Question

Let $a=\hat{i}−2\hat{j}+3\hat{k}$, if $b$ is a vector such that $aâ‹\dots b=âˆ\pounds b{âˆ\pounds }^2$ and $âˆ\pounds a−bâˆ\pounds =\sqrt{7}$, then $âˆ\pounds bâˆ\pounds$ is equal to

• A. 7
• B. 14
• C. $\sqrt{7}$
• D. 21

Answer 1

Answer: (C)

We know that.
$âˆ\pounds a−b{âˆ\pounds }^2=(a−b)â‹\dots (a−b)$
$⇒âˆ\pounds a−b{âˆ\pounds }^2=âˆ\pounds a{âˆ\pounds }^2+âˆ\pounds b{âˆ\pounds }^2−2aâ‹\dots b$
$⇒(\sqrt{7}{)}^2=[{\sqrt{(1{)}^2+(−2{)}^2+(3{)}^2}â}^2+âˆ\pounds b{âˆ\pounds }^2−2âˆ\pounds b{âˆ\pounds }^2$
$⇒7=14−âˆ\pounds b{âˆ\pounds }^2$
$⇒âˆ\pounds b{âˆ\pounds }^2=7$
$⇒âˆ\pounds bâˆ\pounds =\sqrt{7}$