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

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(A) Let the direction ratios of the two lines be $\vec{b_1} = (1, 2, 2)$ and $\vec{b_2} = (2, 2, -1)$.The formula for the angle $	heta$ between two l

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the direction ratios of the two lines be $\vec{b_1} = (1, 2, 2)$ and $\vec{b_2} = (2, 2, -1)$.The formula for 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:$\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$Substituting the values:$\cos 	heta = \left| \frac{1(2) + 2(2) + 2(-1)}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + 2^2 + (-1)^2}} ight|$$\cos 	heta = \left| \frac{2 + 4 - 2}{\sqrt{1 + 4 + 4} \sqrt{4 + 4 + 1}} ight|$$\cos 	heta = \left| \frac{4}{\sqrt{9} \sqrt{9}} ight| = \frac{4}{3 	imes 3} = \frac{4}{9}$Therefore,$	heta = \cos ^{-1}\left(\frac{4}{9}ight)$.

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

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