'+cc'=1$$ "}, {"identifier": "C", "content": "$${a \\over {a'}} + {c \\over {c'}} = - 1$$ "}, {"identifier": "D", "content": "$${a \\over {a'}} + {c \\over {c'}} = 1$$"}] | ["A"] | null | Equation of lines
<br><br>$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$
<br><br>$${{x - b'} \over {a'}} = {y \over 1} = {{z - d'} \over {c'}}$$
<br><br>Line are perpendicular
<br><br>$$ \Rightarrow aa' + 1 + cc' = 0$$ | mcq | aieee-2006 |
2cg5qE8eqUZZH2mZ | maths | 3d-geometry | lines-in-space | The line passing through the points $$(5,1,a)$$ and $$(3, b, 1)$$ crosses the $$yz$$-plane at the point $$\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)$$ . Then | [{"identifier": "A", "content": "$$a=2,$$ $$b=8$$"}, {"identifier": "B", "content": "$$a=4,$$ $$b=6$$ "}, {"identifier": "C", "content": "$$a=6,$$ $$b=4$$"}, {"identifier": "D", "content": "$$a=8,$$ $$b=2$$"}] | ["C"] | null | Equation of line through $$\left( {5,1,a} \right)$$ and
<br><br>$$\left( {3,b,1} \right)$$ is $${{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda $$
<br><br>$$\therefore$$ Any point on this line is a
<br><br>$$\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \alpha } \right)\lambda + a} \right]$$
<br><br>It crosses $$yz$$ plane where $${ - 2\lambda + 5 = 0}$$
<br><br>$$\lambda = {5 \over 2}$$
<br><br>$$\therefore$$ $$\left( {0,\left( {b - 1} \right){5 \over 2} + 1,\left( {1 - a} \right){5 \over 2} + a} \right) = \left( {0,{{17} \over 2},{{ - 17} \over 2}} \right)$$
<br><br>$$ \Rightarrow \left( {b - 1} \right){5 \over 2} + 1 = {{17} \over 2}$$
<br><br>and $$\left( {1 - a} \right){5 \over 2} + a = - {{13} \over 2}$$
<br><br>$$ \Rightarrow b = 4$$ and $$a = 6$$ | mcq | aieee-2008 |
ynbPZFBIbyg94BX0 | maths | 3d-geometry | lines-in-space | If the straight lines $$\,\,\,\,\,$$ $$\,\,\,\,\,$$ $${{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ $$\,\,\,\,\,$$ and$$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2}$$ intersects at a point, then the integer $$k$$ is equal to | [{"identifier": "A", "content": "$$-5$$"}, {"identifier": "B", "content": "$$5$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$-2$$"}] | ["A"] | null | The two lines intersect if shortest distance between them is zero $$i.e.$$
<br><br>$${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$$
<br><br>$$ \Rightarrow \left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$$
<br><br>where $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k,{\overrightarrow b _1} = k\widehat i + 2\widehat j + 3\widehat k$$
<br><br>$$\,\,\,\,\,\,\,{\overrightarrow a _2} = 2\widehat i + 3\widehat j + \widehat k,\,\,{\widehat b_2} = 3\widehat i + k\widehat j + 2\widehat k$$
<br><br>$$ \Rightarrow \left| {\matrix{
1 & 1 & { - 2} \cr
k & 2 & 3 \cr
3 & k & 2 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0$$
<br><br>$$ \Rightarrow - 2{k^2} - 5k + 25 = 0 \Rightarrow k = - 5$$ $$\,\,\,\,$$ or $$\,\,\,\,$$ $${5 \over 2}$$
<br><br>As $$k$$ is an integer, therefore $$k=-5$$ | mcq | aieee-2008 |
wccmQiRxzMjvLNB5 | maths | 3d-geometry | lines-in-space | <b>Statement - 1 :</b> The point $$A(1,0,7)$$ is the mirror image of the point
<br/><br>$$B(1,6,3)$$ in the line : $${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$$
<br/><br><b>Statement - 2 :</b> The line $${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ bisects the line
<br/><br>segment joining $$A(1,0,7)$$ and $$B(1, 6, 3)$$ </br></br></br> | [{"identifier": "A", "content": "Statement -1 is true, Statement -2 is true; Statement -2 is <b>not</b> a correct explanation for Statement -1."}, {"identifier": "B", "content": "Statement -1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false , Statement -2 is true."}, {"identifier": "D", "content": "Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1. "}] | ["A"] | null | The directions ratios of the line segment joining points
<br><br>$$A\left( {1,0,7} \right)\,\,$$ and $$\,\,\,B\left( {1,6,3} \right)$$ are $$0,6, - 4.$$
<br><br>The direction ratios of the given line are $$1,2,3.$$
<br><br>Clearly $$1 \times 0 + 2 \times 6 + 3 \times \left( { - 4} \right) = 0$$
<br><br>So, the given line is perpendicular to line $$AB.$$
<br><br>Also, the mid point of $$A$$ and $$B$$ is $$\left( {1,3,5} \right)$$ which lies on the given line.
<br><br>So, the image of $$B$$ in the given line is $$A$$, because the given line is the perpendicular bisector of line segment joining points $$A$$ and $$B$$, But statement -$$2$$ is not a correct explanation for statement$$-1.$$ | mcq | aieee-2011 |
7cGbYtvbBGU0bEmH | maths | 3d-geometry | lines-in-space | If the line $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}$$ and $${{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}$$ intersect, then $$k$$ is equal to : | [{"identifier": "A", "content": "$$-1$$ "}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$${9 \\over 2}$$"}, {"identifier": "D", "content": "$$0$$ "}] | ["C"] | null | Given lines in vector form are
<br><br>$$\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)$$
<br><br>and $$\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)$$
<br><br>These will intersect if shortest distance between them $$=0$$
<br><br>i.e.$$\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}} \times {\overrightarrow b _2} = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
{3 - 1} & {k + 1} & { - 1} \cr
2 & 3 & 4 \cr
1 & 2 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0$$
<br><br>$$ \Rightarrow k = 9/2$$ | mcq | aieee-2012 |
CXIqtyKExltEbk3B | maths | 3d-geometry | lines-in-space | If the lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar, then $$k$$ can have : | [{"identifier": "A", "content": "any value "}, {"identifier": "B", "content": "exactly one value"}, {"identifier": "C", "content": "exactly two values "}, {"identifier": "D", "content": "exactly three values"}] | ["C"] | null | Given lines will be coplanar
<br><br>If $$\,\,\,\,\left| {\matrix{
{ - 1} & 1 & 1 \cr
1 & 1 & { - k} \cr
k & 2 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$$
<br><br>$$ \Rightarrow k = 0, - 3$$ | mcq | jee-main-2013-offline |
ikcTgrIigYUF6ly3PV0QM | maths | 3d-geometry | lines-in-space | The shortest distance between the lines $${x \over 2} = {y \over 2} = {z \over 1}$$ and
<br/>$${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval : | [{"identifier": "A", "content": "[0, 1)"}, {"identifier": "B", "content": "[1, 2)"}, {"identifier": "C", "content": "(2, 3]"}, {"identifier": "D", "content": "(3, 4]"}] | ["C"] | null | Shortest distance between the lines
<br><br>$${{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}}$$
<br><br>and $${{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}}$$ is
<br><br> $$\left| {{{\left| {\matrix{
{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr
{{a_1}} & {{b_1}} & {{c_1}} \cr
{{a_2}} & {{b_2}} & {{c_2}} \cr
} } \right|} \over {\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}} \right|$$
<br><br>$$ \therefore $$ Shortest distance between two given lines are,
<br><br> $$\left| {{{\left| {\matrix{
{ - 2} & 4 & 5 \cr
2 & 2 & 1 \cr
{ - 1} & 8 & 4 \cr
} } \right|} \over {\sqrt {{{\left( {8 - 8} \right)}^2} + {{\left( { - 1 - 8} \right)}^2} + {{\left( {16 + 2} \right)}^2}} }}} \right|$$
<br><br>= $$\left| {{{ - 36 + 90} \over {\sqrt {405} }}} \right|$$
<br><br>= $${{54} \over {20.1}}$$
<br><br>= 2.68 | mcq | jee-main-2016-online-9th-april-morning-slot |
wufSe9GWqPvxj5A6ZtUib | maths | 3d-geometry | lines-in-space | If the angle between the lines, $${x \over 2} = {y \over 2} = {z \over 1}$$
<br/><br/>and $${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$$ is $${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$$ then p is equal to : | [{"identifier": "A", "content": "$${7 \\over 2}$$ "}, {"identifier": "B", "content": "$${2 \\over 7}$$"}, {"identifier": "C", "content": "$$-$$ $${7 \\over 4}$$"}, {"identifier": "D", "content": "$$-$$ $${4 \\over 7}$$"}] | ["A"] | null | Let $$\theta $$ be the angle between the two lines
<br><br>Here direction cosines of $${x \over 2}$$ = $${y \over 2}$$ = $${z \over 1}$$ are 2, 2, 1
<br><br>Also second line can be written as :
<br><br>$${{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}$$
<br><br>$$ \therefore $$ its direction cosines are 2, $${{P \over 7}}$$, 4
<br><br>Also, cos$$\theta $$ = $${2 \over 3}$$ (Given)
<br><br>$$ \because $$ cos$$\theta $$ $$ = \left| {{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2\sqrt {a_2^2 + b_2^2 + c_2^2} } }}} \right|$$
<br><br>$$ \Rightarrow $$ $${2 \over 3}$$ $$ = \left| {{{\left( {2 \times 2} \right) + \left( {2 \times {P \over 7}} \right) + \left( {1 \times 4} \right)} \over {\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}} \right|$$
<br><br> $$ = {{4 + {{2P} \over 7} + 4} \over {3 \times \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}$$
<br><br>$$ \Rightarrow $$ $${\left( {4 + {P \over 7}} \right)^2} = 20 + {{{P^2}} \over {49}}$$
<br><br>$$ \Rightarrow $$ 16 + $${{8P} \over 7} + {{{P^2}} \over {49}}$$ = 20 + $${{{P^2}} \over {49}}$$
<br><br>$$ \Rightarrow $$ $${{8P} \over 7} = 4$$ $$ \Rightarrow $$ $$P = {7 \over 2}$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
0mlTVuKdlxftvs27Ar3rsa0w2w9jwxk75cq | maths | 3d-geometry | lines-in-space | If the length of the perpendicular from the point ($$\beta $$, 0, $$\beta $$) ($$\beta $$ $$ \ne $$ 0) to the line,
<br/>$${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$$ is $$\sqrt {{3 \over 2}} $$, then
