The angle between the lines whose direction r | vedclass

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Question

(A) Let the direction ratios of the two lines be $\vec{a} = (1, 2, 1)$ and $\vec{b} = (2, -3, 6)$.The angle $	heta$ between two lines with direction

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{2}{7\sqrt{6}}\right)$

Vedclass · Answer

(A) Let the direction ratios of the two lines be $\vec{a} = (1, 2, 1)$ and $\vec{b} = (2, -3, 6)$.The angle $	heta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the formula:$\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$Substituting the given values:$\cos 	heta = \frac{|(1)(2) + (2)(-3) + (1)(6)|}{\sqrt{1^2 + 2^2 + 1^2} \sqrt{2^2 + (-3)^2 + 6^2}}$$\cos 	heta = \frac{|2 - 6 + 6|}{\sqrt{1 + 4 + 1} \sqrt{4 + 9 + 36}}$$\cos 	heta = \frac{2}{\sqrt{6} \sqrt{49}}$$\cos 	heta = \frac{2}{7\sqrt{6}}$Therefore,$	heta = \cos^{-1}\left(\frac{2}{7\sqrt{6}}ight)$.

The angle between the lines whose direction ratios are proportional to $(1, 2, 1)$ and $(2, -3, 6)$ is

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