If a, b, c are mutually perpendicular vectors | vedclass

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(C) Given that $a, b, c$ are mutually perpendicular vectors of equal magnitudes,let $|a| = |b| = |c| = k$,where $k > 0$.Since they are mutually perpen

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

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(C) Given that $a, b, c$ are mutually perpendicular vectors of equal magnitudes,let $|a| = |b| = |c| = k$,where $k > 0$.Since they are mutually perpendicular,their dot products are zero: $a \cdot b = b \cdot c = c \cdot a = 0$.Let $	heta$ be the angle between $a$ and $a + b + c$. The formula for the angle is $\cos 	heta = \frac{a \cdot (a + b + c)}{|a| |a + b + c|}$.First,calculate the numerator: $a \cdot (a + b + c) = a \cdot a + a \cdot b + a \cdot c = |a|^2 + 0 + 0 = k^2$.Next,calculate the magnitude $|a + b + c|$: $|a + b + c|^2 = (a + b + c) \cdot (a + b + c) = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = k^2 + k^2 + k^2 + 0 = 3k^2$.Thus,$|a + b + c| = \sqrt{3}k$.Substituting these into the formula: $\cos 	heta = \frac{k^2}{k \cdot \sqrt{3}k} = \frac{k^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

If $a, b, c$ are mutually perpendicular vectors of equal magnitudes,then the angle between the vectors $a$ and $a + b + c$ is

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