question_answer

Let ABCD be a parallelogram such that \[\overrightarrow{AB}=\vec{q},\,\,\overrightarrow{AD}=\vec{p}\] 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 :   AIEEE  Solved  Paper-2012

A) \[\vec{r}=3\vec{q}-\frac{3(\vec{p}.\,\vec{q})}{(\vec{p}.\,\vec{p})}\vec{p}\]          

B) \[\vec{r}=-\vec{q}+\left( \frac{\vec{p}.\,\vec{q}}{\vec{p}.\,\vec{p}} \right)\vec{p}\]

C) \[\vec{r}=\vec{q}-\left( \frac{\vec{p}.\,\vec{q}}{\vec{p}.\,\vec{p}} \right)\vec{p}\]           

D) \[\vec{r}=-3\vec{q}+\frac{3(\vec{p}.\,\vec{q})}{(\vec{p}.\,\vec{p})}\vec{p}\]

Correct Answer: B

Solution :

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