Let a, b, c in mathbb{R} be such that a^{2} + | vedclass

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(D) Let $\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$.Given $a^{2} + b^{2} + c^{2} = 1$,we have $|\v

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$\frac{2\pi}{3}$

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(D) Let $\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$.Given $a^{2} + b^{2} + c^{2} = 1$,we have $|\vec{p}| = \sqrt{a^{2} + b^{2} + c^{2}} = 1$ and $|\vec{q}| = \sqrt{b^{2} + c^{2} + a^{2}} = 1$.The dot product is $\vec{p} \cdot \vec{q} = ab + bc + ca$.Let $a \cos 	heta = b \cos \left(	heta + \frac{2\pi}{3}ight) = c \cos \left(	heta + \frac{4\pi}{3}ight) = k$.Then $a = \frac{k}{\cos 	heta}$,$b = \frac{k}{\cos(	heta + 2\pi/3)}$,$c = \frac{k}{\cos(	heta + 4\pi/3)}$.Since $a+b+c = k(\sec 	heta + \sec(	heta + 2\pi/3) + \sec(	heta + 4\pi/3)) = 0$ (using the identity for sum of secants),we have $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = 0$.Since $a^{2} + b^{2} + c^{2} = 1$,we get $1 + 2(ab + bc + ca) = 0$,so $ab + bc + ca = -1/2$.Thus,$\cos \phi = \frac{\vec{p} \cdot \vec{q}}{|\vec{p}||\vec{q}|} = \frac{-1/2}{1 	imes 1} = -1/2$.Therefore,$\phi = \frac{2\pi}{3}$.

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