$$\beta $$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "-1"}] | ["D"] | null | $${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} = p\,\,P\left( {\beta ,0,\beta } \right)$$<br><br>
any point on line A = (p, 1, – p – 1)<br><br>
Now, DR of AP $$ \equiv $$ < p – $$\beta $$, 1 – 0, – p – 1 – $$\beta $$ ><br><br>
Which is perpendicular to line so<br><br>
(p – $$\beta $$). 1 + 0.1 – 1(– p – 1 – $$\beta $$) = 0<br><br>
$$ \Rightarrow $$ p – $$\beta $$ + p + 1 + $$\beta $$ = 0<br><br>
$$p = {{ - 1} \over 2}$$<br><br>
Point $$A\left( {{{ - 1} \over 2},1 - {1 \over 2}} \right)$$<br><br>
Now, distance AP = $$\sqrt {{3 \over 2}} $$<br><br>
$$ \Rightarrow A{P^2} = {3 \over 2}$$<br><br>
$$ \Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} + 1 + {\left( {\beta + {1 \over 2}} \right)^2} = {3 \over 2}$$<br><br>
$$ \Rightarrow 2{\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 2}$$<br><br>
$$ \Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 4}$$<br><br>
$$ \Rightarrow \beta = 0, - 1,\left( {\beta \ne 0} \right)$$<br><br>
$$ \therefore \beta = - 1$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
SGw6KlnNG3CtzIgUGr18hoxe66ijvwv9w1o | maths | 3d-geometry | lines-in-space | The vertices B and C of a $$\Delta $$ABC lie on the line,
<br/><br/>$${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$$ such that BC = 5 units. <br/><br/>Then the
area (in sq. units) of this triangle, given that the
point A(1, –1, 2), is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "$$5\\sqrt {17} $$"}, {"identifier": "C", "content": "$$\\sqrt {34} $$"}, {"identifier": "D", "content": "$$2\\sqrt {34} $$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265608/exam_images/ei4rpd9as7gipkdqbmus.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264422/exam_images/cu7zggkrbhyehugwdzx8.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266711/exam_images/szym4mrimiv19nlybmq9.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - 3D Geometry Question 245 English Explanation"></picture>
Area of $$\Delta $$ABC = $${1 \over 2} \times BC \times AD$$<br><br>
Given BC = 5 so we need perpendicular
distance of A from line BC.<br><br>
Let a point D on BC = (3$$\lambda $$ - 2, 1, 4$$\lambda $$)
<br><br>Direction ratio of AD<br><br>
3$$\lambda $$ - 3, 2, 4$$\lambda $$ - 2<br><br>
As AD perpendicular to the BC, so<br><br>
DR of AD. DR of BC = 0
<br><br>
$$(3\lambda - 3)3 + 2(0) + (4\lambda - 2)4 = 0$$<br><br>
$$9\lambda - 9 + 16\lambda - 8 = 0 \Rightarrow \lambda = {{17} \over {25}}$$<br><br>
Hence, $$D = \left( {{1 \over {25}},1,{{68} \over {25}}} \right)$$<br><br>
$$\left| {\overline {AD} } \right| = \sqrt {{{\left( {{1 \over {25}} - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( {{{68} \over {25}} - 2} \right)}^2}} $$<br><br>
$$ \Rightarrow \sqrt {{{\left( {{{ - 24} \over {25}}} \right)}^2} + 4 + {{\left( {{{18} \over {25}}} \right)}^2}} $$<br><br>
$$ \Rightarrow \sqrt {{{{{\left( {24} \right)}^2} + 4{{\left( {25} \right)}^2} + {{\left( {18} \right)}^2}} \over {{{25}^2}}}} $$<br><br>
$$ \Rightarrow \sqrt {{{576 + 2500 + 324} \over {{{25}^2}}}} $$<br><br>
$$ \Rightarrow \sqrt {{{3400} \over {{{25}^2}}}} \Rightarrow {{\sqrt {34} .10} \over {25}} = {{2\sqrt {34} } \over 5}$$<br><br>
Area of triangle = $${1 \over 2} \times \left| {\overline {BC} } \right| \times \left| {\overline {AD} } \right|$$<br><br>
= $${1 \over 2} \times 5 \times {{2\sqrt {34} } \over 5} = \sqrt {34} $$ | mcq | jee-main-2019-online-9th-april-evening-slot |
BUAbYW0qzI2qZL65Fu5S1 | maths | 3d-geometry | lines-in-space | The length of the perpendicular from the point
(2, –1, 4) on the straight line,
<br/><br/>$${{x + 3} \over {10}}$$= $${{y - 2} \over {-7}}$$ = $${{z} \over {1}}$$
is :
| [{"identifier": "A", "content": "less than 2"}, {"identifier": "B", "content": "greater than 4\n"}, {"identifier": "C", "content": "greater than 2 but less than 3"}, {"identifier": "D", "content": "greater than 3 but less than 4"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266241/exam_images/fhdx0dohqlxs327d7ok9.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267798/exam_images/i5tycucchql2hd8qs3c1.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - 3D Geometry Question 252 English Explanation"></picture>
<br>Let $${{x + 3} \over {10}}$$= $${{y - 2} \over {-7}}$$ = $${{z} \over {1}}$$ = $$\lambda $$
<br><br>$$ \therefore $$ General point Q(10$$\lambda $$ - 3, -7$$\lambda $$ + 2, $$\lambda $$).
<br><br>$$\overrightarrow {PQ} $$ = (10$$\lambda $$ - 5)$$\widehat i$$ + (3 - 7$$\lambda $$)$$\widehat j$$ + ($$\lambda $$ - 4)$$\widehat k$$
<br><br>The vector parrallel to the given line is
<br><br>= 10$$\widehat i$$ - 7$$\widehat j$$ + $$\widehat k$$
<br><br>As $$\overrightarrow {PQ} $$ and (10$$\widehat i$$ - 7$$\widehat j$$ + $$\widehat k$$) are perpendicular to the each other.
<br><br>$$\overrightarrow {PQ} $$ . (10$$\widehat i$$ - 7$$\widehat j$$ + $$\widehat k$$) = 0
<br><br>$$ \Rightarrow $$ (10$$\lambda $$ - 5)10 + (3 - 7$$\lambda $$)(-7) + ($$\lambda $$ - 4)(1) = 0
<br><br>$$ \Rightarrow $$ 100$$\lambda $$ – 50 + 49$$\lambda $$ – 21 + $$\lambda $$ – 4 = 0
<br><br>$$ \Rightarrow $$ 150$$\lambda $$ = 75
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${1 \over 2}$$
<br><br>$$ \therefore $$ The point Q = (2, $$ - {3 \over 2}$$, $${1 \over 2}$$)
<br><br>$$ \therefore $$ Length of perpendicular (PQ) = $$\sqrt {0 + {1 \over 4} + {{49} \over 4}} $$
<br><br>= $$\sqrt {{{50} \over 4}} = {5 \over {\sqrt 2 }}$$ = 3.53 | mcq | jee-main-2019-online-8th-april-morning-slot |
IGkJuXOyBfS6DOS0NnzhG | maths | 3d-geometry | lines-in-space | If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then : | [{"identifier": "A", "content": "ab' + bc' + 1 = 0"}, {"identifier": "B", "content": "cc' + a + a' = 0"}, {"identifier": "C", "content": "bb' + cc' + 1 = 0"}, {"identifier": "D", "content": "aa' + c + c' = 0"}] | ["D"] | null | Equation of 1<sup>st</sup> line is
<br><br>$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$
<br><br>Dr's of 1<sup>st</sup> line = ($$a$$, 1 , c)
<br><br>Equation of 2<sup>nd</sup> line is
<br><br>$${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$$
<br><br>Dr's of 2<sup>nd</sup> line = ($$a'$$, c' , 1)
<br><br>Lines are perpendicular, so the dot product of the Dr's of two lines are zero.
<br><br>$$ \therefore $$ $$aa$$' + c + c' = 0 | mcq | jee-main-2019-online-9th-january-evening-slot |
RhDgRy3oAgbUFBBEoo7k9k2k5fommlo | maths | 3d-geometry | lines-in-space | If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through ($$\alpha $$, 7, 1)
is
$$\left( {{5 \over 3},{7 \over 3},{{17} \over 3}} \right)$$, then $$\alpha $$ is equal to______. | [] | null | 4 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267568/exam_images/jagkrfpsaoaaeae9ashz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Mathematics - 3D Geometry Question 235 English Explanation">
<br><br>Direction Ratio of PQ are
<br><br>= ($${5 \over 3} - 1$$, $${7 \over 3} - 0$$, $${{17} \over 3} - 3$$)
<br><br>= (2, 7, 8)
<br><br>Direction ratio of line QA are
<br><br>= ($$\alpha - {5 \over 3}$$, $$7 - {7 \over 3}$$, 1 - $${{17} \over 3}$$)
<br><br>= (3$$\alpha $$ – 5, 14, –14)
<br><br>PQ is perpendicular to line QA
<br><br>$$ \therefore $$ $$\overrightarrow {PQ} .\overrightarrow {QA} $$ = 0
<br><br>$$ \Rightarrow $$ 2(3$$\alpha $$ – 5) + 7.14 + (–14).8 = 0
<br><br>$$ \Rightarrow $$ $$\alpha $$ = 4 | integer | jee-main-2020-online-7th-january-evening-slot |
H7fMwfkoz6SPD9jgNn7k9k2k5gznyfy | maths | 3d-geometry | lines-in-space | The shortest distance between the lines
<br/><br/>$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and
<br/><br/>$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$${7 \\over 2}\\sqrt {30} $$"}, {"identifier": "C", "content": "$$3\\sqrt {30} $$"}, {"identifier": "D", "content": "$$2\\sqrt {30} $$"}] | ["C"] | null | $$\overrightarrow a $$
= < 3, 8, 3 >
<br><br>$$\overrightarrow b $$
= < – 3, – 7, 6 >
<br><br>$$\overrightarrow p $$
= < 3, – 1, 1 >
<br><br>$$\overrightarrow q $$
= < –3, 2, 4 >
<br><br>$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & { - 1} & 1 \cr
{ - 3} & 2 & 4 \cr
} } \right|$$ = < -6, -15, 3 >
<br><br>Shortest distance = $$\left| {{{\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$$
<br><br>= $$\left| {{{\left( { - 6, - 15,3} \right).\left( { - 6, - 15,3} \right)} \over {\sqrt {36 + 225 + 9} }}} \right|$$
<br><br>= $$\left| {{{36 + 225 + 9} \over {\sqrt {36 + 225 + 9} }}} \right|$$
<br><br>= $$\sqrt {270} $$ = $$3\sqrt {30} $$ | mcq | jee-main-2020-online-8th-january-morning-slot |
HngSHjDPG5hOAlG07kjgy2xukfg75pwu | maths | 3d-geometry | lines-in-space | If (a, b, c) is the image of the point (1, 2, -3) in<br/><br/> the line $${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$$, then a + b + c is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "-1"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267424/exam_images/kdcf95puxmjkf83bykfg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Morning Slot Mathematics - 3D Geometry Question 223 English Explanation">
<br><br>Equation of line : $${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$$ = $$\lambda $$ (Assume)
<br><br>A point on line L is
= R(2$$\lambda $$ – 1, –2$$\lambda $$ + 3, –$$\lambda $$)
<br><br>DR's of PR = < 2$$\lambda $$ – 2, –2$$\lambda $$ + 1, –$$\lambda $$ + 3 >
<br><br>$$ \because $$ PR is perpendicular to line L
