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(B) Let $\vec{b}_{1}$ and $\vec{b}_{2}$ be the vectors parallel to the given pair of lines.The direction ratios of the first line are $(2, 2, 1)$,so $

Vedclass · Accepted Answer

$	heta = \cos^{-1}\left(\frac{2}{3}ight)$

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(B) Let $\vec{b}_{1}$ and $\vec{b}_{2}$ be the vectors parallel to the given pair of lines.The direction ratios of the first line are $(2, 2, 1)$,so $\vec{b}_{1} = 2\hat{i} + 2\hat{j} + \hat{k}$.The direction ratios of the second line are $(4, 1, 8)$,so $\vec{b}_{2} = 4\hat{i} + \hat{j} + 8\hat{k}$.Calculate the magnitudes:$|\vec{b}_{1}| = \sqrt{2^{2} + 2^{2} + 1^{2}} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.$|\vec{b}_{2}| = \sqrt{4^{2} + 1^{2} + 8^{2}} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$.Calculate the dot product:$\vec{b}_{1} \cdot \vec{b}_{2} = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18$.If $	heta$ is the angle between the lines,then $\cos 	heta = \frac{|\vec{b}_{1} \cdot \vec{b}_{2}|}{|\vec{b}_{1}| |\vec{b}_{2}|}$.$\cos 	heta = \frac{18}{3 	imes 9} = \frac{18}{27} = \frac{2}{3}$.Therefore,$	heta = \cos^{-1}\left(\frac{2}{3}ight)$.

Find the angle between the following pair of lines:
$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$

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Find the angle between the following pair of lines:$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$