Question

The length of two opposite edges of a tetrahedron are $12$ and $15$ units and the shortest distance between them is $10$ units. If the volume of the tetrahedron is $200$ cubic units, then the angle between the $2$ edges is

• A. $si{n}^{-1}\frac{1}{2}$
• B. $si{n}^{-1}\frac{2}{3}$
• C. $si{n}^{-1}\frac{3}{4}$
• D. $si{n}^{-1}\frac{4}{5}$

Answer 1

Answer: (B)

Let, $ABCD$ be the tetrahedron and position vectors of $A,B,C,D$ be $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ respectively.
Given, $\left|\overrightarrow{b}-\overrightarrow{a}\right|=12\&\left|\overrightarrow{d}-\overrightarrow{c}\right|=15$
Shortest distance $=\left|\frac{\left(\overrightarrow{d}-\overrightarrow{b}\right)\cdot \left(\overrightarrow{b}-\overrightarrow{a}\right)\times \left(\overrightarrow{d}-\overrightarrow{c}\right)}{\left|\left(\overrightarrow{b}-\overrightarrow{a}\right)\times \left(\overrightarrow{d}-\overrightarrow{c}\right)\right|}\right|=10$ ... $\left(1\right)$
Also, volume $=\frac{1}{6}\left|\left(\overrightarrow{d}-\overrightarrow{a}\right)\cdot \left(\overrightarrow{b}-\overrightarrow{a}\right)\times \left(\overrightarrow{c}-\overrightarrow{a}\right)\right|=200$ ... $\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ , we get,
$\left|\left(\overrightarrow{b}-\overrightarrow{a}\right)\times \left(\overrightarrow{d}-\overrightarrow{c}\right)\right|\times 10=6\times 200$
$\left|\overrightarrow{b}-\overrightarrow{a}\right|\left|\overrightarrow{d}-\overrightarrow{c}\right|sin\theta =120$
$sin\theta =\frac{2}{3}\Rightarrow \theta =si{n}^{-1}\frac{2}{3}$