Q. Let $ABCD$ be a parallelogram such that $\overrightarrow{AB} = \vec{q}, \overrightarrow{AD} = \vec{p}$ and $∠BAD$ 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 :

AIEEEAIEEE 2012Vector Algebra Report Error

A

$\vec{r} = 3\vec{q} - \frac{3\left(\vec{p}.\vec{q}\right)}{\left(\vec{p}.\vec{p}\right)}\vec{p}$

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B

$\vec{r} = -\vec{q} +\left( \frac{\vec{p}.\vec{q}}{\vec{p}.\vec{p}}\right)\vec{p}$

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C

$\vec{r} = \vec{q} -\left( \frac{\vec{p}.\vec{q}}{\vec{p}.\vec{p}}\right)\vec{p}$

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D

$\vec{r} =-3 \vec{q} + \frac{3\left(\vec{p}.\vec{q}\right)}{\left(\vec{p}.\vec{p}\right)}\vec{p}$

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Solution:

$\overrightarrow{AX} =\frac{\vec{p}.\vec{q}}{\left|\vec{p}\right|} \frac{\vec{p}}{\left|\vec{p}\right|} = \frac{\vec{p}.\vec{q}}{\left|\vec{p}\right|^{2}}\vec{p}$
$\overrightarrow{BX} = \overrightarrow{BA} + \overrightarrow{AX}$
$= - \vec{q} + \frac{\vec{p}.\vec{q}}{\left|\vec{p}\right|^{2}}\vec{p}$

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