<s>[INST] Analytic Geometry | Two Planes and a Angle | Two Solutions This is me again, I have another problem which I haven't been able to solve, the 
legend goes like this:
Find the equation of the plane that contains the points $P_1(1,0,-1)$, $P_2(2,0,2)$ and forms a 600 ($\theta$) angle with the other plane $2x-2y+z+6=0$ (Two Solutions).  
Here's what I've done:
Since we already know two points, P1$(1, 0, -1)$ and P2$(2,0,2)$ they both satisfy a planes equation:
$$Ax+By+Cz+D=0$$
So, we substitute $x,y,z$ respectively for each point, giving use these two equations


*

*$A(1)+B(0)+C(-1)+D=0$ 

*$A(2)+B(0)+C(2)+D=0$


Simplifying we get:  


*

* $A-C+D=0$ 

* $2A+2C+D=0$


Now, according to the following theorem the angle that is formed by the perpendiculars of to planes is the following:
$$\cos{\theta} = \pm \frac{AA'+BB'+CC'}{\sqrt{A^2+B^2+C^2} \sqrt{(A')^2+(B')^2+(C')^2}}$$ Where $[A,B,C]$ Are the director numbers of the normal in the first plane and $[A',B',C']$ Are the director numbers of the normal in the second plane.
Now, given those and taking the known angle of 60o, we substitute the director numbers of $2x-2y+z+6-0:[A'=2,B'=-2,C'=1]$ and the angle in the last formula, giving us:
$$\cos{60^\circ} = \pm \frac{A(2)-B(-2)+C(1)}{\sqrt{A^2+B^2+C^2} \sqrt{(2)^2+(-2)^2+(1)^2}}$$
Simplifying, we get(positive because $\theta$ is acute):
$$\frac{1}{2}=\frac{2A-2B+C}{3 \sqrt{A^2+B^2+C^2}}$$
Another simplification finally for:
$$3(A^2+B^2+C^2)=(4A-4B+2C)^2$$
Now, we have 3 equations:


*

* $A-C+D=0$ 

* $2A+2C+D=0$

*$3(A^2+B^2+C^2)=(4A-4B+2C)^2$


If we take 1) and 2) and write them as $D(A,C)$ we get:


* 

*$D=C-A$ 

*$D=-2A-2C$


From which we can obtain A in terms of C
$$-2A-2C=C-A$$
$$A=-3C$$
Now we substitute back in 1):
$$(-3C)-C+D=0$$
$$C=\frac{1}{4}D$$
Since I want every director number in terms of D we substitute again but in A since A was in function of C.
$$A=-3C$$
$$A=-3(\frac{1}{4}D)$$
$$A=\frac{-3}{4}D$$
Now, we have found $A(D)$, $C(D)$ we have yet to find $B(D)$ I'll explain why I think and want everything in terms of D, if we take the general form of a planes equation $Ax+By+Cz+D=0$ and substitute $A$ and $C$ with our known values we get:
$$(\frac{-3}{4}D)x+By+(\frac{1}{4}D)z+D=0$$
if we divide by $D \neq 0$ and then multiply by 4 to eliminate fractions:
$$\frac{-3}{4}x+\frac{B}{D}y+\frac{1}{4}z+1=0$$
$$-3x+\frac{4B}{D}y+z+4=0$$
From here it's clear that if we find B in terms of D and B is linear we have the planes equation, so, in 3) $3(A^2+B^2+C^2)=(4A-4B+2C)^2$ we substitute A and C:
$$3((\frac{-3}{4}D)^2+B^2+(\frac{1}{4}D)^2)=(4(\frac{-3}{4})-4B+2(\frac{1}{4}D))^2$$
$$\vdots$$
$$104B^2+80DB+35D^2=0$$
Here is were I get stuck, I get imaginary parts, since I have to use the quadratic formula, so any help is really appreciated, the problem states that there are two solutions, but this is what I have thought of if anyone can supply any solution I'll be very grateful.
Regards.. Tristian
 [/INST] The normal vector to the sought plane is orthogonal to the
line $P_1 P_2$ and at angle $\pi/3$ to the normal to your given plane.
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