The angle between two lines frac{x + 1}{2} = | vedclass

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(D) The direction ratios of the first line are $\vec{a_1} = (2, 2, -1)$.The direction ratios of the second line are $\vec{a_2} = (1, 2, 2)$.The angle

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{4}{9}\right)$

Vedclass · Answer

(D) The direction ratios of the first line are $\vec{a_1} = (2, 2, -1)$.The direction ratios of the second line are $\vec{a_2} = (1, 2, 2)$.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,we get $\cos 	heta = \frac{|(2)(1) + (2)(2) + (-1)(2)|}{\sqrt{2^2 + 2^2 + (-1)^2} \sqrt{1^2 + 2^2 + 2^2}}$.$\cos 	heta = \frac{|2 + 4 - 2|}{\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4}} = \frac{4}{\sqrt{9} \sqrt{9}} = \frac{4}{3 	imes 3} = \frac{4}{9}$.Therefore,$	heta = \cos^{-1}\left(\frac{4}{9}ight)$.

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

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