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

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(B) The direction ratios of the first line are $\vec{b_1} = (4, 1, 8)$.The direction ratios of the second line are $\vec{b_2} = (2, 2, 1)$.The angle $

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{2}{3}\right)$

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(B) The direction ratios of the first line are $\vec{b_1} = (4, 1, 8)$.The direction ratios of the second line are $\vec{b_2} = (2, 2, 1)$.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 $\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$.Substituting the values:$\cos 	heta = \left| \frac{4(2) + 1(2) + 8(1)}{\sqrt{4^2 + 1^2 + 8^2} \sqrt{2^2 + 2^2 + 1^2}} ight|$$\cos 	heta = \left| \frac{8 + 2 + 8}{\sqrt{16 + 1 + 64} \sqrt{4 + 4 + 1}} ight|$$\cos 	heta = \left| \frac{18}{\sqrt{81} \cdot \sqrt{9}} ight| = \frac{18}{9 \cdot 3} = \frac{18}{27} = \frac{2}{3}$.Therefore,$	heta = \cos^{-1}\left(\frac{2}{3}ight)$.

The angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ is

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