Question

$\overset{\rightarrow}{a}$and $\overset{\rightarrow}{b}$are two given vectors. On these vectors as adjacent sides, a parallelogram is constructed. The vector which is the altitude of the parallelogram and perpendicular to $\overset{\rightarrow}{a}$ is given by-

• A. $\overset{\rightarrow}{a}–\overset{\rightarrow}{b}$
• B. $\left( \frac { \vec { a } \cdot \vec { b } } { | \vec { b } | ^ { 2 } } \right)$ $\overset{\rightarrow}{b}–\overset{\rightarrow}{a}$
• C. $\frac{\overset{\rightarrow}{b} \times (\overset{\rightarrow}{a} \times \overset{\rightarrow}{b})}{|\overset{\rightarrow}{b}|^{2}}$
• D. $\overset{\rightarrow}{a} + \overset{\rightarrow}{b}$

Answer 1

Answer: (A)

$\overset{\rightarrow}{a}–\overset{\rightarrow}{b}$

Answer 2

The altitude required is

$\overrightarrow { \mathrm { BN } }$ = – $\overrightarrow { \mathrm { OB } }$

= projection of $\overrightarrow { \mathrm { CB } }$ on $\overrightarrow { \mathrm { OA } }$ –

= ( $\vec { b }$ .)–

= – =$\frac { ( \vec { b } \cdot \vec { a } ) \vec { a } } { | \vec { a } | ^ { 2 } }$ – $\vec { b }$ .