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(B) Given that $a, b, c$ are mutually perpendicular vectors, we have $a \cdot b = b \cdot c = c \cdot a = 0$. Let $|a| = k$. Then $|b| = \frac{1}{2}k$

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{1}{2\sqrt{2}}\right)$

Vedclass · Answer

(B) Given that $a, b, c$ are mutually perpendicular vectors, we have $a \cdot b = b \cdot c = c \cdot a = 0$. Let $|a| = k$. Then $|b| = \frac{1}{2}k$ and $|c| = \frac{\sqrt{3}}{2}k$. First, calculate the magnitude of the vector $a+b+c$: $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 = k^2 + \left(\frac{1}{2}kight)^2 + \left(\frac{\sqrt{3}}{2}kight)^2 = k^2 + \frac{1}{4}k^2 + \frac{3}{4}k^2 = 2k^2$. So, $|a+b+c| = \sqrt{2}k$. Now, let $	heta$ be the angle between $(a+b+c)$ and $b$. Using the dot product formula, $\cos 	heta = \frac{(a+b+c) \cdot b}{|a+b+c| |b|}$. Since $a \cdot b = 0$ and $c \cdot b = 0$, we have $(a+b+c) \cdot b = a \cdot b + b \cdot b + c \cdot b = 0 + |b|^2 + 0 = |b|^2$. Substituting the values: $\cos 	heta = \frac{|b|^2}{|a+b+c| |b|} = \frac{|b|}{|a+b+c|} = \frac{\frac{1}{2}k}{\sqrt{2}k} = \frac{1}{2\sqrt{2}}$. Therefore, $	heta = \cos^{-1}\left(\frac{1}{2\sqrt{2}}ight)$.

If $a, b, c$ are three mutually perpendicular vectors such that the magnitudes of $b$ and $c$ are $1/2$ times and $\sqrt{3}/2$ times that of $a$, respectively, then the angle between the vectors $a+b+c$ and $b$ is

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