Let a→=i^+j^+2k^,  b¯=b1i^+b2j^+2k^ andc→=5i→+j^+2k  be three vectors such that the projection of b→ on a→ is |a→|. If a→+b→ is perpendicularto c→, then |b→|  is equal to

# Let  and$\stackrel{\to }{c}=5\stackrel{\to }{i}+\stackrel{^}{j}+\sqrt{2k}$  be three vectors such that the projection of $\stackrel{\to }{b}$ on $\stackrel{\to }{a}$ is $|\stackrel{\to }{a}|$. If $\stackrel{\to }{a}+\stackrel{\to }{b}$ is perpendicularto $\stackrel{\to }{c},$ then $|\stackrel{\to }{b}|$  is equal to

1. A

$\sqrt{32}$

2. B

$\sqrt{22}$

3. C

4

4. D

6

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### Solution:

Given, projection of $\overline{b}$ on $\stackrel{\to }{a}=\frac{\stackrel{\to }{a}·\stackrel{\to }{b}}{|\stackrel{\to }{a}|}=|\stackrel{\to }{a}|$

Also, $\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)\perp \stackrel{\to }{c}⇒\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)·\stackrel{\to }{c}=0$

$⇒\left(1+{b}_{1}\right)5+\left(1+{b}_{2}\right)1+2\sqrt{2}\left(\sqrt{2}\right)=0$

Solving and

Now, $|\stackrel{\to }{b}|=\sqrt{{b}_{1}^{2}+{b}_{2}^{2}+2}=6$

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