<br><br>$$ \therefore $$ 2(2$$\lambda $$ –2) –2 (–2$$\lambda $$ + 1) –1 (–$$\lambda $$ + 3) = 0
<br><br>$$ \Rightarrow $$ 4$$\lambda $$ – 4 + 4$$\lambda $$ – 2 + $$\lambda $$ – 3 = 0
<br><br>$$ \Rightarrow $$ 9$$\lambda $$ – 9 = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 1
<br><br>$$ \therefore $$ Coordinate of foot of perpendicular = R = (1, 1, –1)
<br><br>As R is the midpoint of line PQ, so
<br><br>$${{a + 1} \over 2} = 1$$ $$ \Rightarrow $$ a = 1
<br><br>$${{b + 2} \over 2} = 1$$ $$ \Rightarrow $$ b = 0
<br><br>$${{c - 3} \over 2} = - 1$$ $$ \Rightarrow $$ c = 1
<br><br>$$ \therefore $$ a + b + c = 2 | mcq | jee-main-2020-online-5th-september-morning-slot |
cy8c71WuyhUNtGF5jA1klrkomrx | maths | 3d-geometry | lines-in-space | Let a, b$$ \in $$R. If the mirror image of the point P(a, 6, 9) with respect to the line <br/><br/>$${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$ is (20, b, $$-$$a$$-$$9), then | a + b |, is equal to : | [{"identifier": "A", "content": "88"}, {"identifier": "B", "content": "90"}, {"identifier": "C", "content": "86"}, {"identifier": "D", "content": "84"}] | ["A"] | null | Given, P(a, 6, 9)<br/><br/>Equation of line $${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$<br/><br/>Image of point P with respect to line is point Q(20, b, $$-$$a $$-$$9)<br/><br/>Mid-point of P and Q = $$\left( {{{a + 20} \over 2},{{6 + b} \over 2},{{ - a} \over 2}} \right)$$<br/><br/>This point lies on line<br/><br/>$$\therefore$$ $${{{{a + 20} \over 2} - 3} \over 7} = {{{{6 + b} \over 2} - 2} \over 5} = {{{{ - a} \over 2} - 1} \over { - 9}}$$<br/><br/>$$ \Rightarrow {{a + 14} \over {14}} = {{b + 2} \over {10}} = {{a + 2} \over {18}}$$<br/><br/>$$ \Rightarrow {{a + 14} \over {14}} = {{a + 2} \over {18}}$$ and $${{b + 2} \over {10}} = {{a + 2} \over {18}}$$<br/><br/>Solving, we get a = $$-$$ 56, b = $$-$$ 32<br/><br/>$$\therefore$$ $$\left| {a + b} \right| = \left| { - 56 - 32} \right| = 88$$ | mcq | jee-main-2021-online-24th-february-evening-slot |
Jp12LbScNdtH0VokX81klrmjwqw | maths | 3d-geometry | lines-in-space | Let $$\lambda$$ be an integer. If the shortest distance between the lines <br/><br/>x $$-$$ $$\lambda$$ = 2y $$-$$ 1 = $$-$$2z and x = y + 2$$\lambda$$ = z $$-$$ $$\lambda$$ is $${{\sqrt 7 } \over {2\sqrt 2 }}$$, then the value of | $$\lambda$$ | is _________. | [] | null | 1 | $${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$$<br><br>$${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$$ ....... (1)<br><br>Point on line = $$\left( {\lambda ,{1 \over 2},0} \right)$$<br><br>$${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$$ ....... (2)<br><br>Point on line = $$(0, - 2\lambda ,\lambda )$$<br><br>Distance between skew lines $$ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$$<br><br>$$\left| {\matrix{
\lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr
2 & 1 & { - 1} \cr
1 & 1 & 1 \cr
} } \right|$$<br><br>$$\overline {\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 1 & { - 1} \cr
1 & 1 & 1 \cr
} } \right|} $$<br><br>$$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$$<br><br>$$ = |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$$<br><br>$$ \Rightarrow |\lambda | = 1$$ | integer | jee-main-2021-online-24th-february-evening-slot |
L0USDlCO6DJtPBCMFT1kls3suts | maths | 3d-geometry | lines-in-space | The equation of the line through the point (0, 1, 2) and perpendicular to the line <br/><br/>$${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}}$$ is : | [{"identifier": "A", "content": "$${x \\over 3} = {{y - 1} \\over { - 4}} = {{z - 2} \\over 3}$$"}, {"identifier": "B", "content": "$${x \\over 3} = {{y - 1} \\over 4} = {{z - 2} \\over { - 3}}$$"}, {"identifier": "C", "content": "$${x \\over { - 3}} = {{y - 1} \\over 4} = {{z - 2} \\over 3}$$"}, {"identifier": "D", "content": "$${x \\over 3} = {{y - 1} \\over 4} = {{z - 2} \\over 3}$$"}] | ["C"] | null | $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}} = \lambda $$<br><br>Any point on this line $$(2\lambda + 1,3\lambda - 1, - 2\lambda + 1)$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264765/exam_images/f91a8wrnbdvn5tmzebxr.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Mathematics - 3D Geometry Question 213 English Explanation"><br><br>Direction ratio of given line $$(2,3, - 2)$$<br><br>Direction ratio of line to be found $$(2\lambda + 1,3\lambda - 2, - 2\lambda - 1)$$<br><br>$$ \therefore $$ $${\overrightarrow d _1}\,.\,{\overrightarrow d _2} = 0$$<br><br>$$ \Rightarrow $$ $$\lambda = 2/17$$<br><br>Direction ratio of line $$(21, - 28, - 21) \equiv (3, - 4, - 3) \equiv ( - 3,4,3)$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
8zfZrqb7fcL0HgCiZX1klt9uh4s | maths | 3d-geometry | lines-in-space | A line 'l' passing through origin is perpendicular to the lines<br/><br/>$${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$<br/><br/>$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$<br/><br/>If the co-ordinates of the point in the first octant on 'l<sub>2</sub>‘ at a distance of $$\sqrt {17} $$ from the point of intersection of 'l' and 'l<sub>1</sub>' are (a, b, c) then 18(a + b + c) is equal to ___________. | [] | null | 44 | $${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$<br><br>$${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $$ D.R. of $${l_1} = 1,2,2$$<br><br>$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$<br><br>$${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow $$ D.R. of $${l_2} = 2,2,1$$<br><br>D.R. of l is $$ \bot $$ to l<sub>1</sub> & k<sub>2</sub><br><br>$$ \therefore $$ D.R. of $$l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$$<br><br>$$ \therefore $$ Equation of $$l:{x \over 2} = {y \over { - 3}} = {z \over 2}$$<br><br>Solving l & l<sub>1</sub><br><br>$$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$$<br><br>$$ \Rightarrow 2\lambda = \mu + 3$$<br><br>$$ - 3\lambda = 2\mu - 1$$<br><br>$$2\lambda = 2\mu + 4$$<br><br>$$ \Rightarrow \mu + 3 = 2\mu + 4$$<br><br>$$\mu = - 1$$<br><br>$$\lambda = 1$$<br><br>$$P(2, - 3,2)$$ {intersection point}<br><br>Let, $$Q(2v + 3,2v + 3,v + 2)$$ be point on l<sub>2</sub><br><br>Now, $$PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $$<br><br>$$ \Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$$<br><br>$$ \Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$$<br><br>$$ \Rightarrow 9{v^2} + 28v + 20 = 0$$<br><br>$$ \Rightarrow 9{v^2} + 18v + 10v + 20 = 0$$<br><br>$$ \Rightarrow (9v + 10)(v + 2) = 0$$<br><br>$$ \Rightarrow v = - 2$$ (rejected), $$ - {{10} \over 9}$$ (accepted)<br><br>$$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$$<br><br>$$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$$<br><br>$$ \therefore $$ $$18(a + b + c)$$<br><br>$$ = 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$$<br><br>$$ = 44$$ | integer | jee-main-2021-online-25th-february-evening-slot |
YtsOhjwJ46iodfXK1u1kmhvwg6d | maths | 3d-geometry | lines-in-space | Let the position vectors of two points P and Q be 3$$\widehat i$$ $$-$$ $$\widehat j$$ + 2$$\widehat k$$ and $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 4$$\widehat k$$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $$-$$1, 2) and ($$-$$2, 1, $$-$$2), respectively. Let lines PR and QS intersect at T. If the vector $$\overrightarrow {TA} $$ is perpendicular to both $$\overrightarrow {PR} $$ and $$\overrightarrow {QS} $$ and the length of vector $$\overrightarrow {TA} $$ is $$\sqrt 5 $$ units, then the modulus of a position vector of A is : | [{"identifier": "A", "content": "$$\\sqrt {171} $$"}, {"identifier": "B", "content": "$$\\sqrt {227} $$"}, {"identifier": "C", "content": "$$\\sqrt {482} $$"}, {"identifier": "D", "content": "$$\\sqrt {5} $$"}] | ["A"] | null | $$\overrightarrow p = 3\widehat i - \widehat j + 2\widehat k$$ & $$\overrightarrow Q = \widehat i + 2\widehat j - 4\widehat k$$<br><br>$${\overrightarrow v _{PR}} = (4, - 1,2)$$ & $${\overrightarrow v _{QS}} = ( - 2,1, - 2)$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265215/exam_images/ugp7hf8znshbglhegrah.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - 3D Geometry Question 203 English Explanation"><br><br>$${L_{PR}}:\overrightarrow r = (3\widehat i - \widehat j + 2\widehat k) + \lambda (4, - 1,2)$$<br><br>$${L_{QS}}:\overrightarrow r = (\widehat i + 2\widehat j - 4\widehat k) + \mu ( - 2,1, - 2)$$
<br><br>Now T on PR = $$\left\langle {3 + 4\lambda , - 1 - \lambda ,2 + 2\lambda } \right\rangle $$
<br><br>Similarly T on QS = (1 $$-$$ 2$$\mu$$, 2 + $$\mu$$, $$-$$4 $$-$$ 2$$\mu$$)<br><br>For $$\lambda$$ & $$\mu$$ : $$\left. \matrix{
3 + 4\lambda = 1 - 2\mu \Rightarrow \mu + 2\lambda = - 1 \hfill \cr
- 1 - \lambda = 2 + \mu \Rightarrow \mu + \lambda = - 3 \hfill \cr} \right\}\matrix{
{\lambda = 2} \cr
{\mu = - 5} \cr
} $$<br><br>$$ \Rightarrow $$ T : (11, $$-$$3, 6)<br><br>D.R. of TA = $${\overrightarrow v _{QS}}$$ $$\times$$ $${\overrightarrow v _{PR}}$$<br><br>= $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{ - 2} & 1 & { - 2} \cr
4 & { - 1} & 2 \cr
} } \right| = 0\widehat i - 4\widehat j - 2\widehat k$$<br><br>$${L_{TA}}:\overrightarrow r = \left( {11\widehat i - 3\widehat j + 6\widehat k} \right) + \lambda ( - 4\widehat j - 2\widehat k)$$<br><br>Now A = (11, $$-$$3 $$-$$4$$\lambda$$, 6 $$-$$ 2$$\lambda$$)<br><br>Given, TA = $$\sqrt 5 $$<br><br>$${( - 3 + 4\lambda + 3)^2} + {(6 + 2\lambda - 6)^2} = 5$$<br><br>$$16{\lambda ^2} + 4{\lambda ^2} = 5$$<br><br>20$$\lambda$$<sup>2</sup> = 5<br><br>$$\lambda$$ = $$\pm$$$${1 \over 2}$$<br><br>$$\matrix{
