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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?

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    You're now calculating the angle $AOB$ instead of $APB$. Try replacing $O$ by $P$ with $P$ on the $x$-axis, i.e. $P(x, 0, 0)$.2012-05-27

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Let the point $P$ be $(x,0,0)$. What is the vector $PA$? What is the vector $PB$? Then use the fact that these two vectors are perpendicular precisely if their "dot product" is $0$.

You will end up with a quadratic equation in $x$. Solve.

If you experience difficulties, please leave a message.

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    Thanks, this helped. Made a logical error of marking $O$ as the point of angle instead of $P$, as pointed out by TMM's comment.2012-05-27