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Solution for The Angle Between the Two Diagonals of a Cube is (A) 30° , (B) 45° ,(C) Cos − 1 ( 1 √ 3 ), (D) Cos − 1 ( 1 3 ). - CBSE (Commerce) Class 12 - Mathematics

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Question

The angle between the two diagonals of a cube is


 

 

  • (a) 30°

  • (b) 45°

  • (c) \[\cos^{- 1} \left( \frac{1}{\sqrt{3}} \right)\]

  • (d) \[\cos^{- 1} \left( \frac{1}{3} \right)\]

Solution

(d) \[\cos^{- 1} \left( \frac{1}{3} \right)\]

\[\text { Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure . Clearly, OP, AR, BS, and CQ are the diagonals of the cube } . \]

\[\text{ Consider the diagonals OP and AR } . \]

\[\text{ Direction ratios of OP and AR are proportional to a - 0, a - 0, a - 0 and 0 - a, a - 0, a - 0, i . e . a, a, a and - a, a, a, respectively } . \]

\[\text { Let }  \theta \text{ be the angle between OP and AR . Then,} \]

\[\cos \theta = \frac{a \times - a + a \times a + a \times a}{\sqrt{a^2 + a^2 + a^2}\sqrt{\left( - a \right)^2 + a^2 + a^2}}\]

\[ \Rightarrow \cos \theta = \frac{- a^2 + a^2 + a^2}{\sqrt{3 a^2}\sqrt{3 a^2}}\]

\[ \Rightarrow \cos \theta = \frac{1}{3} \]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{3} \right) \]

\[\text{ Similarly, the angles between other pairs of the diagonals are equal to } \cos^{- 1} \left( \frac{1}{3} \right) \text{ as the angle between any two diagonals of a cube is } \cos^{- 1} \left( \frac{1}{3} \right) .\]

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Solution The Angle Between the Two Diagonals of a Cube is (A) 30° , (B) 45° ,(C) Cos − 1 ( 1 √ 3 ), (D) Cos − 1 ( 1 3 ). Concept: Direction Cosines and Direction Ratios of a Line.
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