{A:(11, - 3 - 2,6 - 1)} & ; & {A:(11, - 3 + 2,6 + 1)} \cr
{|A|\, = \sqrt {121 + 25 + 25} } & ; & {|A|\, = \sqrt {121 + 1 + 49} } \cr
{ = \sqrt {171} } & ; & {\sqrt {171} } \cr
} $$ | mcq | jee-main-2021-online-16th-march-morning-shift |
ASltiYdrFmn8DXdoyu1kmiwbg93 | maths | 3d-geometry | lines-in-space | If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, <i>l</i> $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L<sub>1</sub> and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 6 }}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$$\\sqrt {{2 \\over 3}} $$"}] | ["A"] | null | (3, 5, 7) lie on given line L<sub>1</sub><br><br>$${{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}$$<br><br>$${{7 - b} \over 4} = 1 \Rightarrow b = 3$$<br><br>$${{3 - a} \over l} = 1 \Rightarrow 3 - a = l$$<br><br>M(4, 3, 8)<br><br>N(3, 5, 7)<br><br>DR'S of MN = (1, $$-$$2, 1)<br><br>MN $$ \bot $$ line L<sub>1</sub><br><br>(1)(l) + ($$-$$2)(3) + 4(1) = 0<br><br>$$ \Rightarrow $$ l = 2<br><br>a = 1<br><br>a = 1, b = 3, l = 2<br><br>$${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$<br><br>$${{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$<br><br>$$A = \, < 1,2,3 > $$<br><br>$$B = \, < 2,4,5 > $$<br><br>$$\overrightarrow {AB} = \, < 1,2,2 > $$<br><br>$$\overrightarrow p = 2\widehat i + 3\widehat j + 4\widehat k$$<br><br>$$\overrightarrow q = 3\widehat i + 4\widehat j + 5\widehat k$$<br><br>$$\overrightarrow p \, \times \overrightarrow q = - \widehat i + 2\widehat j - \widehat k$$<br><br>Shortest distance = $$\left| {{{\overrightarrow {AB} \,.\,(\overrightarrow p \times \overrightarrow q )} \over {|\overrightarrow p \times \overrightarrow q |}}} \right|\, = {1 \over {\sqrt 6 }}$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
1krrpbbk7 | maths | 3d-geometry | lines-in-space | The lines x = ay $$-$$ 1 = z $$-$$ 2 and x = 3y $$-$$ 2 = bz $$-$$ 2, (ab $$\ne$$ 0) are coplanar, if : | [{"identifier": "A", "content": "b = 1, a$$\\in$$R $$-$$ {0}"}, {"identifier": "B", "content": "a = 1, b$$\\in$$R $$-$$ {0}"}, {"identifier": "C", "content": "a = 2, b = 2"}, {"identifier": "D", "content": "a = 2, b = 3"}] | ["A"] | null | Lines are $$x = ay - 1 = z - 2$$<br><br>$$\therefore$$ $${x \over 1} = {{y - {1 \over a}} \over {{1 \over a}}} = {{z - 2} \over 1}$$ .... (i)<br><br>and $$x = 3y - 2 = bz - 2$$<br><br>$$\therefore$$ $${x \over 1} = {{y - {2 \over 3}} \over {{1 \over 3}}} = {{z - {2 \over b}} \over {{1 \over b}}}$$ .... (ii)<br><br>$$\therefore$$ lines are co-planar<br><br>$$\therefore$$ $$\left| {\matrix{
0 & { - {1 \over a} + {2 \over 3}} & { - 2 + {2 \over b}} \cr
1 & {{1 \over a}} & 1 \cr
1 & {{1 \over 3}} & {{1 \over b}} \cr
} } \right| = 0$$<br><br>$$\therefore$$ $$\left| {\matrix{
0 & {{2 \over 3} - {1 \over a}} & {{2 \over b} - 2} \cr
0 & {{1 \over a} - {1 \over 3}} & {1 - {1 \over b}} \cr
1 & {{1 \over 3}} & {{1 \over b}} \cr
} } \right| = 0$$<br><br>$$\therefore$$ $${1 \over a} - {1 \over {ab}} = 0$$<br><br>$$\Rightarrow$$ b = 1 and a $$\in$$ R $$-$$ {0} | mcq | jee-main-2021-online-20th-july-evening-shift |
1krti6m7i | maths | 3d-geometry | lines-in-space | If the shortest distance between the straight lines $$3(x - 1) = 6(y - 2) = 2(z - 1)$$ and $$4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R$$ is $${1 \over {\sqrt {38} }}$$, then the integral value of $$\lambda$$ is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$-$$1"}] | ["A"] | null | $${L_1}:{{(x - 1)} \over 2} = {{(y - 2)} \over 1} = {{(z - 1)} \over 3}\overrightarrow {{r_1}} = 2\widehat i + \widehat j + 3\widehat k$$<br><br>$${L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}} = \widehat i + 2\widehat j + 4\widehat k$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264512/exam_images/catlrcudnwzjdrd5vl2e.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematics - 3D Geometry Question 188 English Explanation"><br>Shortest distance = Projection of $${\overrightarrow a }$$ on $${\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} }$$<br><br>$$ = {{\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right|} \over {\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right|}}$$<br><br>$$\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right| = \left| {\matrix{
1 & {\lambda - 2} & 2 \cr
2 & 1 & 3 \cr
1 & 2 & 4 \cr
} } \right| = \left| {14 - 5\lambda } \right|$$<br><br>$$\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right| = \sqrt {38} $$<br><br>$$\therefore$$ $${1 \over {\sqrt {38} }} = {{\left| {14 - 5\lambda } \right|} \over {\sqrt {38} }}$$<br><br>$$ \Rightarrow \left| {14 - 5\lambda } \right| = 1$$<br><br>$$ \Rightarrow 14 - 5\lambda = 1$$ or $$14 - 5\lambda = - 1$$<br><br>$$ \Rightarrow \lambda = {{13} \over 5}$$ or 3<br><br>$$\therefore$$ Integral value of $$\lambda$$ = 3. | mcq | jee-main-2021-online-22th-july-evening-shift |
1kto6ba4g | maths | 3d-geometry | lines-in-space | The distance of line $$3y - 2z - 1 = 0 = 3x - z + 4$$ from the point (2, $$-$$1, 6) is : | [{"identifier": "A", "content": "$$\\sqrt {26} $$"}, {"identifier": "B", "content": "$$2\\sqrt 5 $$"}, {"identifier": "C", "content": "$$2\\sqrt 6 $$"}, {"identifier": "D", "content": "$$4\\sqrt 2 $$"}] | ["C"] | null | $$3y - 2z - 1 = 0 = 3x - z + 4$$<br><br>$$3y - 2z - 1 = 0$$<br><br>D.R's $$\Rightarrow$$ (0, 3, $$-$$2)<br><br>$$3x - z + 4$$ = 0<br><br>D.R's $$\Rightarrow$$ (3, $$-$$1, 0)<br><br>Let DR's of given line are a, b, c<br><br>Now, 3b $$-$$ 2c = 0 & 3a $$-$$ c = 0<br><br>$$\therefore$$ 6a = 3b = 2c<br><br>a : b : c = 3 : 6 : 9<br><br>Any point on line<br><br>3K $$-$$ 1, 6K + 1, 9K + 1<br><br>Now, 3(3K $$-$$ 1) + 6(6K + 1)1 + 9(9K + 1) = 0<br><br>$$\Rightarrow$$ K = $${1 \over 3}$$<br><br>Point on line $$\Rightarrow$$ (0, 3, 4)<br><br>Given point (2, $$-$$1, 6)<br><br>$$\Rightarrow$$ Distance = $$\sqrt {4 + 16 + 4} = 2\sqrt 6 $$<br><br>Option (c) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l55iy2cg | maths | 3d-geometry | lines-in-space | <p>Let the image of the point P(1, 2, 3) in the line $$L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ be Q. Let R ($$\alpha$$, $$\beta$$, $$\gamma$$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($$\alpha$$ + $$\beta$$ + $$\gamma$$) is equal to __________.</p> | [] | null | 125 | <p>The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.</p>
<p>$$\therefore$$ Let a point on line be $$\lambda$$</p>
<p>$$ \Rightarrow {{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3} = \lambda $$</p>
<p>$$ \Rightarrow P'(3\lambda + 6,\,2\lambda + 1,\,3\lambda + 2)$$</p>
<p>as P' is foot of perpendicular</p>
<p>$$(3\lambda + 5)3 + (2\lambda - 1)2 + (3\lambda - 1)3 = 0$$</p>
<p>$$ \Rightarrow 22\lambda + 15 - 2 - 3 = 0$$</p>
<p>$$ \Rightarrow \lambda = {{ - 5} \over {11}}$$</p>
<p>$$\therefore$$ $$P'\left( {{{51} \over {11}},{1 \over {11}},{7 \over {11}}} \right)$$</p>
<p>Mid-point of $$PP' \equiv \left( {{{{{51} \over {11}} + 1} \over 2},{{{1 \over {11}} + 2} \over 2},{{{7 \over {11}} + 3} \over 2}} \right)$$</p>
<p>$$ \equiv \left( {{{62} \over {22}},{{23} \over {22}},{{40} \over {22}}} \right) \equiv (\alpha ,\beta ,\gamma )$$</p>
<p>$$ \Rightarrow 22(\alpha ,\beta ,\gamma ) = 62 + 23 + 40 = 125$$</p> | integer | jee-main-2022-online-28th-june-evening-shift |
1l56ra7cv | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines <br/><br/>$${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$ and $${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$, is :</p> | [{"identifier": "A", "content": "$${{18} \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${{22} \\over {3\\sqrt 5 }}$$"}, {"identifier": "C", "content": "$${{46} \\over {3\\sqrt 5 }}$$"}, {"identifier": "D", "content": "$$6\\sqrt 3 $$"}] | ["A"] | null | <p>$${L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$</p>
<p>$${L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$</p>
<p>Now, $$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 3 & { - 1} \cr
2 & 1 & 3 \cr
} } \right| = 10\widehat i - 8\widehat j - 4\widehat k$$</p>
<p>and $${\overrightarrow a _2} - {\overrightarrow a _1} = 6\widehat i - 4\widehat j - 4\widehat k$$</p>
<p>$$\therefore$$ $$S.D. = \left| {{{60 + 32 + 16} \over {\sqrt {100 + 64 + 16} }}} \right| = {{108} \over {\sqrt {180} }} = {{18} \over {\sqrt 5 }}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l58a01dt | maths | 3d-geometry | lines-in-space | <p>If the two lines $${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$$ and $${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$$ are perpendicular, then an angle between the lines l<sub>2</sub> and $${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$$ is :</p> | [{"identifier": "A", "content": "$${\\cos ^{ - 1}}\\left( {{{29} \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$${\\sec ^{ - 1}}\\left( {{{29} \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{2 \\over {29}}} \\right)$$"}, {"identifier": "D", "content": "$${\\cos ^{ - 1}}\\left( {{2 \\over {\\sqrt {29} }}} \\right)$$"}] | ["B"] | null | <p>$$\because$$ L<sub>1</sub> and L<sub>2</sub> are perpendicular, so</p>
<p>$$3 \times 1 + ( - 2)\left( {{\alpha \over 2}} \right) + 0 \times 2 = 0$$</p>
<p>$$ \Rightarrow \alpha = 3$$</p>
<p>Now angle between l<sub>2</sub> and l<sub>3</sub>,</p>
