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

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(D) Let $\vec{a}$ and $\vec{b}$ be the direction vectors of the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac

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

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(D) Let $\vec{a}$ and $\vec{b}$ be the direction vectors of the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ respectively.$\vec{a} = 4\hat{i} + \hat{j} + 8\hat{k}$ and $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$.The dot product is $\vec{a} \cdot \vec{b} = (4 	imes 2) + (1 	imes 2) + (8 	imes 1) = 8 + 2 + 8 = 18$.The magnitudes are $|\vec{a}| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$ and $|\vec{b}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Let $	heta$ be the angle between the lines. Then $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{18}{9 	imes 3} = \frac{18}{27} = \frac{2}{3}$.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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