The angle between any two diagonals of a cube is

1. $\cos ^{-1}\left(\frac{1}{3}\right)$
2. $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
3. $\cos ^{-1}\left(\frac{1}{2}\right)$
4. $\cos ^{-1}\left(\frac{2}{3}\right)$

Solution

Diagonal of cube $\mathbf{O P}$ and $\mathbf{A Q}$. $ \begin{aligned} & \mathbf{O P}=a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+a \hat{\mathbf{k}} \\ & \mathbf{A Q}=-a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+a \hat{\mathbf{k}} \end{aligned} $
Angle between diagonal i.e. $ \begin{aligned} \cos \theta & =\frac{-a^2+a^2+a^2}{\sqrt{3 a^2} \sqrt{3 a^2}} \\ \cos \theta & =\frac{a^2}{3 a^2}=\frac{1}{3} \\ \theta & =\cos ^{-1}\left(\frac{1}{3}\right) \end{aligned} $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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