<p>$$\cos \theta = {{1( - 3) + {\alpha \over 2}( - 2) + 2(4)} \over {\sqrt {1 + {{{\alpha ^2}} \over 4} + } 4\sqrt {9 + 4 + 16} }}$$</p>
<p>$$ \Rightarrow \cos \theta = {2 \over {{{29} \over 2}}} \Rightarrow \theta = {\cos ^{ - 1}}\left( {{4 \over {29}}} \right) = {\sec ^{ - 1}}\left( {{{29} \over 4}} \right)$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58gby99 | maths | 3d-geometry | lines-in-space | <p>Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$, $$\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j + \widehat k$$ be three given vectors. Let $$\overrightarrow v $$ be a vector in the plane of $$\overrightarrow a $$ and $$\overrightarrow b $$ whose projection on $$\overrightarrow c $$ is $${2 \over {\sqrt 3 }}$$. If $$\overrightarrow v \,.\,\widehat j = 7$$, then $$\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}] | ["D"] | null | <p>Let $$\overrightarrow v = {\lambda _1}\overrightarrow a + {\lambda _2}\overrightarrow b $$, where $${\lambda _1},\,{\lambda _2} \in R$$.</p>
<p>$$ = ({\lambda _1} + 2{\lambda _2})\widehat i + ({\lambda _1} - 3{\lambda _2})\widehat j + (2{\lambda _1} + {\lambda _2})\widehat k$$</p>
<p>$$\because$$ Projection of $$\overrightarrow v $$ on $$\overrightarrow c $$ is $${2 \over {\sqrt 3 }}$$</p>
<p>$$\therefore$$ $${{{\lambda _1} + 2{\lambda _2} - {\lambda _1} + 3{\lambda _2} + 2{\lambda _1} + {\lambda _2}} \over {\sqrt 3 }} = {2 \over {\sqrt 3 }}$$</p>
<p>$$\therefore$$ $${\lambda _1} + 3{\lambda _2} = 1$$ ..... (i)</p>
<p>and $$\overrightarrow v \,.\,\widehat j = 7 \Rightarrow {\lambda _1} - 3{\lambda _2} = 7$$ ... (ii)</p>
<p>from equation (i) and (ii)</p>
<p>$${\lambda _1} = 4$$, $${\lambda _2} = - 1$$</p>
<p>$$\therefore$$ $$\overrightarrow v = 2\widehat i + 7\widehat j + 7\widehat k$$</p>
<p>$$\therefore$$ $$\overrightarrow v \,.\,(\widehat i + \widehat k) = 2 + 7$$</p>
<p>$$ = 9$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59limf1 | maths | 3d-geometry | lines-in-space | <p>Let l<sub>1</sub> be the line in xy-plane with x and y intercepts $${1 \over 8}$$ and $${1 \over {4\sqrt 2 }}$$ respectively, and l<sub>2</sub> be the line in zx-plane with x and z intercepts $$ - {1 \over 8}$$ and $$ - {1 \over {6\sqrt 3 }}$$ respectively. If d is the shortest distance between the line l<sub>1</sub> and l<sub>2</sub>, then d<sup>$$-$$2</sup> is equal to _______________.</p> | [] | null | 51 | <p>$${{x - {1 \over 8}} \over {{1 \over 8}}} = {y \over { - {1 \over {4\sqrt 2 }}}} = {z \over 0}\,\,\,\,\,\,\,\,\,\,\,\,\,$$ ______L<sub>1</sub></p>
<p>or $${{x - {1 \over 8}} \over 1} = {y \over { - \sqrt 2 }} = {z \over 0}$$ ..... (i)</p>
<p>Equation of L<sub>2</sub></p>
<p>$${{x + {1 \over 8}} \over { - 6\sqrt 3 }} = {y \over 0} = {z \over 8}$$ ...... (ii)</p>
<p>$$d = \left| {{{(\overrightarrow c - \overrightarrow a )\,.\,(\overrightarrow b \times \overrightarrow d )} \over {\left| {\overrightarrow b \times \overrightarrow d } \right|}}} \right|$$</p>
<p>$$ = {{\left( {{1 \over 4}\widehat i} \right)\,.\,\left( {4\sqrt 2 \widehat i + 4\widehat j + 3\sqrt 6 \widehat k} \right)} \over {\sqrt {{{\left( {4\sqrt 2 } \right)}^2} + {4^2} + {{\left( {3\sqrt 6 } \right)}^2}} }}$$</p>
<p>$$ = {{\sqrt 2 } \over {\sqrt {32 + 16 + 54} }} = {1 \over {\sqrt {51} }}$$</p>
<p>$${d^{ - 2}} = 51$$</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l5baeet6 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }$$ and $${{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is $${1 \over {\sqrt 3 }}$$, then the sum of all possible value of $$\lambda$$ is :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "15"}] | ["A"] | null | <p>Let $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k$$</p>
<p>$${\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat k$$</p>
<p>$$\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat k$$</p>
<p>$$\therefore$$ $$\overrightarrow p \times \overrightarrow q = (15 - 4\lambda )\widehat i - (10 - \lambda )\widehat j + 5\widehat k$$</p>
<p>$${\overrightarrow a _2} - {\overrightarrow a _1} = \widehat i + 2\widehat j + 2\widehat k$$</p>
<p>$$\therefore$$ Shortest distance</p>
<p>$$ = \left| {{{(15 - 4\lambda ) - 2(10 - \lambda ) + 10} \over {\sqrt {{{(15 - 4\lambda )}^2} + {{(10 - \lambda )}^2} + 25} }}} \right| = {1 \over {\sqrt 3 }}$$</p>
<p>$$ \Rightarrow 3{(5 - 2\lambda )^2} = {(15 - 4\lambda )^2} + {(10 - \lambda )^2} + 25$$</p>
<p>$$ \Rightarrow 5{\lambda ^2} - 80\lambda + 275 = 0$$</p>
<p>$$\therefore$$ Sum of values of $$\lambda = {{80} \over 5} = 16$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c2ach2 | maths | 3d-geometry | lines-in-space | <p>Let a line having direction ratios, 1, $$-$$4, 2 intersect the lines $${{x - 7} \over 3} = {{y - 1} \over { - 1}} = {{z + 2} \over 1}$$ and $${x \over 2} = {{y - 7} \over 3} = {z \over 1}$$ at the points A and B. Then (AB)<sup>2</sup> is equal to ___________.</p> | [] | null | 84 | Let $A(3 \lambda+7,-\lambda+1, \lambda-2)$ and $B(2 \mu, 3 \mu+7, \mu)$ <br/><br/>So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
<br/><br/>
Clearly $\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}$
<br/><br/>
$\Rightarrow 5 \lambda-3 \mu=-16$
<br/><br/>
And $\lambda-5 \mu=10$
<br/><br/>
From (i) and (ii) we get $\lambda=-5, \mu=-3$
<br/><br/>
So, $A$ is $(-8,6,-7)$ and $B$ is $(-6,-2,-3)$
<br/><br/>
$$
A B=\sqrt{4+64+16} \Rightarrow(A B)^{2}=84
$$ | integer | jee-main-2022-online-24th-june-morning-shift |
1l5c2jfp5 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines <br/><br/>$$\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$$ <br/><br/>and $$\overrightarrow r = \left( { - \widehat j + 2\widehat k} \right) + \mu \left( {\widehat i - \widehat j + \widehat k} \right)$$ is $$\sqrt {{2 \over 3}} $$, then the integral value of a is equal to ___________.</p> | [] | null | 2 | $\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1\end{array}\right|=-a \hat{i}-\hat{j}+(a-1) \hat{k}$ <br/><br/>$\vec{a}_{1}-\vec{a}_{2}=-\hat{i}+\hat{j}+\hat{k}$
<br/><br/>
Shortest distance $=\left|\frac{\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \quad \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^{2}+1+(a-1)^{2}}} \\\\
&\Rightarrow 6\left(a^{2}-2 a+1\right)=2 a^{2}-2 a+2 \\\\
&\Rightarrow \quad(a-2)(2 a-1)=0 \Rightarrow a=2 \text { because } a \in z .
\end{aligned}
$$ | integer | jee-main-2022-online-24th-june-morning-shift |
1l5w1j95q | maths | 3d-geometry | lines-in-space | <p>Consider a triangle ABC whose vertices are A(0, $$\alpha$$, $$\alpha$$), B($$\alpha$$, 0, $$\alpha$$) and C($$\alpha$$, $$\alpha$$, 0), $$\alpha$$ > 0. Let D be a point moving on the line x + z $$-$$ 3 = 0 = y and G be the centroid of $$\Delta$$ABC. If the minimum length of GD is $$\sqrt {{{57} \over 2}} $$, then $$\alpha$$ is equal to ____________.</p> | [] | null | 6 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6dmv11z/765801b8-1a82-4486-8b88-3108e93f1dbd/8269d270-132e-11ed-941a-4dd6502f33e3/file-1l6dmv120.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6dmv11z/765801b8-1a82-4486-8b88-3108e93f1dbd/8269d270-132e-11ed-941a-4dd6502f33e3/file-1l6dmv120.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - 3D Geometry Question 137 English Explanation"></p>
<p>Given, G is the centroid of $$\Delta$$ABC</p>
<p>$$\therefore$$ $$G = \left( {{{0 + \alpha + \alpha } \over 3},\,{{\alpha + 0 + \alpha } \over 3},\,{{\alpha + \alpha + 0} \over 3}} \right)$$</p>
<p>$$ = \left( {{{2\alpha } \over 3},\,{{2\alpha } \over 3},\,{{2\alpha } \over 3}} \right)$$</p>
<p>Also given, D is a point moving on the line $$x + z - 3 = 0 = y$$</p>
<p>Let $$D = (h,\,0,\,k)$$</p>
<p>$$x + z - 3 = 0 \Rightarrow h + k - 3 = 0 \Rightarrow h = 3 - k$$</p>
<p>$$\therefore$$ $$D = (3 - k,\,0,\,k)$$</p>
<p>Now, length of $$GD = d$$</p>
<p>$$ = \sqrt {{{\left( {{{2\alpha } \over 3} - 3 + k} \right)}^2} + {{\left( {{{2\alpha } \over 3}} \right)}^2} + {{\left( {{{2\alpha } \over 3} - k} \right)}^2}} $$</p>
<p>$$ \Rightarrow {d^2} = {\left( {{{2\alpha } \over 3} - 3 + k} \right)^2} + {\left( {{{2\alpha } \over 3}} \right)^2} + {\left( {{{2\alpha } \over 3} - k} \right)^2}$$</p>
<p>Differentiating both side with respect to k, we get</p>
<p>$$2dd' = 2\left( {{{2\alpha } \over 3} - 3 + k} \right) + 0 + 2\left( {{{2\alpha } \over 3} - k} \right) \times ( - 1)$$</p>
<p>For maximum or minimum value of k, $$d' = 0$$</p>
<p>$$\therefore$$ $$0 = 2\left( {{{2\alpha } \over 3} - 3 + k} \right) - 2\left( {{{2\alpha } \over 3} - k} \right)$$</p>
<p>$$ \Rightarrow 2\left( {{{2\alpha } \over 3} - 3 + k} \right) = 2\left( {{{2\alpha } \over 3} - k} \right)$$</p>
<p>$$ \Rightarrow - 3 + k = - k$$</p>
<p>$$ \Rightarrow k = {3 \over 2}$$</p>
<p>$$\therefore$$ $$G{D^2} = {\left( {{{2\alpha } \over 3} - 3 + {3 \over 2}} \right)^2} + {\left( {{{2\alpha } \over 3}} \right)^2} + {\left( {{{2\alpha } \over 3} - {3 \over 2}} \right)^2} = {{57} \over 2}$$</p>
<p>$$ \Rightarrow {{{{(4\alpha - 9)}^2}} \over {36}} + {{16{\alpha ^2}} \over {36}} + {{{{(4\alpha - 9)}^2}} \over {36}} = {{57} \over 2}$$</p>
<p>$$ \Rightarrow {(4\alpha - 9)^2} + 16{\alpha ^2} + {(4\alpha - 9)^2} = 57 \times 18$$</p>
<p>$$ \Rightarrow 16{\alpha ^2} - 72\alpha + 81 + 16{\alpha ^2} + 16{\alpha ^2} - 72\alpha + 81 = 57 \times 18$$</p>
<p>$$ \Rightarrow 48{\alpha ^2} - 144\alpha + 162 = 1026$$</p>
<p>$$ \Rightarrow 24{\alpha ^2} - 72\alpha + 81 - 513 = 0$$</p>
