Find the angle between the body diagonals of a cube

Problem – Find the angle between the body diagonals of a cube.

Solution

Here $\overrightarrow{A}$ and $\overrightarrow{B}$ are body diagonals and $\theta$ is angle between $\overrightarrow{A}$ and $\overrightarrow{B}$

$\overrightarrow{A}$ is given by

$\overrightarrow{A}$ = $\hat{x} +\hat{y}-\hat{z}$

$\left|A\right|$= $\sqrt{1^{2}+1^{2}+(-1)^{2}}$ = $\sqrt{3}$

and $\overrightarrow{B}$ is given by

$\overrightarrow{B}$ = $\hat{x}$ +$\hat{y}$+$\hat{z}$

$\left|B\right|$ = $\sqrt{1^{2} + 1^{2} + 1^{2}}$=$\sqrt{3}$

now angle between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by

$\overrightarrow{A}$ . $\overrightarrow{B}$ = $\left|A\right|\left|B\right|cos(\theta)$

$( \hat{x} +\hat{y}-\hat{z}) . ( \hat{x} +\hat{y}+\hat{z}) = \sqrt{3}\sqrt{3} cos(\theta)$

1+1-1 = 3 cos($\theta$)
$\theta$ = $cos^{-1}(\frac{1}{3})$