Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\mathrm{b}_{1} \hat{\mathrm{i}}+\mathrm{b}_{2} \hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}} \quad$ and $\overrightarrow{\mathrm{c}}=5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}$ 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 $\overrightarrow{\mathrm{c}}$, then $|\overrightarrow{\mathrm{h}}|$ is equal to:
Correct Option: , 4
Projection of $\vec{b}$ on $\vec{a}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}=|\vec{a}|$
$\Rightarrow \mathrm{b}_{1}+\mathrm{b}_{2}=2$ $\ldots(1)$
and $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \perp \overrightarrow{\mathrm{c}} \Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=0$
$\Rightarrow 5 b_{1}+b_{2}=-10$ .......(2)
from (1) and $(2) \Rightarrow b_{1}=-3$ and $b_{2}=5$
then $|\vec{b}|=\sqrt{b_{1}^{2}+b_{2}^{2}+2}=6$
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