An implicit equation for the plane passing through the point (-2,5,-5) that is perpendicular to the line L(t) = <5+2t,3,4> is ...?
a(x-x0) + b(y-y0) + c(z-z0) = 0
The Attempt at a Solution
So in order to find the equation of the plane I would need a normal vector and I got that using the following steps:
Find a point on the line: L(1) = <7, 3, 4>
Get a vector from the given point to the newly found point. (7,3,4) - (-2,5,-5) = (9,-2,9)
Find the normal direction of the line: <2,0,0>
Use the cross product to find the normal vector. (9,-2,9) x <2,0,0> = <0,18,4>
Therefore I have the equation: 18(y-5) + 4(z+5) = 0, but this is wrong. What am I doing wrong?