First slide

Direction ratios and direction cosines

Question

The lengths of the sides of a rectangular parallelepiped are 11,12,13. Then the angle between two diagonals out of four diagonals is

Moderate

A

${Cos}^{- 1} (\frac{96}{\sqrt{232}})$
B

${Cos}^{- 1} (\frac{96}{\sqrt{434}})$
C

${Cos}^{- 1} (\frac{96}{217})$
D

${Cos}^{- 1} (\frac{192}{4 \sqrt{434}})$

Solution

If $\bar{O A}, \bar{O B}, \bar{O C}$ are considered as coordinate axes respectively such that $O A = 13, O B = 12, O C = 11$ ,
Direction ratios of two diagonals $\bar{A F}, \bar{B E}$ are $〈13, - 12, - 11〉 and 〈13, - 12, 11〉$ respectively.
If $θ$ be the angle between the above two diagonals then
$\begin{matrix} \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} \\ = \frac{169 + 144 - 121}{{(\sqrt{169 + 144 + 121})}^{2}} \\ = \frac{192}{434} \\ = \frac{96}{217} \end{matrix}$
Therefore, $θ = \arccos (\frac{96}{217})$

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