Question

Find the angle between any two diagonals of a cube.

Answer 1

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Let $OABCDEFG$ be a cube

with vertices $O(0,0,0)A(a,0,0)B(a,a,0)$

$C(0,a,0)D(0,a,a)E(0,0,a)F(a,0,a)$

$G(a,a,a)$

There are 4 diagonals $OG,CF,AD$ & $BE$

It $A(x_1,y_1,z_1)$ & $B(x_2,y_2,z_2)$ then

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

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

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

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

$=\sqrt{3}a$

$\left|\overline{AD}\right|=\sqrt{-a^2+a^2+a^2}$

$=\sqrt{3}a$

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

Angle between them $co{s}^{-1}\left(\frac{\overline{a}.\overline{b}}{\left|\overline{a}\right|.\left|\overline{b}\right|}\right)$

Angle between 2 diagonals $\overline{OG}$ and $\overline{AD}=co{s}^{-1}\left(\frac{a^2}{\sqrt{3}a.\sqrt{3}a}\right)$

$=co{s}^{-1}\frac{a^2}{3a^2}=co{s}^{-1}\frac{1}{3}$