MCQ
The angle between two diagonals of a cube will be
  • A
    ${\sin ^{ - 1}}1/3$
  • ${\cos ^{ - 1}}1/3$
  • C
    Variable
  • D
    None of these

Answer

Correct option: B.
${\cos ^{ - 1}}1/3$
b
(b) Let the cube be of side $'a'$

$O\,(0,\,\,0,\,\,0),\,\,D\,(a,\,\,a,\,\,a),\,\,B\,(0,\,\,a,\,\,0),\,\,G\,(a,\,\,0,\,\,a)$

Then equation of $OD$ and $BG$ are $\frac{x}{a} = \frac{y}{a} = \frac{z}{a}$ and

$\frac{x}{a} = \frac{{y - a}}{{ - a}} = \frac{z}{a}$ respectively.

Hence, angle between $OD$ and $BG$ is

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

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