MCQ
The angle between two diagonals of a cube will be
- A${\sin ^{ - 1}}1/3$
- ✓${\cos ^{ - 1}}1/3$
- CVariable
- DNone of these
✓
Answer
Correct option: B.
${\cos ^{ - 1}}1/3$
b
(b) Let the cube be of side $'a'$
(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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