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

Question

( A ) $\overrightarrow{r} = -3\overrightarrow{q} + \frac{3(\overrightarrow{p}.\overrightarrow{q})}{(\overrightarrow{p}.\overrightarrow{p})} \overrightarrow{p}$
( B ) $\overrightarrow{r} = 3\overrightarrow{q}- \frac{3(\overrightarrow{p}.\overrightarrow{q})}{(\overrightarrow{p}.\overrightarrow{p})} \overrightarrow{p}$
( C ) $\overrightarrow{r} = -\overrightarrow{q}- \begin{pmatrix}\frac{\overrightarrow{p}.\overrightarrow{q}}{\overrightarrow{p}.\overrightarrow{p}} \end{pmatrix} \overrightarrow{p}$
( D ) $\overrightarrow{r} = -\overrightarrow{q}+ \begin{pmatrix}\frac{\overrightarrow{p}.\overrightarrow{q}}{\overrightarrow{p}.\overrightarrow{p}} \end{pmatrix} \overrightarrow{p}$