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If the position vectors of A, B and C are respectively $2 i - j + k, i - 3 j - 5 k$ and $3 i - 4 j - 4 k,$ then $\cos^{2} A =$

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$\frac{6}{41}$

$\frac{35}{41}$

answer is C.

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Detailed Solution

$AB = - i - 2 j - 6 k AC = i - 3 j - 5 k \cos A = \frac{AB . AC}{|AB| |AC|} = \frac{- 1 + 6 + 30}{\sqrt{1 + 4 + 36} \sqrt{1 + 9 + 25}} = \frac{35}{\sqrt{41} \sqrt{35}} = \sqrt{\frac{35}{41}}$

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