If $a, b, c$ are the lengths of the sides of a rectangular parallelopiped then the angle between two diagonals is

{cos}^{- 1} (\frac{a^{2} + b^{2} + c^{2}}{a + b - c})

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{cos}^{- 1} (\frac{a + b - c}{a^{2} + b^{2} + c^{2}})

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{cos}^{- 1} (\frac{a^{2} - b^{2} - c^{2}}{a^{2} + b^{2} - c^{2}})

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{cos}^{- 1} (\frac{a^{2} + b^{2} - c^{2}}{a^{2} + b^{2} + c^{2}})

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Solution

The correct option is D ${cos}^{- 1} (\frac{a^{2} + b^{2} - c^{2}}{a^{2} + b^{2} + c^{2}})$
Let $O A B C D E F G$ be the rectangular parallelopiped. Let $O A = a, O B = b, O C = c$ be the length of the sides along the $x, y$ and $z$ -axis.
Now, the coordinates of the vertices are $O (0, 0, 0), B (a, 0, 0), C (0, b, 0), A (0, 0, c), D (a, b, 0), E (0, b, c), F (a, 0, c), G (a, b, c)$
Diagonals are $A D, O G, B E, F C .$
Let $θ$ be the angle between OG and AD
$\to O G = a^i + b^j + c^k$
$\to A D = a^i + b^j - c^k$
Now, $cos θ = \frac{\to O G . \to A D}{| \to O G | | \to A D |}$
$\Rightarrow cos θ = \frac{(a^i + b^j + c^k) . (a^i + b^j - c^k)}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{a^{2} + b^{2} + c^{2}}}$
$\Rightarrow cos θ = \frac{a^{2} + b^{2} - c^{2}}{a^{2} + b^{2} + c^{2}}$
$θ = {cos}^{- 1} (\frac{a^{2} + b^{2} - c^{2}}{a^{2} + b^{2} + c^{2}})$

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