The angle between the lines frac{x-3}{1}=frac | vedclass

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(D) The direction ratios of the first line are $\vec{b_1} = (1, 2, 2)$.The direction ratios of the second line are $\vec{b_2} = (3, 2, 6)$.The angle $

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{19}{21}\right)$

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(D) The direction ratios of the first line are $\vec{b_1} = (1, 2, 2)$.The direction ratios of the second line are $\vec{b_2} = (3, 2, 6)$.The angle $	heta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by $\cos 	heta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.Thus,$\cos 	heta = \frac{19}{3 	imes 7} = \frac{19}{21}$.Therefore,$	heta = \cos^{-1}\left(\frac{19}{21}ight)$.

The angle between the lines $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z+4}{2}$ and $\frac{x-5}{3}=\frac{y+2}{2}=\frac{z}{6}$ is . . . . . . .

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