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

Answer 1

Answer: (B)

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