# Finding the distance between a point and a plane using dot products

1. Feb 3, 2012

### haackeDc

I know how I would be able to do this using projection, but am not so sure with dot products.

Do I dot the normal vector with an imaginary point and then figure something out from there?

If the normal is a= <a1,a2,a3>
and the random point is (p1,p2,p3)

If I dot them, I would get a1p1 + a2p2 + a3p3 = 0

And then solve for this point?

I don't see how to do it from here, could anyone help me?

2. Feb 9, 2012

### conquest

(0, 0, 0) solves this!

I think though that you want to find the distance between any given point and any given plane. I think it should work if you shift the plane into the origin ( by setting the euqaiton for the plane to zero). Then shift your point accordingly. then you dot the vector going to the point into the unit normal vector to the plane.

3. Feb 13, 2012

### HallsofIvy

If the plane is Ax+ By+ Cz= D and the point is $(x_0, y_0, z_0)$, you know that <A, B, C> is a normal vector to the plane and so $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$ is a line that passes through $(x_0, y_0, z_0)$ and is perpendicular to the plane. Replace x, y, and z in the equation of the plane with those to solve for the point where the normal line passes through the plane and find the distance from $(x_0, y_0, z_0)$ to that point.

4. Feb 13, 2012

### haackeDc

Question resolved. Thanks for your help everyone.