Find the angle between the lines frac{x-2}{3} | vedclass

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(C) The given equations are not in standard form. We rewrite them as:Line $1$: $\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z-2}{0}$. The direction vector

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$\pi / 2$

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(C) The given equations are not in standard form. We rewrite them as:Line $1$: $\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z-2}{0}$. The direction vector is $\vec{b_1} = 3\hat{i} - 2\hat{j} + 0\hat{k}$.Line $2$: $\frac{x-1}{1} = \frac{y+3/2}{3/2} = \frac{z+5}{2}$. The direction vector is $\vec{b_2} = 1\hat{i} + \frac{3}{2}\hat{j} + 2\hat{k}$.Let $	heta$ be the angle between the lines. Then $\cos 	heta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.$\vec{b_1} \cdot \vec{b_2} = (3)(1) + (-2)(3/2) + (0)(2) = 3 - 3 + 0 = 0$.Since the dot product is $0$,the lines are perpendicular.Therefore,$	heta = \pi / 2$.

Find the angle between the lines $\frac{x-2}{3} = \frac{y+1}{-2}, z=2$ and $\frac{x-1}{1} = \frac{2y+3}{3} = \frac{z+5}{2}$.

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