Let $\ddot{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}, \ddot{\mathrm{b}}=\mathrm{b}_1 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}$ and $\ddot{\mathrm{c}}=5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}$ be three vectors such that the projection vector of $\stackrel{a}{b}$ on $\stackrel{a}{a}$ is $\stackrel{n}{a}$. if $\stackrel{a}{a}+\stackrel{u}{b}$ is perpendicular to $\stackrel{n}{c}$, then $\stackrel{n}{b}$ is equal to:
Select the correct option:
A
$\sqrt{22}$
B
$\sqrt{32}$
C
4
D
6
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Projection of
$$
\text { bon } \mathbf{a}=\frac{\mathbf{b}_2 \mathbf{a}}{|\mathbf{a}|}=\frac{\mathbf{b}_1+\mathbf{b}_2+2}{4}
$$
According to question $\frac{b_1+b_2+2}{2}=\sqrt{1+1+2}=2$
$$
\Rightarrow \mathrm{b}_1+\mathrm{b}_2=2
$$
Also $\stackrel{\mathrm{a}}{\mathrm{a}} \cdot \stackrel{\mathrm{a}}{\mathrm{c}}+\stackrel{\mathrm{a}}{\mathrm{b}} \cdot \stackrel{\mathrm{a}}{\mathrm{c}}=0$
$$
\Rightarrow 8+5 \mathrm{~b}_1+\mathrm{b}_2+2=0
$$
From (1) and (2),
$$
\begin{aligned}
& \mathrm{b}_1=-3, \mathrm{~b}_2=5 \\
\Rightarrow & \mathrm{~b}=3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}
\end{aligned}
$$
$$
|\vec{b}|=\sqrt{9+25+2}=6
$$
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