Perpendicular to a plane given a parametric equation

Question

i'm having a small issue with a certain question.

Given a parametric equation of a plane $x=5-2a-3b$, $y=3-4a+2b$, $z=7-6a-2b$, find a point $P$ on the plane so that the position vector of $P$ is perpendicular to the plane.

How would you go about this for a parametric equation? I think I could convert this to a cartesian equation and dissect an answer that way, but how can I do this without having to convert it?

The hint it gives on the page is that $P$ has the vector $\overrightarrow{OP}$, so I'd imagine the first thing I would do is use the dot product with dummy variables for the $i$, $j$ and $k$ values of $P$. Am I on the right track?

Thanks in advance.

Answer 1

First of all, you need to find a vector in the direction perpendicular to the plane. Your plane is given by $$ v(a,b) = (5,3,7) - (2,4,6)a - (3,-2,2)b $$ You need a direction that is perpendicular to both $(2,4,6)$ and $(3,-2,2)$. One way to do so is to compute the cross-product. We compute $$ (2,4,6) \times (3,-2,2) = (20,14,-16) $$ So, $(20,14,-16)$ is a direction perpendicular to the plane. What we want then is a multiple of this vector that lies on the plane. In particular, find $t$ such that the point $(20t,14t,-16t)$ is a point on the plane described.

It may be useful to use the Cartesian description of the plane, which is $$ 20(x-5) + 14(y-3) -16(z-7) = 0 $$

Answer 2

One approach would be to let $P=(5-2a-3b,3-4a+2b,7-6a-2b)$ and then set

$\vec{OP}\cdot\langle1,2,3\rangle=0$ and $\vec{OP}\cdot\langle-3,2,-2\rangle=0$ and then solve for a and b.