Question

Let $ABCD$ be 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

• A. $\overrightarrow{r}=3\overrightarrow{q}-\frac{3(\overrightarrow{p}.\overrightarrow{q})}{(\overrightarrow{p}.\overrightarrow{q})}\overrightarrow{p}$
• B. $\overrightarrow{r}=-\overrightarrow{q}+\left(\frac{\overrightarrow{p}.\overrightarrow{q}}{\overrightarrow{p}.\overrightarrow{p}}\right)\overrightarrow{p}$
• C. $\overrightarrow{r}=\overrightarrow{q}-\left(\frac{\overrightarrow{p}.\overrightarrow{q}}{\overrightarrow{p}.\overrightarrow{p}}\right)\overrightarrow{p}$
• D. $\overrightarrow{r}=-3\overrightarrow{q}+\frac{3(\overrightarrow{p}.\overrightarrow{q})}{(\overrightarrow{p}.\overrightarrow{q})}\overrightarrow{p}$

Answer 1

The correct option is C $\overrightarrow{r}=\overrightarrow{q}-\left(\frac{\overrightarrow{p}.\overrightarrow{q}}{\overrightarrow{p}.\overrightarrow{p}}\right)\overrightarrow{p}$

If we take

two components of

$\overline{q}$, one would

be along $\overline{p}$ and other along

$\overrightarrow{r}$

$\Rightarrow \overline{r}=\overrightarrow{q}-\left(componentof\overrightarrow{q}along\overrightarrow{p}\right)$

$\overrightarrow{q}-\left(\frac{\overline{q}.\overrightarrow{p}}{|\overrightarrow{p}|}\hat{p}\right)$

$\overrightarrow{r}=\overline{q}-\left(\frac{\overline{q}.\overrightarrow{p}}{|\overrightarrow{p}{|}^2}\overrightarrow{p}\right)$