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 \[-a,a,-a.\] \[\therefore \] \[\cos \theta =\frac{|-{{a}^{2}}+{{a}^{2}}-{{a}^{2}}|}{\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}}\] \[=\frac{{{a}^{2}}}{\sqrt{3}a\sqrt{3}a}\] \[=\frac{{{a}^{2}}}{3{{a}^{2}}}=\frac{1}{3}\] \[\Rightarrow \] \[\cos \theta =\frac{1}{3}\]You need to login to perform this action.
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