Question

Let $\vec{a} = \hat{i} + \hat{j} + \sqrt{2} \hat{k} , \vec{b} = b_1 \hat{i} + b_2 \hat{j} + \sqrt{2} \hat{k}$ and $\vec{c} = 5 \hat{i} + \hat{j} + \sqrt{2} \hat{k}$ be three vectors such that the projection vector of $\vec{b}$ on $\vec{a}$ , If $\vec{a} + \vec{b}$ is perpendicular to $\vec{c}$, then $| \vec{b}|$ is equal to :

• A. $\sqrt{22}$
• B. $4$
• C. $\sqrt{32}$
• D. $6$

Answer 1

The Correct Option is D

Projection of $\vec{b} \vec{a} = \frac{\vec{a} . \vec{b}}{\left|\vec{a}\right|} = \left|\vec{a}\right| $
$ \Rightarrow b_{1} + b_{2} = 2 $ ...(1)
and $ \left(\vec{a} + \vec{b}\right) \bot \vec{c} \Rightarrow \left(\vec{a} + \vec{b}\right) . \vec{c} =0 $
$ \Rightarrow 5b_{1} + b_{2} = -10 $ ....(2)
from (1) and (2) $\Rightarrow \; b_1 = -3$ and $b_2 = 5$
then $ \left|\vec{b}\right| = \sqrt{b_{1}^{2} + b^{2}_{2} + 2} = 6 $