Let a , = , i , + , j , + , 2 k, , , b , = , b_1 i , + , b_2 j , … - Vidyadip

MCQ

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 $\vec a$ is $\vec a$. If $\vec a\, + \vec b$ is perpendicular to $\vec c$ , then $\left| {\vec b} \right|$ is equal to

A
$\sqrt {22}$
B
$4$
C
$\sqrt {32}$
✓
$6$

✓

Answer

Correct option: D.

$6$

d
Projection of $\overrightarrow{\mathrm{b}}$ on $\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|}=|\overrightarrow{\mathrm{a}}|$

$\Rightarrow \mathrm{b}_{1}+\mathrm{b}_{2}=2$ .....$(1)$

and $(\vec{a}+\vec{b}) \perp \vec{c} \Rightarrow(\vec{a}+\vec{b}) \cdot \vec{c}=0$

$\Rightarrow 5 b_{1}+b_{2}=-10$ .....$(2)$

from $ ( 1)$ and $(2) $

$\Rightarrow b_{1}=-3$ and $b_{2}=5$

then $|\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{b}_{1}^{2}+\mathrm{b}_{2}^{2}+2}=6$

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