Question
The angle between the two diagonals of a cube is:
✓
Answer
- $\cos^{-1}\Big(\frac{1}{{3}}\Big)$
Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR, Consider the diagonals OP and AR.
Direction ratios of OP and AR are proportional to a - 0, a - 0, a - 0 and 0 - a, a - 0, a - 0, e.i. a, a, a and -a, a, a, respectivelly.
Let $\theta$ be the angle between OP and AR. Then,
$\cos\theta=\frac{\text{a}\times-\text{a}+\text{a}\times\text{a}+\text{a}\times\text{a}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{(-\text{a})^2+\text{a}^2+\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{-\text{a}+\text{a}^2+\text{a}^2}{\sqrt{3\text{a}^2}\sqrt{3\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{1}{3}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{3}\Big)$
Similarly, the angles between other pairs of the diagonals are equal to $\cos^{-1}\Big(\frac{1}{3}\Big)$ as the angle between any two diagonals.
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