Thread: Vector equation of a line normal to a plane.

1. Vector equation of a line normal to a plane.

Can someone please help me find the vector equation for the line through point R(3,0,-3) normal to the plane
-4x-12y+4z=-24 ?

I have the normal as n = -4i -12j +4k, but am at lost to the next step.

Any help will be greatly appreciated.

2. Re: Vector equation of a line normal to a plane.

Originally Posted by tammmyl
Can someone please help me find the vector equation for the line through point R(3,0,-3) normal to the plane
-4x-12y+4z=-24 ?
I have the normal as n = -4i -12j +4k, but am at lost to the next step.
We can use the normal $\displaystyle i+3j-k$ so the the equation of the line is $\displaystyle <3,0,-3>+t<1,3,-1>$.
Recall that a line is perpendicular to a plane if it is parallel to the normal of the plane.

3. Re: Vector equation of a line normal to a plane.

The obvious point (which I just have to mention) is that when t= 0, this is (3, 0, -3), the given point, and its direction is given by the "slope", <1, 3, -1>, as given by the normal to the plane.