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 can be

• A. ${\cos }^{-1}\frac{2}{7}$
• B. ${\cos }^{-1}\frac{4}{7}$
• C. ${\cos }^{-1}\frac{3}{7}$
• D. ${\cos }^{-1}\frac{1}{7}$

Answer 1

Answer: (A), (C)

${\cos }^{-1}\frac{3}{7}$

Let the sides of parallelopiped of length $3,2,1\text{units}$ are along $x-$axis, $y-$axis and $z-$axis respectively, shown in the figure.
Here, $\overrightarrow{OE}=3\hat{i},\overrightarrow{OA}=2\hat{j},\overrightarrow{OF}=\hat{k}$
So, point $B\equiv (\overrightarrow{OE}+\overrightarrow{EB})=3\hat{i}+2\hat{j}$ and
point $D\equiv (\overrightarrow{OE}+\overrightarrow{ED})=3\hat{i}+\hat{k}$
Now diagonal vector , $\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=3\hat{i}-2\hat{j}+\hat{k}$ and
diagonal vector $\overrightarrow{FB}=\overrightarrow{OB}-\overrightarrow{OF}=3\hat{i}+2\hat{j}-\hat{k}$
So, angle between the two diagonals
$\cos \theta =\frac{\overrightarrow{AD}\cdot \overrightarrow{FB}}{|\overrightarrow{AD}||\overrightarrow{FB}|}\Rightarrow \cos \theta =\frac{9-4-1}{9+4+1}=\frac{2}{7}\Rightarrow \theta ={\cos }^{-1}\frac{2}{7}$
also vector $\overrightarrow{OC}=3\hat{i}+2\hat{j}+\hat{k}$
So, angle between diagonals $\overrightarrow{OC}$ and $\overrightarrow{AD}:$
$\cos \beta =\frac{\overrightarrow{OC}\cdot \overrightarrow{AD}}{|\overrightarrow{OC}|\overrightarrow{AD}|}=\frac{9-4+1}{9+4+1}\Rightarrow \beta ={\cos }^{-1}\frac{3}{7}$
Also vector $\overrightarrow{GE}=3\hat{i}-2\hat{j}-\hat{k}$
So, angle between diagonals $\overrightarrow{GE}$ and $\overrightarrow{AD}={\cos }^{-1}\left(\frac{6}{7}\right)$
Angle between diagonals $\overrightarrow{OC}$ and $\overrightarrow{FB}={\cos }^{-1}\left(\frac{6}{7}\right)$
Angle between diagonals $\overrightarrow{GE}$ and $\overrightarrow{FB}={\cos }^{-1}\left(\frac{3}{7}\right)$
Angle between diagonals $\overrightarrow{OC}$ and $\overrightarrow{GE}={\cos }^{-1}\left(\frac{2}{7}\right)$