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

• A

  $\vec{r}=3 \vec{q}-\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$
• Correct answer
  B

  $\vec{r}=-\overrightarrow{\mathrm{q}}+\left(\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}}{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}}}\right) \overrightarrow{\mathrm{p}}$
• C

  $\vec{r}=\vec{q}-\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}$
• D

  $\vec{r}=-3 \vec{q}+\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$

Solution

$\overrightarrow{\mathrm{AE}}=$ vector component of $\overrightarrow{\mathrm{q}}$ on $\overrightarrow{\mathrm{p}}$ $\overrightarrow{\mathrm{AE}}=\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})} \overrightarrow{\mathrm{p}} \quad \therefore$ From $\triangle \mathrm{ABE} ; \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BE}}=\overrightarrow{\mathrm{AE}}$ $\Rightarrow \overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}=\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{p}}}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})} \quad \Rightarrow \overrightarrow{\mathrm{r}}=-\overrightarrow{\mathrm{q}}+\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}})} \overrightarrow{\mathrm{p}}$

Hence, the correct answer is option: 'B'.