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1lsg90h1z | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\mathrm{b}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{k}$$ be two vectors such that $$|\overrightarrow{\mathrm{a}}|=1, \vec{a} \cdot \vec{b}=2$$ and $$|\vec{b}|=4$$. If $$\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$$, then the angle between $$\vec{b}$$ and $$\vec{c}$$ is equal to:</p> | [{"identifier": "A", "content": "$$\\cos ^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)$$\n"}, {"identifier": "B", "content": "$$\\cos ^{-1}\\left(\\frac{2}{3}\\right)$$\n"}, {"identifier": "C", "content": "$$\\cos ^{-1}\\left(\\frac{2}{\\sqrt{3}}\\right)$$\n"}, {"identifier": "D", "content": "$$\\cos ^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)$$"}] | ["D"] | null | <p>Given $$|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$$</p>
<p>$$\vec{\mathrm{c}}=2(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-3 \vec{\mathrm{b}}$$</p>
<p>Dot product with $$\overrightarrow{\mathrm{a}}$$ on both sides</p>
<p>$$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$$ ..... (1)</p>
<p>Dot product with $$\vec{b}$$ on both sides</p>
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 \quad \text{... (2)}\\
& \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 \\
& |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 \\
& |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) \\
& |\overrightarrow{\mathrm{c}}|^2=4[12]+144 \\
& |\overrightarrow{\mathrm{c}}|^2=48+144 \\
& |\overrightarrow{\mathrm{c}}|^2=192 \\
& \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{c}} \mid} \\
& \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} \\
& \therefore \cos \theta=\frac{-48}{8 \sqrt{3} .4} \\
& \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} \\
& \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luxwcrzd | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Between the following two statements:</p>
<p>Statement I : Let $$\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$$ and $$\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$$. Then the vector $$\vec{r}$$ satisfying $$\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$$ and $$\vec{a} \cdot \vec{r}=0$$ is of magnitude $$\sqrt{10}$$.</p>
<p>Statement II : In a triangle $$A B C, \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2}$$.</p> | [{"identifier": "A", "content": "Both Statement I and Statement II are correct.\n"}, {"identifier": "B", "content": "Both Statement I and Statement II are incorrect.\n"}, {"identifier": "C", "content": "Statement I is correct but Statement II is incorrect.\n"}, {"identifier": "D", "content": "Statement I is incorrect but Statement II is correct."}] | ["D"] | null | <p>$$\begin{aligned}
& \because \quad \forall \text { two vectors } \vec{c} \text { & } \vec{d} \\
& |\vec{c} \times \vec{d}|^2=|\vec{c}|^2|\vec{d}|^2-(\vec{c} \cdot \vec{d})^2 \\
& \text { replacing } \vec{c}=\vec{a} ~\& ~\vec{d}=\vec{r} \\
& \Rightarrow|\vec{a} \times \vec{r}|=|\vec{a}|^2|\vec{r}|^2-(\vec{a} \cdot \vec{r})^2 \\
& \Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}|^2|\vec{r}|^2 \quad(\because \vec{a} \times \vec{r}=\vec{a} \times \vec{b} \text { and } \vec{a} \cdot \vec{r}=0)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow 35=14|\vec{r}|^2 \\
