Question

Prove that the angle between any two diagonals of a cube is given by cos=1/3​​

Answer 1

Consider a cube of side a units, with sides along x, y and z axis and one vertex at origin, as in the figure.

Here AB and CD are two diagonals of the cube, because A and B are diagonally opposite corners, also C and D.

In the figure, coordinates of A is (0, 0, 0) and that of B is (a, a, a). Then vector AB is given by,

$\vec{\rm{AB}}=\rm{\left < a,\ a,\ a\right > }$

Coordinates of C is (a, 0, 0) and that of D is (0, a, a). Then vector CD is given by,

$\vec{\rm{CD}}=\rm{\left < -a,\ a,\ a\right > }$

The diagonals each have a magnitude of a√3.

$|\vec{\rm{AB}}|=|\vec{\rm{CD}}|=\rm{a\sqrt3}$

In the figure θ is the angle between the diagonals.

We take the dot product of the vectors AB and CD to get angle between them.

$\longrightarrow\vec{\rm{AB}}\cdot\vec{\rm{CD}}=|\vec{\rm{AB}}|\cdot|\vec{\rm{CD}}|\cdot\cos\theta$

$\longrightarrow\rm{\left < a,\ a,\ a\right > \cdot\left < -a,\ a,\ a\right > =a\sqrt3\cdot a\sqrt3\cdot \cos\theta}$

$\longrightarrow\rm{-a^2+a^2+a^2=3a^2\cos\theta}$

$\longrightarrow\rm{a^2=3a^2\cos\theta}$

$\longrightarrow\rm{\cos\theta=\dfrac{a^2}{3a^2}}$

$\longrightarrow\rm{\underline{\underline{\cos\theta=\dfrac{1}{3}}}}$

The angle between the diagonals satisfy this equation.

Hence Proved!