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If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are mutually perpendicular vectors of the same magnitude, then…
If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are mutually perpendicular vectors of the same magnitude, then the cosine of the angle between $\mathbf{a}$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}$ is
$\frac{1}{\sqrt{2}}$ $\frac{1}{\sqrt{3}}$ $\frac{1}{2}$ $\frac{\sqrt{3}}{2}$ Solution
(b) Let $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=\lambda$
Now, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})$
$
\begin{aligned}
& =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\
& =\lambda^2+\lambda^2+\lambda^2+2(0+0+0)[\because \mathbf{a} \perp \mathbf{b}, \mathbf{b} \perp \mathbf{c}, \mathbf{c} \perp \mathbf{a}] \\
& =3 \lambda^2
\end{aligned}
$
$
\therefore|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{3} \lambda
$
Now, let $\theta$ be angle between $\mathbf{a}$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}$
$
\begin{aligned}
\therefore \cos \theta & =\frac{\mathbf{a} \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})}{|\mathbf{a}||\mathbf{a}+\mathbf{b}+\mathbf{c}|}=\frac{|\mathbf{a}|^2}{|\mathbf{a}||\mathbf{a}+\mathbf{b}+\mathbf{c}|} \\
& =\frac{\lambda^2}{\lambda+\sqrt{3} \lambda}=\frac{1}{\sqrt{3}} .
\end{aligned}
$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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