If vec{f}, vec{g}, vec{h} are mutually orthog | vedclass

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(B) Given that $\vec{f}, \vec{g},$ and $\vec{h}$ are mutually orthogonal vectors.So,$\vec{f} \cdot \vec{g} = \vec{g} \cdot \vec{h} = \vec{f} \cdot \ve

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$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

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(B) Given that $\vec{f}, \vec{g},$ and $\vec{h}$ are mutually orthogonal vectors.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,let $	heta$ be the angle between $\vec{f}+\vec{g}+\vec{h}$ and $\vec{h}$.Then,$(\vec{f}+\vec{g}+\vec{h}) \cdot \vec{h} = |\vec{f}+\vec{g}+\vec{h}| |\vec{h}| \cos 	heta$.Since $\vec{f} \cdot \vec{h} = 0$ and $\vec{g} \cdot \vec{h} = 0$,we have $(\vec{f}+\vec{g}+\vec{h}) \cdot \vec{h} = |\vec{h}|^2 = k^2$.Also,$|\vec{f}+\vec{g}+\vec{h}|^2 = |\vec{f}|^2 + |\vec{g}|^2 + |\vec{h}|^2 + 2(\vec{f} \cdot \vec{g} + \vec{g} \cdot \vec{h} + \vec{h} \cdot \vec{f}) = k^2 + k^2 + k^2 = 3k^2$.So,$|\vec{f}+\vec{g}+\vec{h}| = \sqrt{3}k$.Substituting these into the dot product formula: $k^2 = (\sqrt{3}k)(k) \cos 	heta$.$\cos 	heta = \frac{k^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

If $\vec{f}, \vec{g}, \vec{h}$ are 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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