Question

Find the angle between the two diagonals of a cube.

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

Answer 1

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

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

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

Similarily$,\overline{AD}=-a\stackrel{\wedge }{i}+a\stackrel{\wedge }{j}+a\stackrel{\wedge }{k}$

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

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

$\overline{OG}.\overline{AD}=(a\stackrel{\wedge }{i}+a\stackrel{\wedge }{j}+a\stackrel{\wedge }{k}).(-a\stackrel{\wedge }{i}+a\stackrel{\wedge }{j}+a\stackrel{\wedge }{k})$

$={(-a)}^2+a^2+a^2=a^2$

Angle between $\overline{OG}and\overline{AD}$

$\cos \theta =\frac{\overline{OG}.\overline{AD}}{\left|\overline{OG}\right|.\left|\overline{AD}\right|}$

$⟹\cos \theta =\frac{a^2}{a\sqrt{3}.a\sqrt{3}}$

$⟹\cos \theta =\frac{1}{3}$

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

$\therefore B)Answer.$