# Planes

• Feb 7th 2013, 07:48 PM
Tutu
Planes
Find the equation of the plane perpendicular to the Z-axis and through (2,-1,4)

I know that this plane will be parallel to the XOY plane. But then how do I find the equation of the plane, how do I know how many x-units, y-units and z-units the plane extends to? Is it 2x-y+4z?

Thank you so much!
• Feb 7th 2013, 08:11 PM
jakncoke
Re: Planes
So you know the dot product? The dot product of 2 perpendicular vectors is 0.

So imagine you already have your plane. Now if you pick any point in your plane, call it (x,y,z), then the vector yo get from (x,y,z) - (2,-1,4) must lie on the plane right? or must be perpendicular to the z-axis. Then the vector (x-2, y--1, z-4) or (x-2,y+1,z-4), should be perpendicular to the z-axis, or the vector (0,0,1). so (x-2,y+1,z-4) * (0,0,1) = 0. which gives z-4 = 0, or z=4.
• Feb 8th 2013, 01:03 AM
Tutu
Re: Planes
Thank you so so much!!!
• Feb 8th 2013, 04:56 AM
HallsofIvy
Re: Planes
Using jackncokes' method, if (x, y, z) lies in the plane through $\displaystyle (x_0, y_0, z_0)$, then the vector $\displaystyle <x- x_0, y- y_0, z-z_0>$ is in he plane. If that plane is perpendicular to vector <A, B, C> then, taking the dot product we must have
$\displaystyle <A, B, C>\cdot <x- x_0, y- y_0, z- z_0>= A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.

The "plane perpendicular to the Z-axis and through (2,-1,4)" is perpendicular to vector <0, 0, 1> and so has equation 0(x- 2)+ 0(y-(-1))+ 1(z- 4)= 0 or just z= 4.

That is a general method for finding the equation of a plane. In this case, it should have been obvious that every point in the "plane perpendicular to the Z-axis" has the same z coordinate and since it is "through (2,-1,4)" that coordinate must be 4. That is, every point in that plane has the property that "z= 4". Since that is a single linear equation in 3 dimensional space, it describes a two dimensional plane.