What is the implicit equation for a plane perpendicular to a line?

[shards5]

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-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?

 

[Mark44]

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-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?

Your normal is correct, <2, 0, 0>, which you can read directly from the equation of the line. The numbers in the normal are the coefficients of t, which are 2, 0, and 0.

Once you have a normal <A, B, C> to the plane, and you know a point (a, b, c) on the plane, the equation of the plane is A(x -a) + B(y - b) + C(z - c) = 0.

Calculating the cross product is a waste of time, since you already have a normal to the plane, namely <2, 0, 0>.