If $\vec{f}, \vec{g}, \vec{h}$ mutually orthogonal vectors of equal magnitudes, then the angle between the vectors $\vec{f}+\vec{g}+\vec{h}$ and $\vec{h}$ is

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Aicharya · Accepted Answer

Given, $\vec{f}, \vec{g}$ and $\vec{h}$ are mutually orthogonal vector So, $\vec{f} \cdot \vec{g}=\vec{g} \cdot \vec{h}=\vec{f} \cdot \vec{h}=0$. Let, $|\vec{f}|=|\vec{g}|=|\vec{h}|=k$ Now, $(\vec{f}+\vec{g}+\vec{h}) \cdot \vec{h}=|\vec{f}+\vec{g}+\vec{h} \| \vec{h}| \cos 	heta$ $\Rightarrow|\vec{h}|^2=|\vec{f}+\vec{g}+\vec{h} \| \vec{h}| \cos 	heta \Rightarrow \cos 	heta=\frac{|\vec{h}|}{|\vec{f}+\vec{g}+\vec{h}|}\ldots(i)$ $\begin{aligned} & 	ext { Since, }|\vec{f}+\vec{g}+\vec{h}|^2=|\vec{f}|^2+|\vec{g}|^2+|\vec{h}|^2+0+0+0 \ & \Rightarrow|\vec{f}+\vec{g}+\vec{h}|=\sqrt{k^2+k^2+k^2}=\sqrt{3} k\end{aligned}$ $\operatorname{By}(\mathrm{i}), \cos 	heta=\frac{k}{\sqrt{3} k} \Rightarrow 	heta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}ight)$

If $\vec{f}, \vec{g}, \vec{h}$ mutually orthogonal vectors of equal magnitudes, then the angle between the…

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