Question

If the lengths of the sides of a rectangular parallelopiped are $3,2,1$, then the angle between two diagonals out of four diagonals is

• A. ${\cos }^{-1}\left(\frac{6}{7}\right)$
• B. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• C. ${\cos }^{-1}\left(\frac{13}{14}\right)$
• D. ${\cos }^{-1}\left(\frac{9}{14}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{6}{7}\right)$

Consider the figure of the rectangular parallel pipe.

The possible four diagonals with their vector equations are:

$OG=2\hat{i}+3\hat{j}+\hat{k}$

$BE=2\hat{i}+3\hat{j}-\hat{k}$

$AD=2\hat{i}-3\hat{j}-\hat{k}$

$FC=2\hat{i}-3\hat{j}+\hat{k}$

Hence $OG.BE=|OG||BE|\cos \theta$

$2\left(2\right)+3\left(3\right)-1=\sqrt{14}\sqrt{14}\cos \theta$

$12=14\cos \theta$

$\cos \theta =\frac{6}{7}$

$\theta ={\cos }^{-1}\left(\frac{6}{7}\right)$