& \Rightarrow|\vec{r}|=\sqrt{\frac{35}{14}}=\sqrt{\frac{5}{2}} \neq \sqrt{10}
\end{aligned}$$</p>
<p>$$\therefore$$ Statement I is incorrect</p>
<p>Statement II is correct</p>
<p>$$\text { (i.e., } \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \text { ) }$$</p>
<p>Proof: $$\because(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \geq 0\quad \text{..... (1)}$$</p>
<p>and $$|\overrightarrow{O A}|^2=|\overrightarrow{O B}|^2=|\overrightarrow{O C}|^2=R^2\quad \text{..... (2)}$$</p>
<p>Now, using (1), we get</p>
<p>$$|\overrightarrow{O A}|^2+|\overrightarrow{O B}|^2+|\overrightarrow{O C}|^2 +2(\overrightarrow{O A} \cdot \overrightarrow{O B}+\overrightarrow{O B} \cdot \overrightarrow{O C}+\overrightarrow{O C} \cdot \overrightarrow{O A}) \geq 0$$</p>
<p>$$
\begin{aligned}
& \Rightarrow 3 R^2+2 R^2(\cos 2 A+\cos 2 B+\cos 2 C) \geq 0 \\
& \Rightarrow \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2}
\end{aligned}
$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luxwe3dl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}$$, where $$\alpha$$ and $$\beta$$ are integers and $$\alpha \beta=-6$$. Let the values of the ordered pair $$(\alpha, \beta)$$, for which the area of the parallelogram of diagonals $$\vec{a}+\vec{b}$$ and $$\vec{b}+\vec{c}$$ is $$\frac{\sqrt{21}}{2}$$, be $$\left(\alpha_1, \beta_1\right)$$ and $$\left(\alpha_2, \beta_2\right)$$. Then $$\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$$ is equal to</p> | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "17"}] | ["C"] | null | <p>Area of parallelogram whose diagonals are $$\vec{a}+\vec{b}$$ and $$\vec{b}+\vec{c}$$ is</p>
<p>$$\begin{aligned}
& =\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\
& =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}| \\
& =\frac{1}{2}|-2 \beta \hat{i}-2 \hat{j}+(\alpha+\beta) \hat{k}|
\end{aligned}$$</p>
<p>$$=\frac{1}{2} \sqrt{4 \beta^2+4+(\alpha+\beta)^2}$$</p>
<p>Which is given $$\frac{\sqrt{21}}{2}$$</p>
<p>$$\begin{array}{ll}
\therefore & 4 \beta^2+4+(\alpha+\beta)^2=21 \\
\Rightarrow & (\alpha+\beta)^2+4 \beta^2=17 \\
\Rightarrow & \alpha^2+5 \beta^2+2 \alpha \beta=17 \\
\Rightarrow & \alpha^2+5 \beta^2=29 \\
\therefore & (\alpha, \beta) \in\{(3,2),(-3,-2),(-3,2),(3,-2)\} \\
\because & \alpha \beta=-6 \\
\therefore & (\alpha, \beta) \in\{(-3,2),(3,-2)\} \\
\therefore & \alpha_1^2+\beta_1^2-\alpha_2 \beta_2 \\
= & 9+4-(-6)=19
\end{array}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luy6z4lq | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let three vectors ,$$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$$ form a triangle such that $$\vec{c}=\vec{a}-\vec{b}$$ and the area of the triangle is $$5 \sqrt{6}$$. If $$\alpha$$ is a positive real number, then $$|\vec{c}|^2$$ is equal to:</p> | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "10"}] | ["A"] | null | <p>To solve this, let's start with the given vector equation:</p>
<p>$$\vec{c} = \vec{a} - \vec{b}$$</p>
<p>Given vectors are:</p>
<p>$$\overrightarrow{\mathrm{a}} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}$$</p>
<p>$$\overrightarrow{\mathrm{b}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$$</p>
<p>Then, the vector $$\overrightarrow{\mathrm{c}}$$ is:</p>
<p>$$\overrightarrow{\mathrm{c}} = (\alpha \hat{i} + 4 \hat{j} + 2 \hat{k}) - (5 \hat{i} + 3 \hat{j} + 4 \hat{k})$$</p>
