IIT JEE has been asking novel interesting questions at ancillary levels for the Engineering Entrance Test in India. Some of these math questions are quite thought provoking. In the just held (05-June-2023) exam they asked:
The question
Let P be a plane $\sqrt{3}x+2y+3z=16$ and let the set
$S=\{\alpha \vec i+\beta \vec j+ \gamma \vec j $, $\alpha^2+\beta^2+\gamma^2=1$, the perpendicular distance of point $(\alpha, \beta, \gamma)$ from the plane P be $\frac{7}{2}.\}$
Let vectors $\vec u, \vec v, \vec w $ are three distinct vectors in $S$ such that $|\vec u-\vec v|= |\vec v- \vec w|=|\vec u-\vec w|.$ Find the volume of the parallelopiped determined by vectors $\vec u, \vec v$ and $\vec w$.
My proposed solution:
Note that vectors $\vec u, \vec v, \vec w$ are unit vectors additionally they are mutually equally inclined to each other $\hat u. \hat v = \hat v. \hat w=\hat u. \hat w=\cos \theta$ (let). The perpendicular distance connection gives is $$\sqrt{3}\alpha +2 \beta+ 3 \gamma=2,30......(*)$$
Visualize, $\vec r= \sqrt{3}\vec i+ 2\vec j+3 \vec k$, then $(*)$ is equivalent to $\vec r. \hat u= \vec r. \hat v= \vec r. \hat w= 2~ (\text{choose}) \implies \cos \phi = \frac{1}{2}.$
Here, $\vec r$ is equally inclined at angle $\phi$ to each of $\hat u, \hat v, \hat w. $, so we can write $$\hat r= \lambda [ \hat u +\hat v+\hat w]\implies 1=\lambda^2[3+6 \cos \theta].$$ Taking dot product by $\hat u$ in above, we get $$\cos \phi=\sqrt{\frac{1+2\cos \theta}{3}}.....(**).$$ From here we get $\cos \theta =-\frac{1}{8}.$ Finally, the volume $V$ of the parallelopiped is given by $$V=[\hat u, \hat v, \hat w]=(1-\cos \theta)\sqrt{1+2\cos \theta}=\frac{9\sqrt{3}}{16}.$$
Is this answer correct? What could be other ways of finding it.