The angle between the line joining points $(2, 1, - 3)$ , $(- 3, 1, 7)$ and the line passing through $(- 1, 0, 4)$ and parallel to the line $\frac{x - 1}{3} = \frac{y}{4} = \frac{z + 3}{5}$ is

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$\cos^{- 1} (\frac{1}{\sqrt{10}})$

$\cos^{- 1} (\frac{1}{5 \sqrt{10}})$

$\cos^{- 1} (\frac{7}{5 \sqrt{10}})$

$\cos^{- 1} (\frac{3}{5 \sqrt{10}})$

answer is A.

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Detailed Solution

d.r’s of $(2, 1, - 3) & (- 3, 1, 7)$ are $(- 5, 0, 10)$
d.r’s of line $(3, 4, 5)$
$\cos θ = \frac{| a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} |}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} = \frac{| - 15 + 0 + 50 |}{\sqrt{25 + 100} \sqrt{9 + 16 + 25}} = \frac{35}{\sqrt{125} . \sqrt{50}} \cos θ = \frac{7}{5 \sqrt{10}} θ = \cos^{- 1} (\frac{7}{5 \sqrt{10}})$