<p>$$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + (4 - 3) \hat{j} + (2 - 4) \hat{k}$$</p>
<p>$$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + 1 \hat{j} - 2 \hat{k}$$</p>
<p>The area of the triangle formed by vectors $$\vec{a}$$ and $$\vec{b}$$ is given by the magnitude of the cross product of $$\vec{a}$$ and $$\vec{b}$$, divided by 2:</p>
<p>$$\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = 5 \sqrt{6}$$</p>
<p>This implies:</p>
<p>$$|\vec{a} \times \vec{b}| = 10 \sqrt{6}$$</p>
<p>Let's find $$\vec{a} \times \vec{b}$$:</p>
<p>$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix} = \hat{i}(4 \cdot 4 - 2 \cdot 3) - \hat{j}(\alpha \cdot 4 - 2 \cdot 5) + \hat{k}(\alpha \cdot 3 - 4 \cdot 5)$$</p>
<p>$$\vec{a} \times \vec{b} = \hat{i}(16 - 6) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$$</p>
<p>$$\vec{a} \times \vec{b} = \hat{i}(10) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$$</p>
<p>Magnitude of $$\vec{a} \times \vec{b}$$:</p>
<p>$$|\vec{a} \times \vec{b}| = \sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2}$$</p>
<p>We know:</p>
<p>$$\sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2} = 10 \sqrt{6}$$</p>
<p>Squaring both sides:</p>
<p>$$100 + (4\alpha - 10)^2 + (3\alpha - 20)^2 = 600$$</p>
<p>$$100 + 16\alpha^2 - 80\alpha + 100 + 9\alpha^2 - 120\alpha + 400 = 600$$</p>
<p>$$25\alpha^2 - 200\alpha + 600 = 600$$</p>
<p>$$25\alpha^2 - 200\alpha = 0$$</p>
<p>$$\alpha^2 - 8\alpha = 0$$</p>
<p>$$\alpha(\alpha - 8) = 0$$</p>
<p>Since $$\alpha$$ is a positive real number:</p>
<p>$$\alpha = 8$$</p>
<p>Then, the vector $$\vec{c}$$ is:</p>
<p>$$\vec{c} = (8 - 5)\hat{i} + 1\hat{j} - 2\hat{k}$$</p>
<p>$$\vec{c} = 3\hat{i} + 1\hat{j} - 2\hat{k}$$</p>
<p>Magnitude squared of $$\vec{c}$$ is:</p>
<p>$$|\vec{c}|^2 = 3^2 + 1^2 + (-2)^2$$</p>
<p>$$|\vec{c}|^2 = 9 + 1 + 4$$</p>
<p>$$|\vec{c}|^2 = 14$$</p>
<p>Therefore, the correct option is:</p>
<p>Option A: 14</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
luy6z5a5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}$$ and $$\overrightarrow{O C}=3 \vec{b}$$, where $$O$$ is the origin. If the area of the parallelogram with adjacent sides $$\overrightarrow{O A}$$ and $$\overrightarrow{O C}$$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $$O A B C$$ is equal to:</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "38"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "40"}] | ["C"] | null | <p>$$\begin{aligned}
& 6|\vec{a} \times \vec{b}|=15 \\
& \Rightarrow|\vec{a} \times \vec{b}|=\frac{5}{2}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3cizlj/32ca14d9-c7d9-4752-a76d-bc023959c231/e1a4ec70-1043-11ef-9f6c-75804a813f04/file-jaoe38c1lw3cizlk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lw3cizlj/32ca14d9-c7d9-4752-a76d-bc023959c231/e1a4ec70-1043-11ef-9f6c-75804a813f04/file-jaoe38c1lw3cizlk.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Vector Algebra Question 16 English Explanation 1"></p>
<p>Area of quadrilateral $$O A B C$$</p>
<p>$$=$$ area of $$\triangle O A C+$$ area of $$\triangle A B C$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3cka9k/5af601d2-c9f5-428f-ad0f-ea9fdc1cb8f2/05b19780-1044-11ef-9f6c-75804a813f04/file-jaoe38c1lw3cka9l.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lw3cka9k/5af601d2-c9f5-428f-ad0f-ea9fdc1cb8f2/05b19780-1044-11ef-9f6c-75804a813f04/file-jaoe38c1lw3cka9l.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Vector Algebra Question 16 English Explanation 2"></p><p>$$\begin{aligned}
