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...
Best Answer
$$p:\begin{cases} x + y - 2z - 2 = 0\\ 2x - y + 3z - 1 = 0\\ \end{cases} $$ 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 $$\pi:13 x+4 y-5 z=12$$