Question

Let $a =\hat{ i }-2 \hat{ j }+3 \hat{ k }$, if $b$ is a vector such that $a \cdot b =| b |^{2}$ and $| a - b |=\sqrt{7}$, then $| b |$ is equal to

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

Answer 1

Answer: (C)

We know that.
$| a - b |^{2}=( a - b ) \cdot( a - b ) $
$\Rightarrow | a - b |^{2}=| a |^{2}+| b |^{2}-2 a \cdot b $
$\Rightarrow (\sqrt{7})^{2}=\left[\left.\sqrt{(1)^{2}+(-2)^{2}+(3)^{2}}\right|^{2}+| b |^{2}-2| b |^{2}\right.$
$\Rightarrow 7 =14-| b |^{2} $
$\Rightarrow | b |^{2}=7$
$ \Rightarrow | b |=\sqrt{7}$