& =\frac{15}{2}+\frac{1}{2}|(\overrightarrow{A B} \times \overrightarrow{B C})| \\
& =\frac{15}{2}+\frac{1}{2}|(4 \vec{a}+5 \vec{b}) \times(6 \vec{a}+2 \vec{b})| \\
& =\frac{15}{2}+\frac{1}{2}|(22 \vec{a} \times \vec{b})| \\
& =\frac{15}{2}+11 \times \frac{5}{2} \\
& =\frac{70}{2}=35
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxdun | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\mathrm{ABC}$$ be a triangle of area $$15 \sqrt{2}$$ and the vectors $$\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k}$$ and $$\overrightarrow{\mathrm{AC}}=6 \hat{i}+\mathrm{d} \hat{j}-2 \hat{k}, \mathrm{~d}>0$$. Then the square of the length of the largest side of the triangle $$\mathrm{ABC}$$ is _________.</p> | [] | null | 54 | <p>Area of triangle $$A B C=15 \sqrt{2}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{1}{2}|\overline{A B} \times \overline{A C}|=15 \sqrt{2} \quad \text{.... (i)}\\
& \quad \overline{A B} \times \overline{A C}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -7 \\
6 & d & -2
\end{array}\right| \\
& =(7 d-4) \hat{i}-40 \hat{j}+(d-12) \hat{k} \quad \text{... (ii)}
\end{aligned}$$</p>
<p>From (i) and (ii) $$5 d^2-8 d-4=0$$</p>
<p>$$\Rightarrow d=\frac{-}{5}$$ (Rejected) or $$d=2$$</p>
<p>Also, $$\overline{A B}+\overline{B C}=\overline{A C}$$</p>
<p>$$\begin{aligned}
& \Rightarrow a+1=6 \Rightarrow a=5 \\
& b+2=d \Rightarrow b=0
\end{aligned}$$</p>
<p>and $$c-7=-2 \Rightarrow c=5$$</p>
<p>$$|\overline{A B}|=\sqrt{54},|\overline{A C}|=\sqrt{44},|\overline{B C}|=\sqrt{50}$$</p>
<p>Largest side has length of $$\sqrt{54}$$ units</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lv3vefae | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})$$.
If $$(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$$, then $$|\vec{c}|^2$$ is equal to:</p> | [{"identifier": "A", "content": "1600"}, {"identifier": "B", "content": "1618"}, {"identifier": "C", "content": "1627"}, {"identifier": "D", "content": "1609"}] | ["B"] | null | <p>$$\begin{aligned}
& \left.\begin{array}{l}
\vec{a}=4 \hat{i}-\hat{j}+\hat{k} \\
\vec{b}=11 \hat{i}-\hat{j}+\hat{k}
\end{array}\right] \\
& \vec{a} \cdot \vec{b}=44+1+1=46 \\
& |\vec{a}|^2=18,\left|\vec{b}^2\right|=123 \\
& (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \\
& (2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670 \\
& (\vec{a}+\vec{b}) \times c=(2 \vec{a}-3 \vec{b}) \times \vec{c} \\
& (-\vec{a}+4 \vec{b}) \times \vec{c}=0 \\
& \vec{c}=\lambda(4 \vec{b}-\vec{a}) \\
& (2 \vec{a}+3 \vec{b}) \cdot \lambda(4 \vec{b}-\vec{a})=1670 \\
& \lambda\left(5 \vec{a} \cdot \vec{b}-2|\vec{a}|^2+12|\vec{b}|^2\right)=1670 \\
& \lambda=\frac{1670}{5 \times 46-2 \times 18+12 \times 123} \\
& \lambda=1 \\
& \vec{c}=4 \vec{b}-\vec{a} \\
& =4(11 \hat{i}-\hat{j}+\hat{k})-(4 \hat{i}-\hat{j}+\hat{k}) \\
& =40 \hat{i}-3 \hat{j}+3 \hat{k} \\
& \left|\vec{c}^2\right|=1600+9+9 \\
& =1618 \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv5gt1qa | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$$ and $$\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$$ be three given vectors. If $$\vec{r}$$ is a vector such that $$\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$$ and $$\vec{r} \cdot(\vec{b}-\vec{c})=0$$, then $$\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$$ is equal to __________.</p> | [] | null | 569 | <p>$$\begin{aligned}
