Let ABCD be a parallelogram such that AB= q, AD= p and BAD be an acute angle.

Question

Let $ABCD$ be a parallelogram such that $\overrightarrow{AB}=\vec q$, $\overrightarrow{AD}=\vec p$ and $\angle BAD$ be an acute angle. If $\vec r$ is the vector that coincides with the altitude directed from the vertex $B$ to the side $AD$, then $\vec r$ is given by:

Accepted Answer

Let $X$ be the foot of the perpendicular from $B$ to line $AD$. Then $\overrightarrow{AX}=\dfrac{\vec p\cdot\vec q}{|\vec p|^2}\vec p$ (projection of $\overrightarrow{AB}$ onto $\vec p$).

$$\vec r = \overrightarrow{BX} = \overrightarrow{BA}+\overrightarrow{AX} = -\vec q+\frac{\vec p\cdot\vec q}{\vec p\cdot\vec p}\vec p$$

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