Given vectors $\to a = \frac{1}{2}^i + \frac{\sqrt{3}}{2}^j,$ $\to b = \frac{\sqrt{3}}{2}^j + \frac{1}{2}^k$ and $\to c =^i + 2^j + 3^k .$ Then the volume of a parallelepiped with three coterminous edges as $\to u = (\to a \cdot \to a) \to a + (\to a \cdot \to b) \to b + (\to a \cdot \to c) \to c,$ $\to v = (\to a \cdot \to b) \to a + (\to b \cdot \to b) \to b + (\to b \cdot \to c) \to c$ and $\to w = (\to a \cdot \to c) \to a + (\to b \cdot \to c) \to b + (\to c \cdot \to c) \to c$ equals

(2 cos 30^{\circ} - sin 30^{\circ})^{3}

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(2 cos 60^{\circ} - sin 60^{\circ})^{3}

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(2 cos 45^{\circ} - sin 60^{\circ})^{3}

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(2 sin 45^{\circ} - sin 60^{\circ})^{3}

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Solution

The correct option is A $(2 cos 30^{\circ} - sin 30^{\circ})^{3}$
Volume of parallelepiped
$= | [\to u \to v \to w] |$
$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a \cdot \to a & \to a \cdot \to b & \to a \cdot \to c \to a \cdot \to b & \to b \cdot \to b & \to b \cdot \to c \to a \cdot \to c & \to b \cdot \to c & \to c \cdot \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ [\to a \to b \to c] ∣ ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ [\to a \to b \to c]^{3} ∣ ∣ ∣$

Now, $[\to a \to b \to c] = ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 0 & \frac{\sqrt{3}}{2} & \frac{1}{2} 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$\Rightarrow [\to a \to b \to c] = \frac{1}{4} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & \sqrt{3} & 0 0 & \sqrt{3} & 1 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow [\to a \to b \to c] = \frac{1}{4} [1 (3 \sqrt{3} - 2) - \sqrt{3} (0 - 1)]$
$= \frac{1}{4} [4 \sqrt{3} - 2] = \sqrt{3} - \frac{1}{2}$
$= 2 cos 30^{\circ} - sin 30^{\circ}$
Hence, the volume of parallelepiped is $(2 cos 30^{\circ} - sin 30^{\circ})^{3}$ cu. units.

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