Question

The angle between two diagonals of a cube is

• A. $30^{\circ }$
• B. $45^{\circ }$
• C. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer 1

Answer: (C)

Let edge of a cube be 1 unit.
The diagonals of a cube are $OA$ and $BC$.
So, DR's of diagonals $OA$ are $(1,1,1)$ and $BC$ are $(0-1,1,1)$, i.e., $(-1,1,1)$

Now, angle between diagonals,
$\cos \theta =\frac{1(-1)+1(1)+1(1)}{\sqrt{1^2+1^2+1^2}\sqrt{(-1{)}^2+1^2+1^2}}$
$=\frac{1}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+1^2+1^2}}$
$=\frac{1}{\sqrt{3}\sqrt{3}}=\frac{1}{3}$
$\Rightarrow \theta ={\cos }^{-1}(1/3)$