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Let ABCD be a parallelogram such that $\underset{AB}{\rightarrow}$ $= \vec{q}$ , $\underset{AD}{\rightarrow}$ $= \vec{p}$ 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
Option: 1
$\vec{r}= 3\vec{q}-\frac{3\left ( \vec{p}.\vec{q} \right )}{\left ( \vec{p}.\vec{p} \right )}\vec{p}$

Option: 2
$\vec{r}= -\vec{q}+\left ( \frac{\vec{p}.\vec{q}}{\vec{p}.\vec{p}} \right )\vec{p}$

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

Option: 4
$\vec{r}= -3\vec{q}+\frac{3\left ( \vec{p}.\vec{q} \right )}{\left ( \vec{p}.\vec{p} \right )}\vec{p}$

Answers (1)

best_answer

$\vec{r}=$ $\underset{BA}{\rightarrow}$ + $\underset{AQ}{\rightarrow}$

$=-\vec{q}+projection\; of\;$ $\underset{BA}{\rightarrow}$ across $\underset{AD}{\rightarrow}$

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

As learnt in concept

Projection of vector b on vector a -

$\vec{b}\cos \Theta = \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |}$

- wherein

dot product

Posted by

SANGALDEEP SINGH

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