Question

The angle between two diagonals of a cube is

• A. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

Answer: (B)

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

Let $OABCDEFG$ be a cube with vertices as below

$O(0,0,0),A(a,0,0),B(a,a,0),C(0,a,0),D(0,a,a),E(0,0,a),F(a,0,a),G(a,a,a)$

There are four diagonals $OG,CF,AD$ and $BE$ for the cube.

Let us consider any two say $OG$ and $AD$

We know that if $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ are two points in space then

$\overrightarrow{AB}=(x_2-x_1)i+(y_2-y_1)j+(z_2-z_1)k$

$\Rightarrow \overrightarrow{OG}=(a-0)i+(a-0)j+(a-0)k=ai+aj+ak$

and $\overrightarrow{AD}=(0-a)i+(a-0)j+(a-0)k=-ai+aj+ak$.

Therefore $|\overrightarrow{OG}|=\sqrt{a^2+a^2+a^2}=a\sqrt{3}$ and $|\overrightarrow{AD}|=\sqrt{(-a{)}^2+a^2+a^2}=a\sqrt{3}$

$\overrightarrow{OG}\cdot \overrightarrow{AD}=-a^2+a^2+a^2=a^2$

We know that angle between two vectors $\overrightarrow{a},\overrightarrow{b}$ is given by $\theta =co{s}^{-1}\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$

Thus angle between two diagonals $\overrightarrow{OG}$ and $\overrightarrow{AD}$ is

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