question_answer

Angle between any two diagonals of a cube is

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\[a,\,\,a,\,\,a\]. \[\therefore \]  \[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\]                 \[=\frac{{{a}^{2}}-{{a}^{2}}-{{a}^{2}}}{a\sqrt{3}\cdot a\sqrt{3}}=\frac{1}{3}\] \[\Rightarrow \]               \[\theta ={{\cos }^{-1}}\left( \frac{1}{3} \right)\]