Find equation of a plane, that contains a point, is perpendicular to plane and parallel to a line.

Question

I have to solve this problem from Analytic Geometry, but I'm pretty much stuck.

Find equation of plane π, that contains point A, is perpendicular to plane ρ, and is parallel to line p.

Point:

A = [1, 1, 1]

Plane:

ρ:  x - 2y + z - 3 = 0

Line:

p:  x + y - 2z - 2 = 0
   2x - y + 3z - 1 = 0

We should get 3 equations in total, to get the plane π.

We get the first equation from point A:

A = [1, 1, 1] = [x, y, z] -> xa + yb + zc + d= 0
a + b + c + d= 0

The second one is the normal vector of plane ρ:

n = (1, -2, 1) -> xa + yb + zc = 0
a - 2b + c = 0

And the third from the vector line p (this is where I get stuck):

 x + y - 2z - 2 = 0
2x - y + 3z - 1 = 0
Some mind blowing calculation...

Answer 1

$$p:\begin{cases} x + y - 2z - 2 = 0\\ 2x - y + 3z - 1 = 0\\ \end{cases} $$ Let $x=t$ we get $$p:\begin{cases} y - 2z = -t+2\\ - y + 3z= 1-2t\\ \end{cases} $$ Add the equations to have $z=3-3t$

Plug in the first equation and solve $y = 8-7t$

Line $p$ has parametric equation $[t, 8 - 7 t, 3 - 3 t]$

So the direction vector is $\vec{u}=[1,-7,-3]$

The parallel plane $\pi$ has normal vector $\vec{n}=[a,b,c]$. $\pi$ is parallel to the line $p$ if $\vec{n}$ is perpendicular to $\vec{u}$ that is $\vec{n}\cdot \vec{u}=0\to a-7b-3c=0$

$\pi$ must be perpendicular to $\rho$ which has normal vector $\vec{v}=[1,-2,1]$, thus $\vec{n}\cdot \vec{v}=0\to a-2b+c=0$ $$ \begin{cases} a-7b-3c=0\\ a-2b+c=0\\ \end{cases}\to \begin{cases} 7b+3c=a\\ 2b-c=a\\ \end{cases} $$ Solution is $\left[a,\frac{4 a}{13}, -\frac{5a}{13}\right]$

Plane $\pi$ has equation

$a(x-1)+b(y-1)+c(z-1)=0$

Plug the values of $b,c$

$$a(x-1)+\frac{4 a}{13}(y-1)-\frac{5a}{13}(z-1)=0$$ cancel $a$ and get $$\pi:13 x+4 y-5 z=12$$

Answer 2

If you know how the cross product works, you only need to calculate two cross products.

The line $p$ is given as the intersection of two planes: you could convert this into a vector (or parametric) equation but you're only interested in its direction. The cross product of the normal vectors of both planes gives you exactly that: $$\left(1,1,-2\right) \times \left(2,-1,3\right) = \left(1,-7,-3\right)$$ Since $\pi$ has to be parallel to $p$, a normal vector of $\pi$ should be perpendicular to this vector. On the other hand, since $\pi$ has to be perpendicular to $\rho$, their normal vectors should be perpendicular too. Hence, a normal vector of $\pi$ should be perpendicular to $\left(1,-2,1\right)$ and $\left(1,-7,-3\right)$, so it is given by their cross product: $$\left(1,-2,1\right) \times \left(1,-7,-3\right) = \left(13,4,-5\right)$$ Now you have a point $A=(1,1,1)$ and a normal vector $\left(13,4,-5\right)$ so the plane is given by: $$13(x-1)+4(y-1)-5(z-1)=0 \iff 13 x + 4 y - 5 z - 12 = 0$$