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(C) The direction ratios (d.r.s.) of the line joining the points $P(2,1,-3)$ and $Q(-3,1,7)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (-3-2, 1-1, 7

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The direction ratios (d.r.s.) of the line joining the points $P(2,1,-3)$ and $Q(-3,1,7)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (-3-2, 1-1, 7-(-3)) = (-5, 0, 10)$.The direction ratios of the line parallel to $\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$ are $(3, 4, 5)$.Let $	heta$ be the acute angle between these two lines. The formula for the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is $\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$.Substituting the values:$\cos 	heta = \left| \frac{(-5)(3) + (0)(4) + (10)(5)}{\sqrt{(-5)^2 + 0^2 + 10^2} \sqrt{3^2 + 4^2 + 5^2}} ight|$$\cos 	heta = \left| \frac{-15 + 0 + 50}{\sqrt{25 + 100} \sqrt{9 + 16 + 25}} ight|$$\cos 	heta = \frac{35}{\sqrt{125} \sqrt{50}} = \frac{35}{5\sqrt{5} \cdot 5\sqrt{2}} = \frac{35}{25\sqrt{10}} = \frac{7}{5\sqrt{10}}$.Therefore,$	heta = \cos ^{-1}\left(\frac{7}{5 \sqrt{10}}ight)$.

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

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