& \vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k} \\
& \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k} \\
& \vec{c}=17 \hat{i}-2 \hat{j}+\hat{k} \\
& \vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a} \\
& (\vec{r}-(\vec{b}+\vec{c})) \times \vec{a}=0 \\
& \Rightarrow \vec{r}=(\vec{b}+\vec{c})+\lambda \vec{a} \\
& \vec{r}=(20 \hat{i}+5 \hat{j}-12 \hat{k})+\lambda(9 \hat{i}-13 \hat{j}+25 \hat{k}) \\
& =(20+9 \lambda) \hat{i}+(5-13 \lambda) \hat{j}+(25 \lambda-12) \hat{k}
\end{aligned}$$</p>
<p>Now $$\vec{r} \cdot(\vec{b}-\vec{c})=0$$</p>
<p>$$\vec{r} \cdot(-14 \hat{i}+9 \hat{j}-14 \hat{k})=0$$</p>
<p>Now</p>
<p>$$\begin{aligned}
& -14(20+9 \lambda)+9(5-13 \lambda)-14(25 \lambda-12)=0 \\
& -593 \lambda-67=0 \\
& \lambda=-\frac{67}{593} \\
& \therefore \vec{r}=(\vec{b}+\vec{c})-\frac{67}{593} \vec{a} \\
& \frac{|593 \vec{r}+67 \vec{a}|^2}{|593|^2}=|\vec{b}+\vec{c}|^2=|20 \hat{i}+5 \hat{j}-12 \hat{k}|^2 \\
& =569
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lv7v4g3o | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>If $$\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)$$ and $$\mathrm{D}(-1,-4,-2)$$ are the vertices of a quadrilateral ABCD, then its area is :</p> | [{"identifier": "A", "content": "$$24 \\sqrt{7}$$\n"}, {"identifier": "B", "content": "$$48 \\sqrt{7}$$\n"}, {"identifier": "C", "content": "$$24 \\sqrt{29}$$\n"}, {"identifier": "D", "content": "$$12 \\sqrt{29}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgfpgex/f708b26f-9a53-4063-a888-8e61f5ee5884/8151b490-1776-11ef-bded-616e4abb66b9/file-1lwgfpgey.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgfpgex/f708b26f-9a53-4063-a888-8e61f5ee5884/8151b490-1776-11ef-bded-616e4abb66b9/file-1lwgfpgey.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Vector Algebra Question 6 English Explanation"></p>
<p>Area of quadrilateral $$A B C D=$$ area of $$\triangle A B C+$$ area of $$\triangle A D C$$</p>
<p>In $$\triangle A B C$$</p>
<p>$$\begin{aligned}
& \overrightarrow{A B}=4,8,-8 \\
& \overrightarrow{B C}=-2,-3,-4 \\
& \overrightarrow{A B} \times \overrightarrow{B C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & 8 & -8 \\
-2 & -3 & -4
\end{array}\right| \\
& =-56 \hat{i}+32 \hat{j}+4 \hat{k}
\end{aligned}$$</p>
<p>Area of $$\triangle A B C=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|$$</p>
<p>$$\begin{aligned}
& =\frac{1}{2} \sqrt{56^2+32^2+4^2} \\
& =\frac{1}{2} \sqrt{4176}=\frac{12 \sqrt{29}}{2}=6 \sqrt{29}
\end{aligned}$$</p>
<p>In $$\triangle A D C=\overrightarrow{A D}=-2,-3,-4$$</p>
<p>$$\overrightarrow{D C}=4,8,-8$$</p>
<p>$$\overrightarrow{A D} \times \overrightarrow{D C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & -4 \\ 4 & 8 & -8\end{array}\right|$$</p>
<p>$$=56 \hat{i}-32 \hat{j}-4 k$$</p>
<p>$$\text { Area of } \frac{1}{2}|\overrightarrow{A D} \times \overrightarrow{D C}|$$</p>
<p>$$\begin{aligned}
& =\frac{1}{2} \sqrt{4176} \\
& =6 \sqrt{29}
\end{aligned}$$</p>
<p>Area of $$A B C D=6 \sqrt{29}+6 \sqrt{29}$$</p>
<p>$$=12 \sqrt{29}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
lv7v3oem | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$$.
