Question

Let $\overrightarrow{a}=(2+\sin \theta )\hat{i}+\cos \theta \hat{j}+\sin 2\theta \hat{k}$, $\overrightarrow{b}=\sin (\theta +\frac{2\pi }{3})\hat{i}+\cos (\theta +\frac{2\pi }{3})\hat{j}+\sin (2\theta +\frac{4\pi }{3})\hat{k}$ and $\overrightarrow{c}=\sin (\theta -\frac{2\pi }{3})\hat{i}+\cos (\theta -\frac{2\pi }{3})\hat{j}+\sin (2\theta -\frac{4\pi }{3})\hat{k}$ be three vectors where $\theta \in (0,\frac{\pi }{2})$. The maximum volume of the tetrahedron whose coterminous edges are given by the vectors $2\overrightarrow{b}\times \overrightarrow{c},3\overrightarrow{c}\times \overrightarrow{a}$ and $\overrightarrow{a}\times 4\overrightarrow{b}$ is

Answer 1

$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\left|\begin{array}{ccc}2+\sin \theta & \cos \theta & \sin 2\theta \\ \sin (\theta +\frac{2\pi }{3})& \cos (\theta +\frac{2\pi }{3})& \sin (2\theta +\frac{4\pi }{3})\\ \sin (\theta -\frac{2\pi }{3})& \cos (\theta -\frac{2\pi }{3})& \sin (2\theta -\frac{4\pi }{3})\end{array}\right|$

$R_1\to R_1+R_2+R_3,=\left|\begin{array}{ccc}2+\sin \theta +2\sin \theta \cos \frac{2\pi }{3}& \cos \theta +2\cos \theta \cos \frac{2\pi }{3}& \sin 2\theta +2\sin 2\theta \cos \frac{4\pi }{3}\\ \sin (\theta +\frac{2\pi }{3})& \cos (\theta +\frac{2\pi }{3})& \sin (2\theta +\frac{4\pi }{3})\\ \sin (\theta -\frac{2\pi }{3})& \cos (\theta -\frac{2\pi }{3})& \sin (2\theta -\frac{4\pi }{3})\end{array}\right|$

$=\left|\begin{array}{ccc}2& 0& 0\\ \sin (\theta +\frac{2\pi }{3})& \cos (\theta +\frac{2\pi }{3})& \sin (2\theta +\frac{4\pi }{3})\\ \sin (\theta -\frac{2\pi }{3})& \cos (\theta -\frac{2\pi }{3})& \sin (2\theta -\frac{4\pi }{3})\end{array}\right|$

$=2[\cos (\theta +\frac{2\pi }{3})\sin (2\theta -\frac{4\pi }{3})-\sin (2\theta +\frac{4\pi }{3})\cos (\theta -\frac{2\pi }{3})]=\sin (3\theta -\frac{2\pi }{3})-\sin (2\pi -\theta )-\sin (3\theta +\frac{2\pi }{3})-\sin (\theta +2\pi )=\sin (3\theta -\frac{2\pi }{3})-\sin (3\theta +\frac{2\pi }{3})=-\left(2\cos 3\theta \sin \frac{2\pi }{3}\right)=-\sqrt{3}\cos 3\theta$
The maximum value occurs when $3\theta =\pi \Rightarrow \theta =\frac{\pi }{3}$
${\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}_{\text{max}}=\sqrt{3}$

Required volume
$=\frac{1}{6}[2\overrightarrow{b}\times \overrightarrow{c}3\overrightarrow{c}\times \overrightarrow{a}\overrightarrow{a}\times 4\overrightarrow{b}]=\frac{1}{6}\times 24[\overrightarrow{b}\times \overrightarrow{c}\overrightarrow{c}\times \overrightarrow{a}\overrightarrow{a}\times \overrightarrow{b}]=4{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2$

Maximum volume of the tetrahedron
$=4\times 3=12$