question_answer

The cosine of the angle between any two diagonals of a cube is:

A) \[\frac{1}{3}\]                      

B)         \[\frac{1}{2}\]                     

C)  \[\frac{2}{3}\]                     

D)        \[\frac{1}{\sqrt{3}}\]

Correct Answer: A

Solution :

Let the sides of a cube be a. \[\therefore \] Coordinate of vertices are \[O(0,0,0),D(a,a,a),B(0,a,0)\]and\[G(a,0,a)\] The direction ratios of the diagonals OD and BG are \[a-0,\text{ }a-0,a-0\]and \[0-a,a,0-a\] or \[a,a,a\]and \[-a,a,-a.\] \[\therefore \]      \[\cos \theta =\frac{|-{{a}^{2}}+{{a}^{2}}-{{a}^{2}}|}{\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}}\]                 \[=\frac{{{a}^{2}}}{\sqrt{3}a\sqrt{3}a}\]                 \[=\frac{{{a}^{2}}}{3{{a}^{2}}}=\frac{1}{3}\] \[\Rightarrow \]    \[\cos \theta =\frac{1}{3}\]