Question

If $\overline{a}$ and $\overrightarrow{b}$ are two non-zero perpendicular vectors, then a vector $\overrightarrow{y}$ satisfying equations $\overrightarrow{a}$ . $\overrightarrow{y}$ = c (c scalar) and $\overrightarrow{a}$ $\times$ $\overline{y}$ = $\overline{b}$ is

• A. ${\left|\overline{a}\right|}^2(c\overline{a}-(\overline{a}\times \overrightarrow{b}$))
• B. ${\left|\overrightarrow{a}\right|}^2(c\overrightarrow{a}-(\overrightarrow{a}\times \overrightarrow{b}$))
• C. $\frac{1}{{\left|\overrightarrow{a}\right|}^2}(c\overrightarrow{a}-(\overrightarrow{a}\times \overrightarrow{b}$))
• D. $\frac{1}{{\left|\overrightarrow{a}\right|}^2}(c\overrightarrow{a}-(\overrightarrow{a}\times \overline{b}$))

Answer 1

The correct option is D $\frac{1}{{\left|\overrightarrow{a}\right|}^2}(c\overrightarrow{a}-(\overrightarrow{a}\times \overline{b}$))

$\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{0}$ as $\overrightarrow{a}$ & $4\overrightarrow{b}$ are the vectors

and $\overrightarrow{a}\times \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|$

Now $\overrightarrow{a}\cdot \overrightarrow{y}=c$

and $\overrightarrow{a}\times \overrightarrow{y}=\overrightarrow{b}$

$\Rightarrow \overrightarrow{a}\times (\overrightarrow{a}\times \overrightarrow{y})=\overrightarrow{a}\times \overrightarrow{b}$

$\Rightarrow (\overrightarrow{a}\cdot \overrightarrow{y})\overrightarrow{a}-(\overrightarrow{a}\cdot \overrightarrow{a})\overrightarrow{y}=\overrightarrow{a}\times \overrightarrow{b}$
$\Rightarrow c\overrightarrow{a}-|\overrightarrow{a}{|}^2\overrightarrow{y}=c\overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{b}$ $[as\overrightarrow{a}\cdot \overrightarrow{y}=c]$

$\Rightarrow |\overrightarrow{a}{|}^2\overrightarrow{y}=c\overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{b}$

$\Rightarrow \overrightarrow{y}=\frac{1}{|\overrightarrow{a}{|}^2}[c\overrightarrow{a}-(\overrightarrow{a}\times \overrightarrow{b}\left)\right]$