Let $\overrightarrow{a}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}, \overline{b}=b_1\hat{i}+b_2\hat{j}+\sqrt{2}\hat{k}$ and

$\overrightarrow{c}=5\overrightarrow{i}+\hat{j}+\sqrt{2k}$  be three vectors such that the 

projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\left|\overrightarrow{a}\right|$. If $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular

to $\overrightarrow{c},$ then $\left|\overrightarrow{b}\right|$  is equal to 

Solution:

Given, projection of $\overline{b}$ on $\overrightarrow{a}=\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{a}\right|}=\left|\overrightarrow{a}\right|$

$\Rightarrow \frac{b_1+b_2+2}{\sqrt{4}}=\sqrt{4}\Rightarrow b_1+b_2=2$

Also, $(\overrightarrow{a}+\overrightarrow{b})\perp \overrightarrow{c}\Rightarrow (\overrightarrow{a}+\overrightarrow{b})·\overrightarrow{c}=0$

$\Rightarrow \left(1+b_1\right)5+\left(1+b_2\right)1+2\sqrt{2}\left(\sqrt{2}\right)=0$

$\Rightarrow 5b_1+b_2=-10$

Solving $\text{ (i) }$and $\text{ (ii), we get }{b}_1=-3\text{ and }{b}_2=5$

Now, $\left|\overrightarrow{b}\right|=\sqrt{b_1^2+b_2^2+2}=6$