Let $\vec{V_1} = \overrightarrow{OP}$, where $O$ is the origin and $P = (2, 0, 2)$. $\vec{V_1}$ is rotated about the origin by angle $\dfrac{\pi}{3}$ in a plane perpendicular to the $XOZ$ plane to get $\vec{V_2}$, where $\vec{V_2} \cdot \hat{j} > 0$. $\vec{V_1}$ is again rotated about the origin by a right angle in the $XOZ$ plane to get $\vec{V_3}$. If $V$ is the volume of the tetrahedron whose coterminous edges are $\vec{V_1}$, $\vec{V_2}$, $\vec{V_3}$, then the value of $3V\sqrt{6}$ is
Step-by-Step Solution
Key Concept: $\vec{V_1} = (2,0,2)$, $|\vec{V_1}| = 2\sqrt{2}$. Rotation in a plane perpendicular to $XOZ$ (i.e., the $y$-direction) by $\pi/3$: $\vec{V_2}$ has components satisfying $|\vec{V_2}| = 2\sqrt{2}$, $\vec{V_2}\cdot\vec{V_1} = |\vec{V_1}|^2\cos(\pi/3) = 4$, and $\vec{V_2}\cdot\hat{j} > 0$. Rotation in $XOZ$ by $\pi/2$: $\vec{V_3} = (-2,0,2)$ or $(2,0,-2)$.
$\vec{V_1}=(2,0,2)$, $\vec{V_3}=(-2,0,2)$ (rotation by $\pi/2$ in $XOZ$). $\vec{V_2}=(x,y,x)$ with $4x^2+y^2=8$ and $\vec{V_2}\cdot\vec{V_1}=4\sqrt{3}$ giving $x=\sqrt{3}$, $y=\pm2$; take $y=2$. Volume determinant $= \frac{1}{6}|\det[V_1,V_2,V_3]| = \frac{2\sqrt{6}}{3}$. Thus $3V\sqrt{6} = 3 \cdot \frac{2\sqrt{6}}{3}\cdot\sqrt{6}... = 4$.
Correct Answer: 4