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If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three mutually perpendicular vectors such that the magnitudes of $\mathbf{b}$ and $\mathbf{c}$ are $1 / 2$ times and $\sqrt{3} / 2$ times that of $\mathbf{a}$, respectively, then the angle between the vectors $\mathbf{a}+\mathbf{b}+\mathbf{c}$ and $\mathbf{b}$ is
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Verified Answer
The correct answer is:
$\cos ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)$
$\begin{aligned}
& \text { Given, }|\mathbf{b}|=\frac{1}{2}|\mathbf{a}| \text { and }|\mathbf{c}|=\frac{\sqrt{3}}{2}|\mathbf{a}| \\
& \text { and } \quad \begin{aligned}
\mathbf{a} \cdot \mathbf{b} & =\mathbf{b} \cdot \mathbf{c}=\mathbf{a} \cdot \mathbf{c}=0 \\
|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 & =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2=4|\mathbf{b}|^2+|\mathbf{b}|^2+3|\mathbf{b}|^2=8|\mathbf{b}|^2 \\
\Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}| & =2 \sqrt{2}|\mathbf{b}|
\end{aligned}
\end{aligned}$
Angle between $(\mathbf{a}+\mathbf{b}+\mathbf{c})$ and $\mathbf{b}$ is
$\begin{aligned}
\cos \theta & =\frac{(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot \mathbf{b}}{|\mathbf{a}+\mathbf{b}+\mathbf{c}||\mathbf{b}|} \Rightarrow \cos \theta=\frac{|\mathbf{b}|^2}{2 \sqrt{2}|\mathbf{b}|^2}=\frac{1}{2 \sqrt{2}} \\
\theta & =\cos ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)
\end{aligned}$
& \text { Given, }|\mathbf{b}|=\frac{1}{2}|\mathbf{a}| \text { and }|\mathbf{c}|=\frac{\sqrt{3}}{2}|\mathbf{a}| \\
& \text { and } \quad \begin{aligned}
\mathbf{a} \cdot \mathbf{b} & =\mathbf{b} \cdot \mathbf{c}=\mathbf{a} \cdot \mathbf{c}=0 \\
|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 & =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2=4|\mathbf{b}|^2+|\mathbf{b}|^2+3|\mathbf{b}|^2=8|\mathbf{b}|^2 \\
\Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}| & =2 \sqrt{2}|\mathbf{b}|
\end{aligned}
\end{aligned}$
Angle between $(\mathbf{a}+\mathbf{b}+\mathbf{c})$ and $\mathbf{b}$ is
$\begin{aligned}
\cos \theta & =\frac{(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot \mathbf{b}}{|\mathbf{a}+\mathbf{b}+\mathbf{c}||\mathbf{b}|} \Rightarrow \cos \theta=\frac{|\mathbf{b}|^2}{2 \sqrt{2}|\mathbf{b}|^2}=\frac{1}{2 \sqrt{2}} \\
\theta & =\cos ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)
\end{aligned}$
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