If $$\vec{a} \cdot \vec{c}=130$$, then $$\vec{b} \cdot \vec{c}$$ is equal to __________.</p> | [] | null | 30 | <p>$$(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$$</p>
<p>$$\begin{aligned}
\Rightarrow \quad & \vec{b} \times \vec{c}+2(\vec{a} \times \vec{c})=0 \\
& (\vec{b}+2 \vec{a}) \times \vec{c}=0 \\
& \vec{c}=\lambda(\vec{b}+2 \vec{a}) \\
& \vec{c} \cdot \vec{a}=130 \Rightarrow \lambda=1 \\
& \vec{c}=4 \hat{i}-7 \hat{j}+15 \hat{k} \\
& \vec{b} . \vec{c}=30
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s2007 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$$ and $$\vec{c}$$ be three vectors such that $$(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$$. If $$\vec{a} \cdot \vec{c}=-29$$, then $$\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$$ is equal to:</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "12"}] | ["C"] | null | <p>$$\begin{gathered}
(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}+\vec{a})=0 \\
\Rightarrow \quad \vec{c}+\hat{i}=\lambda(\vec{a}+\vec{b}+\hat{i}+\vec{a}) \\
=\lambda(2 \vec{a}+\vec{b}+\hat{i}) \\
\quad=\lambda(7 \hat{i}+8 \hat{j}) \\
\Rightarrow \quad \vec{c}=(7 \lambda-1) \hat{i}+8 \lambda \hat{j} \\
\quad \vec{c} \cdot \vec{a}=-29 \\
\Rightarrow \quad 14 \lambda-2+40 \lambda=-29 \\
\Rightarrow \quad 54 \lambda=-27 \\
\Rightarrow \quad \lambda=-\frac{1}{2}
\end{gathered}$$</p>
<p>$$\begin{aligned}
& \therefore \quad \vec{c}=\left(\frac{-7}{2}-1\right) \hat{i}-4 \hat{j}=\frac{-9}{2} \hat{i}-4 \hat{j} \\
& \vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=9-4=5
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lv9s20g2 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Consider three vectors $$\vec{a}, \vec{b}, \vec{c}$$. Let $$|\vec{a}|=2,|\vec{b}|=3$$ and $$\vec{a}=\vec{b} \times \vec{c}$$. If $$\alpha \in\left[0, \frac{\pi}{3}\right]$$ is the angle between the vectors $$\vec{b}$$ and $$\vec{c}$$, then the minimum value of $$27|\vec{c}-\vec{a}|^2$$ is equal to:</p> | [{"identifier": "A", "content": "124"}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "121"}, {"identifier": "D", "content": "105"}] | ["A"] | null | <p>$$\begin{aligned}
& \vec{a}=\vec{b} \times \vec{c} \\
& |\vec{a}|=2,|\vec{b}|=3
\end{aligned}$$</p>
<p>$$\vec{a} \cdot \vec{b}=0$$ and $$\vec{a} \cdot \vec{c}=0$$</p>
<p>$$\begin{aligned}
& |\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a} \\
& =4+|\vec{c}|^2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& |\vec{a}|=|\vec{b} \times \vec{c}|=|\vec{b}| \sin \alpha|\vec{c}| \\
& \Rightarrow \sin \alpha|\vec{c}|=\frac{2}{3} \\
& \Rightarrow \sin ^2 \alpha=\frac{4}{9|\vec{c}|^2} \\
& \Rightarrow|\vec{c}|^2=\frac{4}{9 \sin ^2 \alpha} \\
& \Rightarrow|\vec{c}-\vec{a}|^2=4+\frac{4}{9 \sin ^2 \alpha}
\end{aligned}$$</p>
<p>For $$|\vec{c}-\vec{a}|^2$$ to be minimum for $$\alpha \in\left[0, \frac{\pi}{3}\right]$$</p>
<p>$$\begin{gathered}
\sin \alpha=\frac{\sqrt{3}}{2} \\
27|\vec{c}-\vec{a}|^2=27\left[4+\frac{4.4}{9.3}\right] \\
=27\left[\frac{124}{27}\right]=124
\end{gathered}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294hw | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$$. Then the square of the projection of $$\vec{a}$$ on $$\vec{b}$$ is:</p> | [{"identifier": "A", "content": "$$\\frac{1}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{5}$$\n"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\frac{2}{3}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \\
