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. $\sqrt{7}$
• B. $\sqrt{3}$
• C. 7
• D. 3

Answer 1

Answer: (A)

$\sqrt{7}$

Answer 2

The correct answer is A:$\sqrt{7}$
Given, $a =\hat{ i }-2 \hat{ j }+3 \hat{ k }$
$a \cdot b =| b |^{2}$
and $| a - b |=\sqrt{7}$
$\Rightarrow| a - b |^{2}=7$
$\Rightarrow| a |^{2}+| b |^{2}-2 a \cdot b =7$
$\Rightarrow(\sqrt{1+4+9})^{2}+| b |^{2}-2| b |^{2}=7$
$\Rightarrow 14-| b |^{2}=7$
$\Rightarrow | b |^{2}=7$
$\Rightarrow | b |=\sqrt{7}$