Find a vector equation for the line through the point $(3,-8,-8)$ perpendicular to these vectors 
Find a vector equation for the line through the point $(3,-8,-8)$ perpendicular to these vectors $u=\langle 2,2,-1\rangle$ and $v=\langle -9,-8,-3\rangle$.

I'm fairly new to vector equations. Would I find a component vector of $u$ and $v$ and write the parametric equation for it and then use the formula $r(t) = r_0 + tv$?
 A: Here's a sketch.
The best way to approach this is by thinking about what fundamentally defines a line: a vector and a point. In this case, we have the point, but instead of a vector in the direction of the line, we have two vectors perpendicular to the line. However, if we take the cross product of these two vectors, it will by definition be perpendicular to both of them and thus be in the direction of the line itself. Taking the cross product gives $\langle -14, 15, 2\rangle$. Then, the line can be parametrized using standard techniques as 
$$\langle 3,-8,-8\rangle+t\cdot \langle-14, 15, 2\rangle=\langle 3-14t, -8+15t, -8+2t\rangle.$$ 
A: Let $P = (3,-8,-8)$ be the point given above, $u$ and $v$ be defined as above, and $O = (0,0,0)$ (origin).  Let $O \rightarrow P$ denote the distance from the origin to $P$.  The cross product of two vectors results in a perpendicular vector.  Hence $u \times v = (-14,15,2)$.  Thus the equation for the line through the point $P$ perpendicular to $u$ and $v$ is 
$r(t) = O\rightarrow P + t(-14,15,2) = (3,-8,-8) + t(-14,15,2)$ 
where $O \rightarrow P = P - O$.