& \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i} \\
& \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & -1 \\
1 & 1 & 0
\end{array}\right| \\
&=\hat{i}(1)-\hat{j}(1)+\hat{k}(2-1) \\
&=\hat{i}-\hat{j}+\hat{k}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& (\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 1 \\
1 & 0 & 0
\end{array}\right| \\
& \quad=\hat{i}(0)-\hat{j}(-1)+\hat{k}(1) \\
& \quad=\hat{j}+\hat{k}
\end{aligned}$$</p>
<p>$$\left( {(\overrightarrow a \times (\widehat i + \widehat j) \times \widehat i} \right) \times \widehat i = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right|$$</p>
<p>$$\begin{aligned}
& =\hat{i}(0)-\hat{j}(-1)+\hat{k}(-1) \\
\vec{b}= & \hat{j}-\hat{k}
\end{aligned}$$</p>
<p>Projection of $$\vec{a}$$ on $$\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$</p>
<p>$$=\frac{2}{\sqrt{2}}=\sqrt{2}$$</p>
<p>Square of projection $$=2$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
lvb294j7 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$$. If $$\overrightarrow{\mathrm{c}}$$ is a is vector such that $$|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$$ and the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$ is $$60^{\circ}$$, then $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ is equal to:</p> | [{"identifier": "A", "content": "$$\\frac{3}{2} \\sqrt{6}$$\n"}, {"identifier": "B", "content": "$$\\frac{9}{2}(6-\\sqrt{6})$$\n"}, {"identifier": "C", "content": "$$\\frac{9}{2}(6+\\sqrt{6})$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{2} \\sqrt{3}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& |(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 60^{\circ} \\
& \left|\begin{array}{ccc}
i & j & k \\
6 & 1 & -1 \\
1 & 1 & 0
\end{array}\right|=i(1)-j(1)+k(5) \\
& =i-j+5 k \\
& |\vec{a} \times \vec{b}|=\sqrt{1+1+25}=\sqrt{27} \\
& |\vec{c}-\vec{a}|=2 \sqrt{2} \\
& c^2+a^2-2 a c=8 \\
& c^2-12 c+30=0 \\
& c=\frac{12+\sqrt{24}}{2}=6+\sqrt{6} \\
& \Rightarrow|(\vec{a} \times \vec{b}) \times \vec{c}|=\sqrt{27} \times(6+\sqrt{6}) \times \frac{\sqrt{3}}{2} \\
& =\frac{a}{2}(6+\sqrt{6}) \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
lvc57b8v | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$$ and a vector $$\vec{c}$$ be such that $$\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$$. If $$\vec{a} \cdot \vec{c}=13$$, then $$(24-\vec{b} \cdot \vec{c})$$ is equal to _______.</p> | [] | null | 46 | <p>Let $$\hat{i}+8 \hat{j}+13 \hat{k}=\vec{u}$$</p>
<p>Given $$\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\vec{u}$$</p>
<p>$$\begin{gathered}
\Rightarrow \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\vec{u} \\
(\vec{a}+\vec{b}) \times c=\vec{u}-\vec{a} \times \vec{b}
\end{gathered}$$</p>
<p>Taking cross product with $$\vec{a}$$ on both sides</p>
<p>$$\vec{a} \times((\vec{a}+\vec{b}) \times \vec{c})=\vec{a} \times(\vec{u}-\vec{a} \times \vec{b})$$</p>
<p>$$\Rightarrow \quad \vec{c} \cdot\left(\vec{a}^2+\vec{a} \cdot \vec{b}\right)=13(\vec{a}+\vec{b})-\vec{a} \times \vec{u} +(\vec{a} \cdot \vec{b}) \cdot \vec{a}-\vec{a}^2 \vec{b}\quad$$ $$\{\because \vec{a} . \vec{c}=13\}$$</p>
<p>Putting the values, $$\vec{c}=(-1,-1,3)$$</p>
<p>$$\vec{b}.\vec{c}=-22$$</p>
<p>$$\Rightarrow 24-\vec{b} \cdot \vec{c}=46$$</p> | integer | jee-main-2024-online-6th-april-morning-shift |
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