Question

The angle between any two diagonals of cube are:

• A. ${\cos }^{-1}\left(\frac{1}{2}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{1}{3}\right)$

Let $OABCDEFG$ be a cube with vertices as below

$O(0,0,0),$A\left(a,0,0\right)$,$B\left(a,a,0\right)$,$C\left(0,a,0\right)$,

$D(0,a,a)$, $E(0,0,a)$, $F(a,0,a)$ and $G(a,a,a)$

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

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

$\overrightarrow{OG}=(a-0)\hat{i}+(a-0)\hat{j}+(a-0)\hat{k}=a\hat{i}+b\hat{j}+c\hat{k}$

$\overrightarrow{AD}=(0-a)\hat{i}+(a-0)\hat{j}+(a-0)\hat{k}=-a\hat{i}+b\hat{j}+c\hat{k}$

$\left|\overrightarrow{OG}\right|=\sqrt{a^2+a^2+a^2}=a\sqrt{3}$

$\left|\overrightarrow{AD}\right|=\sqrt{{(-a)}^2+a^2+a^2}=a\sqrt{3}$

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

We know that angle between any two vectors $\overrightarrow{a}$ and $\overrightarrow{b}={\cos }^{-1}\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}\right)$

$\Rightarrow$ Angle between the two diagonals $\overrightarrow{OG}$ and $\overrightarrow{AD}$

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

$={\cos }^{-1}\left(\frac{a^2}{3a^2}\right)$

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