Question

The angle between any two diagonals of a cube is

• A. $cos\theta =\frac{\sqrt{3}}{2}$
• B. $cos\theta =\frac{1}{\sqrt{2}}$
• C. $cos\theta =\frac{1}{3}$
• D. $theta=\frac{1}{\sqrt{6}}$

Answer 1

The correct option is C $cos\theta =\frac{1}{3}$

DC’s of diagonals OP and AQ are
$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}and-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
If angle between them is $\theta$, then
$cos\theta =\left(\frac{1}{\sqrt{3}}\right)(-\frac{1}{\sqrt{3}})+\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right)$
$=\frac{1}{3}$