Find the acute angle between the line joining | vedclass

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Question

(A) Let the direction ratios of the line joining the points $(2, 1, -3)$ and $(-3, 1, 7)$ be $(a_1, b_1, c_1)$.$a_1 = -3 - 2 = -5$,$b_1 = 1 - 1 = 0$,$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the direction ratios of the line joining the points $(2, 1, -3)$ and $(-3, 1, 7)$ be $(a_1, b_1, c_1)$.$a_1 = -3 - 2 = -5$,$b_1 = 1 - 1 = 0$,$c_1 = 7 - (-3) = 10$.So,the direction vector is $\vec{v_1} = (-5, 0, 10)$.The second line is parallel to the line $\frac{x - 1}{3} = \frac{y}{4} = \frac{z + 3}{5}$,so its direction ratios $(a_2, b_2, c_2)$ are $(3, 4, 5)$.So,the direction vector is $\vec{v_2} = (3, 4, 5)$.The angle $	heta$ between the two lines is given by $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.$\cos 	heta = \frac{|(-5)(3) + (0)(4) + (10)(5)|}{\sqrt{(-5)^2 + 0^2 + 10^2} \sqrt{3^2 + 4^2 + 5^2}}$.$\cos 	heta = \frac{|-15 + 0 + 50|}{\sqrt{25 + 0 + 100} \sqrt{9 + 16 + 25}}$.$\cos 	heta = \frac{35}{\sqrt{125} \sqrt{50}} = \frac{35}{5\sqrt{5} \cdot 5\sqrt{2}} = \frac{35}{25\sqrt{10}} = \frac{7}{5\sqrt{10}}$.Therefore,$	heta = \cos^{-1}\left(\frac{7}{5\sqrt{10}}ight)$.

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

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