The angle between the lines frac{x + 4}{1} = | vedclass

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(C) The direction ratios of the first line are $\vec{b_1} = (1, 2, 3)$.The direction ratios of the second line are $\vec{b_2} = (3, -2, 1)$.The angle

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The direction ratios of the first line are $\vec{b_1} = (1, 2, 3)$.The direction ratios of the second line are $\vec{b_2} = (3, -2, 1)$.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:$\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$Substituting the values:$\cos 	heta = \frac{|(1)(3) + (2)(-2) + (3)(1)|}{\sqrt{1^2 + 2^2 + 3^2} \sqrt{3^2 + (-2)^2 + 1^2}}$$\cos 	heta = \frac{|3 - 4 + 3|}{\sqrt{1 + 4 + 9} \sqrt{9 + 4 + 1}}$$\cos 	heta = \frac{2}{\sqrt{14} \sqrt{14}} = \frac{2}{14} = \frac{1}{7}$Therefore,$	heta = \cos^{-1}\left(\frac{1}{7}ight)$.

The angle between the lines $\frac{x + 4}{1} = \frac{y - 3}{2} = \frac{z + 2}{3}$ and $\frac{x}{3} = \frac{y - 1}{-2} = \frac{z}{1}$ is

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