The volume of the parallelepiped whose coterminous edges are represented by the vectors 2b→×c→, 3c→×a→ and 4a→×b→, where a→=(1+sin θ)i^+cosθj^+sin2θk^, b→=sin(θ+2π3)i^+cos(θ+2π3)j^+sin(2θ+4π3)k^, c→=sin(θ−2π3)i^+cos(θ−2π3)j^+sin(2θ−4π3)k^ is 18 cubic units, then the values of θ , in the interval (0,π2) can be

The volume of the parallelepiped whose coterminous edges are represented by the vectors 2b→×c→, 3c→×a→ and 4a→×b→, where a→=(1+sin θ)i^+cosθj^+sin2θk^, b→=sin(θ+2π3)i^+cos(θ+2π3)j^+sin(2θ+4π3)k^, c→=sin(θ−2π3)i^+cos(θ−2π3)j^+sin(2θ−4π3)k^ is 18 cubic units, then the values of θ , in the interval (0,π2) can be

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The volume of the parallelepiped whose coterminous edges are represented by the vectors
$2 \vec{b} \times \vec{c}, 3 \vec{c} \times \vec{a} a n d 4 \vec{a} \times \vec{b}, w h e r e \vec{a} = (1 + \sin θ) \hat{i} + \cos θ \hat{j} + \sin 2 θ \hat{k},$ $\vec{b} = \sin (θ + \frac{2 π}{3}) \hat{i} + \cos (θ + \frac{2 π}{3}) \hat{j} + \sin (2 θ + \frac{4 π}{3}) \hat{k},$ $\vec{c} = \sin (θ - \frac{2 π}{3}) \hat{i} + \cos (θ - \frac{2 π}{3}) \hat{j} + \sin (2 θ - \frac{4 π}{3}) \hat{k}$
is 18 cubic units, then the values of $θ$ , in the interval $(0, \frac{π}{2})$ can be

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$\frac{π}{3}$

$\frac{π}{9}$

$\frac{2 π}{9}$

$\frac{4 π}{9}$

answer is C.

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Detailed Solution

Volume = $| [2 \vec{b} \times \vec{c} 3 \vec{c} \times \vec{a} 4 \vec{a} \times \vec{b}] | = 18$
24 ${[\vec{a} \vec{b} \vec{c}]}^{2} = 18 \Rightarrow | [\vec{a} \vec{b} \vec{c}] | = \frac{\sqrt{3}}{2}$
Now, ${[\vec{a} \vec{b} \vec{c}]}^{2} = | \begin{matrix} (1 + \sin θ) & \cos θ & \sin 2 θ \\ \sin (θ + \frac{2 π}{3}) & \cos (θ + \frac{2 π}{3}) & \sin (2 θ + \frac{4 π}{3}) \\ \sin (θ - \frac{2 π}{3}) & \cos (θ - \frac{2 π}{3}) & \sin (2 θ - \frac{4 π}{3}) \end{matrix} |$
Applying $R_{1} \to R_{1} + R_{2} + R_{3}$ we get after simplification,
$| [\vec{a} \vec{b} \vec{c}] | = \frac{\sqrt{3}}{2} | \cos 3 θ | = \frac{\sqrt{3}}{2} \Rightarrow \cos 3 θ = \pm 1 \Rightarrow 3 θ = π \Rightarrow θ = \frac{π}{3}$