<p>$$ \Rightarrow 24{\alpha ^2} - 72\alpha - 432 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} - 3\alpha - 18 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} - 6\alpha + 3\alpha - 18 = 0$$</p>
<p>$$ \Rightarrow \alpha (\alpha - 6) + 3(\alpha - 6) = 0$$</p>
<p>$$ \Rightarrow (\alpha - 6)(\alpha + 3) = 0$$</p>
<p>$$ \Rightarrow \alpha = 6,\, - 3$$</p>
<p>Given, $$\alpha > 0$$</p>
<p>$$\therefore$$ Possible value of $$\alpha = 6$$.</p> | integer | jee-main-2022-online-30th-june-morning-shift |
1l6f2sskl | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :</p> | [{"identifier": "A", "content": "$$2 \\sqrt{29}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{37}{29}}$$"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{29}}{2}$$"}] | ["A"] | null | <p>$${L_1}:{{x + 7} \over 6} = {{y - 6} \over 7} = {{z - 0} \over 1}$$</p>
<p>Any point on it $${\overrightarrow a _1}( - 7,6,0)$$ and L<sub>1</sub> is parallel to $${\overrightarrow b _1}( - 6,7,1)$$</p>
<p>$${L_2}:{{x - 7} \over { - 2}} = {{y - 2} \over 1} = {{z - 6} \over 1}$$</p>
<p>Any point on it $${\overrightarrow a _2}(7,2,6)$$ and L<sub>2</sub> is parallel to $${\overrightarrow b _2}( - 2,1,1)$$</p>
<p>Shortest distance between L<sub>1</sub> and L<sub>2</sub></p>
<p>$$ = \left| {{{({{\overrightarrow a }_2} - {{\overrightarrow a }_1})\,.\,({{\overrightarrow b }_1} \times {{\overrightarrow b }_2})} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}} \right| = \left| {{{( - 14,4, - 6)\,.\,(3,2,4)} \over {\sqrt {9 + 4 + 16} }}} \right|$$</p>
<p>$$ = 2\sqrt {29} $$.</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6gio0a1 | maths | 3d-geometry | lines-in-space | <p>The length of the perpendicular from the point $$(1,-2,5)$$ on the line passing through $$(1,2,4)$$ and parallel to the line $$x+y-z=0=x-2 y+3 z-5$$ is :</p> | [{"identifier": "A", "content": "$$\\sqrt{\\frac{21}{2}}$$"}, {"identifier": "B", "content": "$$\\sqrt{\\frac{9}{2}}$$"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{73}{2}}$$"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbsbor/276c38c8-004d-4184-95dc-ab0e753739ef/40a706c0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbsbos.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbsbor/276c38c8-004d-4184-95dc-ab0e753739ef/40a706c0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbsbos.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Morning Shift Mathematics - 3D Geometry Question 131 English Explanation"> </p>
<p>The line $$x + y - z = 0 = x - 2y + 3z - 5$$ is parallel to the vector</p>
<p>$$\overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 1 & { - 1} \cr
1 & { - 2} & 3 \cr
} } \right| = (1,4, - 3)$$</p>
<p>Equation of the line through $$P(1,2,4)$$ and parallel to $$\overrightarrow b $$</p>
<p>$${{x - 1} \over 1} = {{y - 2} \over { - 4}} = {{z - 4} \over { - 3}}$$</p>
<p>Let $$N \equiv (\lambda + 1, - 4\lambda + 2, - 3\lambda + 4)$$</p>
<p>$$\overrightarrow {QN} = (\lambda , - 4\lambda + 4, - 3\lambda - 1)$$</p>
<p>$$\overrightarrow {QN} $$ is perpendicular to $$\overrightarrow {b} $$</p>
<p>$$ \Rightarrow (\lambda , - 4\lambda + 4, - 3\lambda - 1)\,.\,(1,4, - 3) = 0$$</p>
<p>$$ \Rightarrow \lambda = {1 \over 2}$$</p>
<p>Hence $$\overrightarrow {QN} = \left( {{1 \over 2},2,{{ - 5} \over 2}} \right)$$ and $$\left| {\overrightarrow {QN} } \right| = \sqrt {{{21} \over 2}} $$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6gjpve9 | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be two points on the line $$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$$ at a distance $$\sqrt{26}$$ from the point $$P(4,2,7)$$. Then the square of the area of the triangle $$P Q R$$ is ___________.</p> | [] | null | 153 | <p>$$L:{{x + 1} \over 2} = {{y + 2} \over 3} = {{2 - 1} \over 2}$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nc1t7n/ab3c067b-907d-4d61-90d4-40c5dd219f33/4878de40-2c50-11ed-9dc0-a1792fcc650d/file-1l7nc1t7o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nc1t7n/ab3c067b-907d-4d61-90d4-40c5dd219f33/4878de40-2c50-11ed-9dc0-a1792fcc650d/file-1l7nc1t7o.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Morning Shift Mathematics - 3D Geometry Question 132 English Explanation"> </p>
<p>Let $$T(2t - 1,\,3t - 2,\,2t + 1)$$</p>
<p>$$\because$$ $$PT\,{ \bot ^r}\,QR$$</p>
<p>$$\therefore$$ $$2(2t - 5) + 3(3t - 4) + 2(2t - 6) = 0$$</p>
<p>$$17t = 34$$</p>
<p>$$\therefore$$ $$t = 2$$</p>
<p>So $$T(3,4,5)$$</p>
<p>$$\therefore$$ $$PT = \sqrt {1 + 4 + 4} = 3$$</p>
<p>$$\therefore$$ $$QT = \sqrt {26 - 9} = \sqrt {17} $$</p>
<p>$$\therefore$$ Area of $$\Delta PQR = {1 \over 2} \times 2\sqrt {17} \times 3 = 3\sqrt {17} $$</p>
<p>$$\therefore$$ Square of $$ar(\Delta PQR) = 153$$.</p> | integer | jee-main-2022-online-26th-july-morning-shift |
1l6kkgbw1 | maths | 3d-geometry | lines-in-space | <p>If the length of the perpendicular drawn from the point $$P(a, 4,2)$$, a $$>0$$ on the line $$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$$ is $$2 \sqrt{6}$$ units and $$Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$$ is the image of the point P in this line, then $$\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$$ is equal to :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "14"}] | ["B"] | null | <p>$$\because$$ PR is perpendicular to given line, so</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qa9n3e/85e3a21e-e14a-4db6-947b-e473de8b2f29/a9e898a0-2def-11ed-a744-1fb8f3709cfa/file-1l7qa9n3f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qa9n3e/85e3a21e-e14a-4db6-947b-e473de8b2f29/a9e898a0-2def-11ed-a744-1fb8f3709cfa/file-1l7qa9n3f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Evening Shift Mathematics - 3D Geometry Question 124 English Explanation"></p>
<p>$$2(2\lambda - 1 - a) + 3(3\lambda - 1) - 1( - \lambda - 1) = 0$$</p>
<p>$$ \Rightarrow a = 7\lambda - 2$$</p>
<p>Now,</p>
<p>$$\because$$ $$PR = 2\sqrt 6 $$</p>
<p>$$ \Rightarrow {( - 5\lambda + 1)^2} + {(3\lambda - 1)^2} + {(\lambda + 1)^2} = 24$$</p>
<p>$$ \Rightarrow 5{\lambda ^2} - 2\lambda - 3 = 0 \Rightarrow \lambda = 1$$ or $$ - {3 \over 5}$$</p>
<p>$$\because$$ $$a > 0$$ so $$\lambda = 1$$ and $$a = 5$$</p>
<p>Now $$\sum\limits_{i = 1}^3 {{\alpha _i} = 2} $$ (Sum of co-ordinate of R) $$-$$ (Sum of coordinates of P)</p>
<p>$$ = 2(7) - 11 = 3$$</p>
<p>$$a + \sum\limits_{i = 1}^3 {{\alpha _i} = 5 + 3 = 8} $$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1ldomd3je | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines <br/><br/>$${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$$ and <br/><br/>$${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$$ is :</p> | [{"identifier": "A", "content": "$$7\\sqrt 3 $$"}, {"identifier": "B", "content": "$$5\\sqrt 3 $$"}, {"identifier": "C", "content": "$$4\\sqrt 3 $$"}, {"identifier": "D", "content": "$$6\\sqrt 3 $$"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{L}_1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\\\
& \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\\\
& \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\\\
& \mathrm{~L}_2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\\\
& \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\\\
& \overrightarrow{r_2}=\hat{i}+4 \hat{j}-5 \hat{k} \\\\
& \overrightarrow{r_1} \times \overrightarrow{r_2}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -3 \\
1 & 4 & -5
\end{array}\right| \\\\
& =2 \hat{i}+2 \hat{j}+2 \hat{k}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Shortest distance (d) }=\left|\frac{\left(\overrightarrow{r_1} \times \overrightarrow{r_2}\right) \cdot\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)}{\left|\overrightarrow{r_1} \times \overrightarrow{r_2}\right|}\right| \\\\
& =\frac{36}{2 \sqrt{3}}=6 \sqrt{3}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldpsmm9a | maths | 3d-geometry | lines-in-space | <p>Let the shortest distance between the lines
<br/><br/>$$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$$ and
<br/><br/>$$L_{1}: x+1=y-1=4-z$$ be $$2 \sqrt{6}$$. If $$(\alpha, \beta, \gamma)$$ lies on $$L$$,
<br/><br/>then which of the following is NOT possible?</p> | [{"identifier": "A", "content": "$$\\alpha+2 \\gamma=24$$"}, {"identifier": "B", "content": "$$2 \\alpha+\\gamma=7$$"}, {"identifier": "C", "content": "$$\\alpha-2 \\gamma=19$$"}, {"identifier": "D", "content": "$$2 \\alpha-\\gamma=9$$"}] | ["A"] | null | $\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$
<br/><br/>$$
\begin{aligned}
& \frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1} \\\\
& \vec{a}_{1}=5 \hat{i}+\lambda \hat{j}-\lambda \hat{k}_{,} \vec{a}_{2}=-\hat{i}+\hat{j}+4 \hat{k} \\\\
& \vec{a}_{1}-\vec{a}_{2}=6 \hat{i}+(\lambda-1) \hat{j}-(\lambda+4) \hat{k} \\\\
& \vec{b}_{1}=-2 \hat{i}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\
& \vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 0 & 1 \\
1 & 1 & -1
\end{array}\right| \\
& =-\hat{i}-\hat{j}-2 \hat{k}
\end{aligned}
$$
<br/><br/>$\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot \vec{b}_{1} \times \vec{b}_{2}=-6+1-\lambda+2 \lambda+8=\lambda+3$
<br/><br/>and $\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{6}$
<br/><br/>$\because \frac{|\lambda+3|}{\sqrt{6}}=2 \sqrt{6}$
<br/><br/>$\therefore \lambda=9, \because \lambda \geq 0$
<br/><br/>$\therefore \quad L: \frac{x-5}{-2}=\frac{y-9}{0}=\frac{z+9}{1}=k$
<br/><br/>$\therefore \quad \alpha=-2 k+5, \beta=9, \gamma=k-9$
<br/><br/>Here $k$ is real then
<br/><br/>$\alpha+2 \gamma=-13 \neq 24$.
