If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors equally inclined to each other at an angle $\frac{π}{3}$ , then $[\vec{b} + \vec{c} \vec{c} + \vec{a} \vec{a} + \vec{b}] =$

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$\frac{1}{\sqrt{2}}$

$\frac{1}{2}$

$\sqrt{2}$

answer is D.

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Detailed Solution

$| \vec{a} | = | \vec{b} | = | \vec{c} | = 1$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \frac{1}{2}$

$[\vec{a} \vec{b} \vec{c}]^{2} = |\begin{array}{lll} \vec{a} \cdot \vec{a} \vec{a} \cdot \vec{b} \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} \vec{b} \cdot \vec{b} \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} \vec{c} \cdot \vec{b} \vec{c} \cdot \vec{c} \end{array}| = |\begin{array}{ccc} 1 & \frac{1}{2} & \frac{1}{2} \\ 1 & 2 & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{array}| = \frac{1}{2}$

$\begin{array}{l} [\vec{a} \vec{b} \vec{c}] = \frac{1}{\sqrt{2}} \\ [\vec{b} + \vec{c} \vec{c} + \vec{a} \vec{a} + \vec{b}] = 2 [\vec{a} \vec{b} \vec{c}] = \sqrt{2} \end{array}$