Let $A =(-2, 3, -2)$ and $B =(-6, -1, 1)$. For which points ($P$) on the $x$-axis, $\angle APB= 90^\circ $?
I figured, since $P$ is supposed to be on the $x$-axis, the $y$ and $z$ coordinates must be zero. By multiplicating $\vec{OA}$ with $\vec{OB}$, I thought this would get me the possible points:
Basic formula: $\vec{OA}/(||\vec{OA}||) * \vec{OB}/||\vec{OB}|| = \cos{90°}$ <= this is just zero.
This does not work. How to proceed?