Find All planes that contain the two points P(2,-1,3), Q(1,0,1) and is parallel to the line x(t)=(2,2,-3)+t(-1,1,-2).
I thought I'd approach it by getting the vector PQ, but PQ is the same vector as the one given in the parallel line. So because the two points and the line are also parallel, this means that the plane can basically only rotate around the two points right? Geometrically, I thought that made sense. I just can't figure out how to get a formula. Help is appreciated.
Best Answer
The line does not tell us anything. So you are right, the two points $(2,-1,3)$ and $(1,0,1)$ being on the plane are the only conditions.
This means
$$A(x-2)+B(y+1)+C(z-3)=0$$
$$Ax+By+Cz=2A-B+3C$$
And
$$A(x-1)+By+C(z-1)=0$$
$$Ax+By+Cz=A+C$$
Combine these two equations to see what you get.
$$A+C=2A-B+3C$$
$$-A=-B+2C$$
$$A=B-2C$$
So the planes are of the form,
$$(B-2C)x+By+Cz=B-C$$
With $B,C \in \mathbb{R}$.