$O A, O B, O C$ are the sides of a rectangular parallelopiped(where O is the origin) whose diagonals are $O O^{'}, A A^{'}, B B^{'}$ and $C C^{'} . D$ is the centre of rectangle $A C^{'} O^{'} B^{'}$ and $D^{'}$ is centre of rectangle $O^{'} A^{'} C B^{'}$ . If sides $O A, O B, O C$ are in ratio 1 : 2 : 3 and $∠ D O D^{'} = α$ and ${cos}^{2} α = \frac{a}{b}$ , (where $a$ and $b$ are coprime numbers) then the value of $a + b$ is

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Solution

Now it is given that $O A, O B, O C$ are in ratio 1 : 2 : 3
Let us consider that
$| - - \to O A | = x$
then $| - - \to O B | = 2 x$
and $| - - \to O C | = 3 x$
Now $- - \to O D$ will be the resultant of sum of vectors $- - \to O A + - - \to A D$
(where $- - \to A D = - - \to A E + - - \to E D = \frac{2 x}{2} + \frac{3 x}{2})$
$- - \to O D = - - \to O A + - - \to A D$
$- - \to O D = - - \to O A + - - \to A E + - - \to E D$

$= x^i + x^j + \frac{3 x}{2}^k$
Similarly $- - \to O D^{'}$ will be the resultant of sum of vectors $- - \to O C + - - \to C D^{'}$

(where $- - \to C D^{'} = - - \to C F + - - \to F D^{'} = \frac{x}{2} + \frac{2 x}{2})$
$- - \to O D^{'} = - - \to O C + - - \to C D^{'}$
$- - \to O D^{'} = - - \to O C + - - \to C F + - - \to F D^{'}$
$= 3 x^k + \frac{x}{2}^i + x^j$

Now the angle between $- - \to O D$ and $- - \to O D^{'}$ can be calculated using dot product

$cos α = \frac{- - \to O D . - - \to O D^{^{'}}}{| - - \to O D | | - - \to O D^{^{'}} |} = \frac{(x^i + x^j + \frac{3 x}{2}^k) . (\frac{x}{2}^i + x^j + 3 x^k)}{\sqrt{x^{2} + x^{2} + (\frac{3 x}{2})^{2}} \sqrt{(\frac{x}{2})^{2} + x^{2} + (3 x)^{2}}}$
$cos α = \frac{6}{\frac{\sqrt{17 \times 41}}{4}}$

$c o s^{2} α = \frac{576}{697} = \frac{a}{b}$
$a + b = 1273$

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