<br/><br/>But all other are in terms of $k$ hence possible. | mcq | jee-main-2023-online-31st-january-morning-shift |
ldr0kpgp | maths | 3d-geometry | lines-in-space | Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to : | [] | null | 158 | <p>$$L:{{x - 2} \over 1} = {{y - 3} \over { - 1}} = {{z - 1} \over { - 1}} = \lambda $$</p>
<p>Any point on L can be taken as</p>
<p>$$B(\lambda + 2, - \lambda + 3, - \lambda + 1)$$</p>
<p>Let $$A(5,3,8)$$</p>
<p>So, $$AB\,.\,(\widehat i - \widehat j - \widehat k) = 0$$</p>
<p>$$[(\lambda - 3)\widehat i - \lambda \widehat j - (\lambda + 7)\widehat k]\,.\,[\widehat i - \widehat j - \widehat k] = 0$$</p>
<p>$$\lambda - 3 + \lambda + \lambda + 7 = 0$$</p>
<p>$$\therefore$$ $$\lambda = {{ - 4} \over 3}$$</p>
<p>$$\overrightarrow {AB} = {{13} \over 3}\widehat i + {4 \over 3}\widehat i - {{17} \over 3}\widehat k$$</p>
<p>$$|\overrightarrow {AB} | = \sqrt {{{169} \over 9} + {{16} \over 9} + {{289} \over 9}} $$</p>
<p>$$ = {{\sqrt {474} } \over 3} = \alpha $$</p>
<p>$$3{\alpha ^2} = {{474} \over 9} \times 3 = 158$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldsfqrb6 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$$ and $${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$$ is :</p> | [{"identifier": "A", "content": "$$2\\sqrt3$$"}, {"identifier": "B", "content": "$$3\\sqrt3$$"}, {"identifier": "C", "content": "$$4\\sqrt3$$"}, {"identifier": "D", "content": "$$5\\sqrt3$$"}] | ["C"] | null | <p>$${\overrightarrow r _1} = \widehat i - 8\widehat j + 4\widehat k$$</p>
<p>$${\overrightarrow r _2} = \widehat i + 2\widehat j + 6\widehat k$$</p>
<p>$$\overrightarrow a = 2\widehat i - 7\widehat j + 5\widehat k$$</p>
<p>$$\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k$$</p>
<p>S.D. $$ = {{\left| {\matrix{
{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{
{\overrightarrow a } & {\overrightarrow b } \cr
} } \cr
} } \right|} \over {\left| {\overrightarrow a \times \overrightarrow b } \right|}}$$</p>
<p>$$\left[ {\matrix{
{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{
{\overrightarrow a } & {\overrightarrow b } \cr
} } \cr
} } \right] = \left| {\matrix{
0 & { - 10} & { - 2} \cr
2 & { - 7} & 5 \cr
2 & 1 & { - 3} \cr
} } \right|$$</p>
<p>$$\therefore$$ $$10( - 16) - 2(16) = - 192$$</p>
<p>$$\left| {\left[ {\matrix{
{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{
{\overrightarrow a } & {\overrightarrow b } \cr
} } \cr
} } \right]} \right| = 192$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & { - 7} & 5 \cr
2 & 1 & { - 3} \cr
} } \right| = 16\widehat i + 16\widehat j + 16\widehat k$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = 16\sqrt 3 $$</p>
<p>S.D. $$ = {{192} \over {16\sqrt 3 }} = 4\sqrt 3 $$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsx6ir2 | maths | 3d-geometry | lines-in-space | <p>Let the co-ordinates of one vertex of $$\Delta ABC$$ be $$A(0,2,\alpha)$$ and the other two vertices lie on the line $${{x + \alpha } \over 5} = {{y - 1} \over 2} = {{z + 4} \over 3}$$. For $$\alpha \in \mathbb{Z}$$, if the area of $$\Delta ABC$$ is 21 sq. units and the line segment $$BC$$ has length $$2\sqrt{21}$$ units, then $$\alpha^2$$ is equal to ___________.</p> | [] | null | 9 | <p>A. $\left(\mathrm{O}_{1} 2, \alpha\right)$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lekt8kqs/c8535748-1d32-4d4a-8b11-6aa2b4c12893/a97e5c40-b582-11ed-b66a-d501f51666d6/file-1lekt8kqt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lekt8kqs/c8535748-1d32-4d4a-8b11-6aa2b4c12893/a97e5c40-b582-11ed-b66a-d501f51666d6/file-1lekt8kqt.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - 3D Geometry Question 94 English Explanation"></p>
$\left|\frac{1}{2} \cdot 2 \sqrt{21}.\right| \begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3\end{array}\left|\frac{1}{\sqrt{25+4+9}}\right|=21 \sqrt{21}$
<br><br>
$\sqrt{(2 \alpha+5)^{2}+(2 \alpha+20)^{2}+(2 \alpha-5)^{2}}=\sqrt{21} \sqrt{38}$
<br><br>
$\Rightarrow 12 \alpha^{2}+80 \alpha+450=798$
<br><br>
$\Rightarrow 12 \alpha^{2}+80 \alpha-348=0$
<br><br>
$\Rightarrow \alpha=3 \Rightarrow \alpha^{2}=9$ | integer | jee-main-2023-online-29th-january-morning-shift |
1ldu4b5nl | maths | 3d-geometry | lines-in-space | <p>The foot of perpendicular of the point (2, 0, 5) on the line $${{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}$$ is ($$\alpha,\beta,\gamma$$). Then, which of the following is NOT correct?</p> | [{"identifier": "A", "content": "$$\\frac{\\alpha}{\\beta}=-8$$"}, {"identifier": "B", "content": "$$\\frac{\\alpha \\beta}{\\gamma}=\\frac{4}{15}$$"}, {"identifier": "C", "content": "$$\\frac{\\beta}{\\gamma}=-5$$"}, {"identifier": "D", "content": "$$\\frac{\\gamma}{\\alpha}=\\frac{5}{8}$$"}] | ["C"] | null | <p>$$
\mathrm{L}: \frac{\mathrm{x}+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text { (let) }
$$</p><p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef5vgns/9610cbeb-a0ae-4272-89c8-8cbf8a31b61b/5f9aceb0-b267-11ed-9d4d-b96eca78f2e5/file-1lef5vgnx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lef5vgns/9610cbeb-a0ae-4272-89c8-8cbf8a31b61b/5f9aceb0-b267-11ed-9d4d-b96eca78f2e5/file-1lef5vgnx.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Evening Shift Mathematics - 3D Geometry Question 92 English Explanation"></p>Let foot of perpendicular is<br><br>
$$
\begin{aligned}
& \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \\\\
& \overrightarrow{\mathrm{PA}}=(3-2 \lambda) \hat{\mathrm{i}}-(5 \lambda+1) \hat{\mathrm{j}}+(6+\lambda) \hat{\mathrm{k}} \\\\
& \text { Direction ratio of line } \Rightarrow \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\\\
& \text { Now, } \Rightarrow \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{b}}=0 \\\\
& \Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0 \\\\
& \Rightarrow \lambda=\frac{-1}{6} \\\\
& \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv \mathrm{P}(\alpha, \beta, \gamma) \\\\
& \Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3} \\\\
& \Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6} \\\\
& \Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6}
\end{aligned}
$$
| mcq | jee-main-2023-online-25th-january-evening-shift |
1ldu52orl | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$x+1=2y=-12z$$ and $$x=y+2=6z-6$$ is :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\frac{5}{2}$$"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $L_{1}: \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}$
<br/><br/>
$$
\begin{aligned}
& L_{2}: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}} \\\\
& \text { S.D }=\left|\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{7}\right| \\\\
& =\left|\frac{-2-6-6}{7}\right|=2 \text { units }
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldu68afw | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {{z - 2} \over 0}$$ is $$\alpha$$, then 28$$\alpha^2$$ is equal to ____________.</p> | [] | null | 18 | Points $(1,2,3)$ and $(2,3,4)$
<br/><br/>
$L_{1}: \frac{(x-1)}{1}=\frac{(y-2)}{1}=\frac{(2-3)}{1}$
<br/><br/>
$L_{2}: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$
<br/><br/>
$\vec{b}_{1}=\hat{i}+\hat{j}+\hat{k}$
<br/><br/>
$\vec{b}_{2}=2 \hat{i}-\hat{j}+0 \hat{k}$<br/><br/>
$$
\begin{aligned}
& \overrightarrow{a_{1}}-\overrightarrow{a_{2}}=0 \hat{i}-3 \hat{j}-\hat{k} \\\\
& d=\left|\frac{\left(\bar{a}_{1}-\bar{a}_{2}\right) \cdot\left(n_{1} \times n_{2}\right)}{\left|n_{1} \times n_{2}\right|}\right| \\\\
&=\left|\frac{6-3}{\sqrt{9+1+4}}\right|=\frac{3}{\sqrt{14}}=\alpha \\\\
& 28 \alpha^{2}=\frac{28 \times 9}{14}=18
\end{aligned}
$$
| integer | jee-main-2023-online-25th-january-evening-shift |
1ldv1rqg2 | maths | 3d-geometry | lines-in-space | <p>The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\sqrt{14}$$"}, {"identifier": "C", "content": "$$\\sqrt6$$"}, {"identifier": "D", "content": "$$2\\sqrt3$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lebz24vh/fd89cb9d-00d8-44b0-a190-8609bc533a04/5f4f5dd0-b0a6-11ed-8017-e30f07067392/file-1lebz24vi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lebz24vh/fd89cb9d-00d8-44b0-a190-8609bc533a04/5f4f5dd0-b0a6-11ed-8017-e30f07067392/file-1lebz24vi.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Morning Shift Mathematics - 3D Geometry Question 89 English Explanation"><br><br>
$\overrightarrow{A P}=7 \hat{i}+4 \hat{j}-5 \hat{k} \Rightarrow|\overrightarrow{A P}|=\sqrt{49+16+25}=\sqrt{90} A N$
<br><br>
$=$ projection of $\overrightarrow{A P}$ on $\vec{b}=\overrightarrow{A P} \cdot \vec{b}=\frac{21+12+5}{\sqrt{19}}=\frac{38}{\sqrt{19}}$
<br><br>
$(P N)^{2}=(A P)^{2}-(A N)^{2}=90-76=14 \Rightarrow P N=\sqrt{14}$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldv1wbyd | maths | 3d-geometry | lines-in-space | <p>Consider the lines $$L_1$$ and $$L_2$$ given by</p>
<p>$${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$$</p>
<p>$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$.</p>
<p>A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is</p> | [{"identifier": "A", "content": "$$4\\sqrt3$$"}, {"identifier": "B", "content": "$$2\\sqrt6$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$3\\sqrt2$$"}] | ["B"] | null | Let,
<br/><br/>
$$
\begin{aligned}
& P \equiv(2 \lambda+1, \lambda+3,2 \lambda+2) \text { and } Q(\mu+2,2 \mu+2 \text {, } 3 \mu+3) \\\\
& \text { d.r's of } P Q \equiv<2 \lambda-\mu-1, \lambda-2 \mu+1,2 \lambda-3 \mu-1> \\\\
& \therefore \quad \frac{2 \lambda-\mu-1}{1}-\frac{\lambda-2 \mu-1}{-1}=\frac{2 \lambda-3 \mu-1}{-2} \\\\
& \therefore \quad-2 \lambda+\mu+1=\lambda-2 \mu+1 \text { and }-2 \lambda+4 \mu-2= \\\\
& -2 \lambda+3 \mu+1 \\\\
& \Rightarrow 3 \lambda-3 \mu=0 \text { and } \mu=3 \\\\
& \therefore \quad \lambda=\pm 3 \text { and } \mu=3 \\\\
& \therefore \quad P \equiv(7,6,8) \text { and } Q(5,8,12) \\\\
& \therefore|P O|=\sqrt{2^{2}+2^{2}+4^{2}}=\sqrt{24}=2 \sqrt{6}
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwxg3ot | maths | 3d-geometry | lines-in-space | <p>If the shortest between the lines $${{x + \sqrt 6 } \over 2} = {{y - \sqrt 6 } \over 3} = {{z - \sqrt 6 } \over 4}$$ and $${{x - \lambda } \over 3} = {{y - 2\sqrt 6 } \over 4} = {{z + 2\sqrt 6 } \over 5}$$ is 6, then the square of sum of all possible values of $$\lambda$$ is :</p> | [] | null | 384 | Shortest distance between the lines<br/><br/> $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$<br/><br/> $\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6<br/><br/>
Vector along line of shortest distance<br/><br/> $=\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k}$ (its magnitude is $\sqrt{6}$ )<br/><br/>
Now $\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=\pm 6$ <br/><br/>$\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}$<br/><br/>
So, square of sum of these values
<br/><br/>$$
=(10 \sqrt{6}-2 \sqrt{6})^2=(8 \sqrt{6})^2=384
$$ | integer | jee-main-2023-online-24th-january-evening-shift |
1ldybxlrv | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$ and $${{x - 6} \over 3} = {{1 - y} \over 2} = {{z + 8} \over 0}$$ is equal to ________</p> | [] | null | 14 | <p>For $${L_1}:$$$${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$</p>
<p>$$\therefore$$ Point on line is $$A(2,-1,6)$$</p>
<p>Parallel vector to this line is,</p>
<p>$$\overrightarrow p = 3\widehat i + 2\widehat j + 2\widehat k$$</p>
<p>For $${L_2}:$$$${{x - 6} \over 3} = {{y - 1} \over -2} = {{z + 8} \over 0}$$</p>
<p>$$\therefore$$ Point on line is $$B(6,1,-8)$$</p>
<p>Parallel vector to this line is,</p>
<p>$$\overrightarrow q = 3\widehat i - 2\widehat j + 0\widehat k$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2m97s7/2871d106-b3b3-42e5-b023-ab35f716d2cf/3fb28b70-ab81-11ed-88fd-4faade8d5a9f/file-1le2m97s8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2m97s7/2871d106-b3b3-42e5-b023-ab35f716d2cf/3fb28b70-ab81-11ed-88fd-4faade8d5a9f/file-1le2m97s8.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Morning Shift Mathematics - 3D Geometry Question 81 English Explanation"></p>
<p>Now shortest distance between this two lines is,</p>
<p>$$ = \left| {{{\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$$</p>
<p>Now, $$\overrightarrow {AB} = 4 \widehat i + 2\widehat j -14\widehat k$$</p>
<p>$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 2 & 2 \cr
3 & -2 & 0 \cr
} } \right|$$</p>
<p>$$ = 4\widehat i +6\widehat j -12\widehat k$$</p>
<p>$$\therefore$$ $|\vec{p} \times \vec{q}|=\sqrt{16+36+144}=\sqrt{196}=14$</p>
<p>and $$\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right) = + 16 + 12 + 168 = +196$$</p>
<p>$$\therefore$$ Shortest distance $$ = \left| {{{ 196} \over {14 }}} \right| = 14$$</p> | integer | jee-main-2023-online-24th-january-morning-shift |
lgnyep2q | maths | 3d-geometry | lines-in-space | Let $\mathrm{S}$ be the set of all values of $\lambda$, for which the shortest distance between<br/><br/> the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is 13. Then $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ is equal to : | [{"identifier": "A", "content": "306"}, {"identifier": "B", "content": "304"}, {"identifier": "C", "content": "308"}, {"identifier": "D", "content": "302"}] | ["A"] | null | Given the two lines :
<br/><br/>$$
\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1} \\\\
\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}
$$
<br/><br/>Let's find the direction vectors of these lines: $\vec{d_1} = \langle 0, 4, 1 \rangle$ and $\vec{d_2} = \langle 3, -4, 0 \rangle$.
