Find the scalar equation

Question

Find the scalar equation for the plane passing through the point $P = (-4,-3,3) $ and containing the line $L$ defined by:

$x = -7-2t$,$y = -2-6t$,$z = 2 + 4t$

What do I do?

Answer 1

First of all find two points on your line.

For example you may take $A(-7,-2,2)$ and $B(-9,-8,6).$

The Vectors $AP =<3,-1,1>$, and $BP=<5,5,-3>$ are on your plane.

find the normal vector to your plane by finding the cross product of $AP$ and $BP$.

I found $N = <-2, 14,20>.$

Thus the equation of the plane is $$-2(x+4)+14(y+3)+20(z-3)=0$$

This equation simplifies to $$ -2x+14y+20z=26$$