The angle between the two lines frac{x-2}{2} | vedclass

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(C) First,rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Line $1$: $\frac{x-2}{2} = \frac{y

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$\sin^{-1}\left(\sqrt{\frac{25}{238}}\right)$

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(C) First,rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Line $1$: $\frac{x-2}{2} = \frac{y-2}{-3} = \frac{z-1}{2}$. The direction vector is $\vec{v_1} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.Line $2$: $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-3}$. The direction vector is $\vec{v_2} = 2\hat{i} + 1\hat{j} - 3\hat{k}$.The cosine of the angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.$\vec{v_1} \cdot \vec{v_2} = (2)(2) + (-3)(1) + (2)(-3) = 4 - 3 - 6 = -5$.$|\vec{v_1}| = \sqrt{2^2 + (-3)^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$.$|\vec{v_2}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.$\cos 	heta = \frac{|-5|}{\sqrt{17} \sqrt{14}} = \frac{5}{\sqrt{238}}$.Using $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{25}{238} = \frac{213}{238}$.Thus,$\sin 	heta = \sqrt{\frac{213}{238}}$,which implies $	heta = \sin^{-1}\left(\sqrt{\frac{213}{238}}ight)$.

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

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