Question

The angle between any two diagonals of a cube is:

• A. $co{s}^{-1}\left(\frac{1}{2}\right)$
• B. $co{s}^{-1}\left(\frac{1}{3}\right)$
• C. $co{s}^{-1}\left(\frac{1}{4}\right)$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$co{s}^{-1}\left(\frac{1}{3}\right)$

Take one corner of the cube, O as the origin and OA, OB, OC - the three edges through it, as the axes.
Let, OA = OB = OC = a
The co-ordinates of the various corners are shown in the figure.



The four diagonals are AL, BM, CN and OP
The DRs of the diagonal AL are 0 - a, a - 0, a - 0,
i.e., -a,a,a
The DRs of diagonal OP are a - 0, a - 0, a - 0 i.e., a, a, a
If $\theta$ is the angle between them, then

$cos\theta =\frac{|(-a)\left(a\right)+\left(a\right)\left(a\right)+\left(a\right)\left(a\right)|}{\sqrt{a^2+a^2+a^2}\sqrt{a^2+a^2+a^2}}cos\theta =\frac{a^2}{3a^2}=\frac{1}{3}\Rightarrow \theta =co{s}^{-1}\left(\frac{1}{3}\right)$