
vector/plane proof
Prove that if $\displaystyle a, b, c$ are not all zero, then the equation: $\displaystyle ax+by+cz+d=0$ represents a plane and $\displaystyle \langle a,b,c\rangle$ is a normal vector to the plane.
Hint if a $\displaystyle \neq$0, then $\displaystyle ax+by+cz+d=0$ is equivalent to $\displaystyle a(x+\frac{d}{d})+b(y0)+c(z0)=0$
My Calc III teacher loves proofs, but I'm not real good with them. How can I prove this? I'm not even sure how to approach it. Can anyone help? Thanks!

Well, the usual equation for a plane looks like this:
$\displaystyle \mathbf{n}\cdot(\mathbf{x}\mathbf{x}_{0})=0,$
where $\displaystyle \mathbf{n}$ is a unit vector normal to the surface, $\displaystyle \mathbf{x}$ is a vector to the arbitrary point on the plane, and $\displaystyle \mathbf{x}_{0}$ is a vector to any particular point on the plane.
Can you get your equation to look like mine? If so, you're done.

$\displaystyle n \cdot (xx_{0})=0$
$\displaystyle n=\langle a, b, c\rangle$
$\displaystyle \langle a, b, c\rangle \cdot (xx_{0})=0$
$\displaystyle \langle a,b,c\rangle\langle x,y,z\rangle\langle a,b,c\rangle\langle x_{0},y_{0},z_{0}\rangle=0$
$\displaystyle ax+by+czax_{0}by_{0}cz_{0}=0$
$\displaystyle ax+by+cz+d=0$
I know that $\displaystyle ax_{0}by_{0}cz_{0}=d$, but is this enough to prove it? Where did you get the usual equation for a plane from? Thanks!

It is simply using the fact that two vectors are perpendicular if and only if their dot product is 0. Let $\displaystyle (x_0, y_0, z_0)$ be any fixed point and $\displaystyle (x, y, z)$ be any other point. Then the vector connecting them is [tex]<x x_0, y y_0, z z_0>[tex] and the "set of all points $\displaystyle (x, y, z)$ such that the vector [tex]<x x_0, y y_0, z z_0>[tex] is perpendicular to $\displaystyle <a, b, c>$ (the tangent plane to that vector containing $\displaystyle (x_0, y_0, z_0)$) must satisfy
$\displaystyle <a, b, c>\cdot<x x_0, yy_0, zz_0>= 0$.