<br/><br/>Now, let's find the cross product of the direction vectors, which will give a vector that is perpendicular to both lines :
<br/><br/>$\vec{n} = \vec{d_1} \times \vec{d_2} = \langle 4, 3, -12 \rangle$
<br/><br/>Let's find the vector connecting a point on line 1 to a point on line 2 :
<br/><br/>$\vec{c} = \langle 2\lambda, 3, -12 \rangle$
<br/><br/>The shortest distance between the two lines is the projection of $\vec{c}$ onto $\vec{n}$ :
<br/><br/>$d = \left|\frac{\vec{c} \cdot \vec{n}}{|\vec{n}|}\right| = \left|\frac{(2\lambda)(4) + (3)(3) - (12)(-12)}{\sqrt{16 + 9 + 144}}\right|$
<br/><br/>We are given that the shortest distance is 13 :
<br/><br/>$13 = \left|\frac{8\lambda + 153}{13}\right|$
<br/><br/>$|8\lambda + 153| = 169$
<br/><br/>We have two cases :
<br/><br/>1. $8\lambda + 153 = 169$
<br/><br/>$\lambda = \frac{16}{8}$
<br/><br/>2. $8\lambda + 153 = -169$
<br/><br/>$\lambda = \frac{-322}{8}$
<br/><br/>Now, let's calculate $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ :
<br/><br/>$8\left|\frac{16}{8} + \frac{-322}{8}\right| = 8\left|\frac{-306}{8}\right| = 306$
<br/><br/>Thus, the correct answer is Option A : 306. | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgowajui | maths | 3d-geometry | lines-in-space | <p>The line, that is coplanar to the line $$\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{x+1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{4}$$"}, {"identifier": "B", "content": "$$\\frac{x+1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{5}$$"}, {"identifier": "C", "content": "$$\\frac{x-1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{5}$$"}, {"identifier": "D", "content": "$$\\frac{x+1}{1}=\\frac{y-2}{2}=\\frac{z-5}{5}$$"}] | ["B"] | null | <p>Given two lines:</p>
<p>$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$</p>
<p>and </p>
<p>$\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$</p>
<p>These lines are coplanar if the determinant of the matrix</p>
<p>$$
\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix}
$$ = 0</p>
<p>Now let's apply this condition to the given problem. The given line is :</p>
<p>$\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$</p>
<p>So, the coordinates of any point on this line are $(-3, 1, 5)$ and the direction ratios are $(-3, 1, 5)$.</p>
<p>Now, let's calculate the determinants for each option and check which one equals zero.</p>
<p>For Option A :</p>
<p>$\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{4}$ </p>
<p>The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 4)$.</p>
<p>The determinant is :</p>
<p>$$
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{vmatrix}
$$</p>
<p>Applying the formula :</p>
<p>= $2(1 \times 4 - 5\times2) - 1((-3\times4) - (5\times-1)) + 0((-3\times2) - (1\times-1))$</p>
<p>= $2(4 - 10) - 1(-12 - (-5)) + 0(-6 - (-1))$</p>
<p>= $2(-6) - 1(-7) + 0(-5)$</p>
<p>= $-12 + 7 + 0$</p>
<p>= $-5$ </p>
<p>The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.</p>
<p>For Option B, we have :</p>
<p>$\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$</p>
<p>The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.</p>
<p>$$
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
$$</p>
<p>Applying the formula, we get :</p>
<p>= $2(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$
</p>
<p>= $2(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$</p>
<p>= $2(-5) - 1(-10) + 0(-5)$</p>
<p>= $-10 + 10 + 0$</p>
<p>= $0$</p>
So, the determinant for Option B equals zero, which confirms that the line in Option B is coplanar with the given line.
<p>For Option C :</p>
<p>$\frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ </p>
<p>The coordinates of any point on this line are $(1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.</p>
<p>The determinant is:</p>
<p>$$
\begin{vmatrix}
1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
4 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{vmatrix}
$$</p>
<p>Applying the formula :</p>
<p>= $4(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$
</p>
<p>= $4(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$</p>
<p>= $4(-5) - 1(-10) + 0(-5)$</p>
<p>= $-20 + 10 + 0$</p>
<p>= $-10$</p>
<p>The determinant for Option C is not equal to zero, so this line is not coplanar with the given line.</p>
<p>For Option D :</p>
<p>$\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ </p>
<p>The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(1, 2, 5)$.</p>
<p>The determinant is :</p>
<p>$$
\begin{vmatrix}
-1 - (-3) & 2 - 1 & 5 - 5 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{vmatrix}
=
\begin{vmatrix}
2 & 1 & 0 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{vmatrix}
$$</p>
<p>Applying the formula :</p>
<p>= $2(1*5 - 5*2) - 1((-3*5) - (5*1)) + 0((-3*2) - (1*1))$
</p>
<p>= $2(5 - 10) - 1(-15 - 5) + 0(-6 - 1)$</p>
<p>= $2(-5) - 1(-20) + 0(-7)$
</p>
<p>= $-10 + 20 + 0$</p>
<p>= $10$</p>
<p>So the determinant for Option D is not equal to zero, which means the line in Option D is not coplanar with the given line.</p> | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgrebnuy | maths | 3d-geometry | lines-in-space | <p>Let the lines $$l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$$ and $$l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$$ be coplanar. If the point $$\mathrm{P}(a, b, c)$$ on $$l_{1}$$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "8"}] | ["C"] | null | $$
\begin{aligned}
& (3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}-2)+\mu(\mathrm{x}-3 \mathrm{y}+2 \mathrm{z}-13)=0 \\\\
& 3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0 \\\\
& 9-4 \mu=0 \\\\
& \mu=\frac{9}{4} \\\\
& 4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0 \\\\
& -100+4 \alpha-54+18 \alpha=0 \\\\
& \Rightarrow \alpha=7 \\\\
& \text { Let } \mathrm{P} \equiv(3 \lambda-5, \lambda-4,-2 \lambda+7) \\\\
& \text { Direction ratio of } \mathrm{PQ}(3 \lambda-1, \lambda-1,-2 \lambda+5) \\\\
& \text { But PQ } \perp \ell_1 \\\\
& \Rightarrow 3(3 \lambda-1)+1 \cdot(\lambda-1)-2(-2 \lambda+5)=0 \\\\
& \Rightarrow \lambda=1 \\\\
& \mathrm{P}(-2,-3,5) \Rightarrow|\mathrm{a}|+|\mathrm{b}|+|\mathrm{c}|=10
\end{aligned}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lguwwp0s | maths | 3d-geometry | lines-in-space | <p>Let a line $$l$$ pass through the origin and be perpendicular to the lines</p>
<p>$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and</p>
<p>$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$$.</p>
<p>If $$\mathrm{P}$$ is the point of intersection of $$l$$ and $$l_{1}$$, and $$\mathrm{Q}(\propto, \beta, \gamma)$$ is the foot of perpendicular from P on $$l_{2}$$, then $$9(\alpha+\beta+\gamma)$$ is equal to _____________.</p> | [] | null | 5 | We have,<br/>
<p>$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and</p>
<p>$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$$.</p>
Let direction ratio of line $l$ be $a, b$ and $c$ Equation of line $l$
<br/><br/>$$
\begin{aligned}
\vec{r} & =(0 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+0 \hat{\mathbf{k}})+\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \\\\
& =\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}})
\end{aligned}
$$
<br/><br/>As, line $l$ is perpendicular to $l_1$ and $l_2$,
<br/><br/>$$
a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 2 & 3 \\
2 & 2 & 1
\end{array}\right|=-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}
$$
<br/><br/>$\therefore$ Equation of line $l: \vec{r}=\delta(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$
<br/><br/>As, $P$ is the intersecting point of $l$ and $l_1$
<br/><br/>$$
-4 \delta=1+\lambda, 5 \delta=-11+2 \lambda,-2 \delta=-7+3 \lambda
$$
<br/><br/>After solving the above three equation, we get
<br/><br/>$$
\delta=-1 \text { and } \lambda=3
$$
<br/><br/>$\therefore$ Co-ordinate of point $P$ is $(4,-5,2)$.
<br/><br/>$Q$ is a point on line $l_2$
<br/><br/>Let co-ordinate of $Q$ be $(-1+2 \mu, 2 \mu, 1+\mu)$
<br/><br/>$$
\begin{gathered}
\overrightarrow {PQ}=(-5+2 \mu) \hat{\mathbf{i}}+(2 \mu+5) \hat{\mathbf{j}}+(\mu-1) \hat{\mathbf{k}} \\\\
\overrightarrow {PQ} \cdot(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \left[\because \overrightarrow {PQ} \perp l_2\right]\\\\
2(-5+2 \mu)+2(2 \mu+5)+\mu-1=0 \\\\
9 \mu-1=0 \Rightarrow \mu=\frac{1}{9}
\end{gathered}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \alpha=-1+\frac{2}{9}=\frac{-7}{9}, \beta=2 \times \frac{1}{9}=\frac{2}{9}, r=1+\frac{1}{9}=\frac{10}{9} \\\\
& \text { Hence, } 9(\alpha+\beta+\gamma)=9\left(-\frac{7}{9}+\frac{2}{9}+\frac{10}{9}\right)=5
\end{aligned}
$$ | integer | jee-main-2023-online-11th-april-morning-shift |
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