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## Homework Statement

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 ...?

## Homework Equations

a(x-x

_{0}) + b(y-y

_{0}) + c(z-z

_{0}) = 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?