
Planes
Find the equation of the plane perpendicular to the Zaxis 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 xunits, yunits and zunits the plane extends to? Is it 2xy+4z?
Thank you so much!

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 zaxis. Then the vector (x2, y1, z4) or (x2,y+1,z4), should be perpendicular to the zaxis, or the vector (0,0,1). so (x2,y+1,z4) * (0,0,1) = 0. which gives z4 = 0, or z=4.

Re: Planes

Re: Planes
Using jackncokes' method, if (x, y, z) lies in the plane through $\displaystyle (x_0, y_0, z_0)$, then the vector $\displaystyle <x x_0, y y_0, zz_0>$ 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 Zaxis 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 Zaxis" 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.