The problem:
Find the point that lies on the line that is the intersection of two planes:
plane one: $$3x-2y+z = 2$$
plane two: $$x+y+2z = 4$$
The point also needs to be the same distance away from points $A(1,-1,-1),B(1,-1,3)$
My try:
First I found the direction of the line that is the intersection of two planes. Here I used cross product between $(3,-2,1)$ and $(1,1,2)$, which are the normals of the two planes. I got $s = (-1,-1,1)$ That is the direction of the intersection line.
What I did next was I solved the system of equations: $$3x-2y+z-2 = 0\\x+y+2z-4 = 0$$
I got: $z = 7y +2$ and $x = \frac{-5y}{3}$
Here I chose $y$ to be equal to $1$ $\implies y = 1 \implies x = \frac{-5}{3} \implies z = 9$
So the equation of the intersection line is: $(1,\frac{-5y}{3}, 9) + t(-1,-1,1) = (x,y,z)$
Then I wanted to find the plane where all the points that have the same distance from $A(1,-1,-1),B(1,-1,3)$ are located. it is trivial this plane's normal will be the direction vector $AB$, thus the normal is $n=(0,0,4)$
Then I wanted to position the plane($O$…origin) : $$OA + \frac{1}{2}AB = (1,-1,-1) +\frac{1}{2}(1,-1,3) = (\frac{3}{2},\frac{-3}{2},\frac{1}{2}) = \text{\{This is the point to position the plane\}}$$
So the equation of this plane is: $4z = 4*\frac{1}{2}=2$
I got $z_0$ form the last coordinate from the point.
Then I Solved for intersection between the first line(intersection between plane one and two) and this last plane. I got the wrong answer: $x = \frac{19}{2}, y = \frac{41}{6}, z = \frac{1}{2}$ I got this results by inserting parametrically defined $z$ into the plane equation. My answer is wrong.
The correct solution is: $T(1,1,1)$
Where did I go wrong?
Best Answer
One error is in "I got: $z = 7y +2$ and $x = \frac{-5y}{3}$".
Hint.
You are doing the problem in a much more complicated way than necessary.
The second condition implies that the point lies in the plane that is perpendicular to the vector $(0,0,4)$, and contains the point $(1,-1,1)$. This plane is given by $ z=1 $.
Now you have two equations in $x$ and $y$: $$ 3x-2y=1,\quad x+y=2 $$