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

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

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

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

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

JEE Main · 2012 · 29 Apr 2012 · Offline · Vector Algebra

Let ABCD be a parallelogram such that

AB = q, AD = p

and

∠ BAD

be an acute angle. If

r

is the vector that coincides with the altitude directed from the vertex

B

to the side

A D

, then

r

is given by

Answer: (B)

Let

E

be the foot of the perpendicular from

B

onto side

A D

. The vector

A E

is the projection of

q = A B

onto

p = A D

, namely

A E = \frac{p \cdot q}{p \cdot p} p

. Since

A B + B E = A E

and the required altitude vector is

r = B E

, it follows that

r = - q + \frac{p \cdot q}{p \cdot p} p

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option