#### 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 : Option: 1 $\sqrt{22}$Option: 2 4Option: 3 $\sqrt{32}$Option: 4 6

Given, the projection vector of $\vec{b}$on $\vec{a}$ is $\vec{a}$  ,

$\frac{\vec{b}.\vec{a}}{\left| {\vec{a}} \right|}=\left| {\vec{a}} \right| \\ \vec{a}.\vec{b}={{\left| {\vec{a}} \right|}^{2}} \\ \vec{a}.\vec{b}={{b}_{1}}+{{b}_{2}}+2 \\ \left| {\vec{a}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{(\sqrt{2})}^{2}}} \\ \left| {\vec{a}} \right|=2 \\ {{b}_{1}}+{{b}_{2}}+2=4 \\ {{b}_{1}}+{{b}_{2}}=2 \\$

If $\vec{a}+\vec{b}$ is perpendicular to $\vec{c},$

$(\vec{a}+\vec{b}).\vec{c}=0 \\ ((1+{{b}_{1}})\hat{i}+(1+{{b}_{2}})\hat{j}+2\sqrt{2}\hat{k}).(5\hat{i}+\hat{j}+\sqrt{2}\hat{k})=0 \\ 5(1+{{b}_{1}})+(1+{{b}_{2}})+4=0 \\ 10+5{{b}_{1}}+{{b}_{2}}=0 \\$

Solving the equations,
${{b}_{1}}+{{b}_{2}}=2 \\ 5{{b}_{1}}+{{b}_{2}}=-10 \\$

We get,
${{b}_{1}}=-3 \\ {{b}_{2}}=5 \\ \left| {\vec{b}} \right|=\sqrt{b_{1}^{2}+b_{2}^{2}+2} \\\\ \left| {\vec{b}} \right|=\sqrt{9+25+2} \\\\ \left| {\vec{b}} \right|=\sqrt{36} \\\\ \left| {\vec{b}} \right|=6 \\\\$