A) \[\frac{\pi }{3}\]
B) \[{{\cos }^{-1}}\left( \frac{1}{3} \right)\]
C) \[{{\cos }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]
D) None of these
Correct Answer: B
Solution :
Let the side of a cube be unit, diagonals of a cube are \[AL\] and\[OP\]. \[\therefore \]\[DR's\]of\[AL\]are\[a,\,\,-a,\,\,-a\]and\[DR's\]of\[OP\]are\[a,\,\,a,\,\,a\]. \[\therefore \] \[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] \[=\frac{{{a}^{2}}-{{a}^{2}}-{{a}^{2}}}{a\sqrt{3}\cdot a\sqrt{3}}=\frac{1}{3}\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left( \frac{1}{3} \right)\]You need to login to perform this action.
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