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

Answer 1

Answer: (A)

\left( \frac { \vec { a } \cdot \vec { b } } { | \vec { a } | ^ { 2 } } \right)$$\overset{\rightarrow}{a}–\overset{\rightarrow}{b}

Answer 2

The altitude required is $\overset{\rightarrow}{BN}$

$\overset{\rightarrow}{BN}$ = $\overset{\rightarrow}{ON}$– $\overset{\rightarrow}{OB}$

= projection of $\overset{\rightarrow}{CB}$ on $\overset{\rightarrow}{OA}$ –$\overset{\rightarrow}{b}$

= ($\overset{\rightarrow}{b}$.$\overset{됀}{a}$)$\overset{됀}{a}$–$\overset{\rightarrow}{b}$

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