If theta is the angle between the lines frac{ | vedclass

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(B) Step $1$: Rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x-1

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$	heta = \frac{\pi}{3}$

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(B) Step $1$: Rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x-1}{2} = \frac{y - (-3/2)}{2} = \frac{z - (-2)}{0}$. The direction vector is $\vec{v_1} = 2\hat{i} + 2\hat{j} + 0\hat{k}$.Step $2$: For the second line: $x=1$ implies the direction ratio for $x$ is $0$. Thus, $\frac{x-1}{0} = \frac{y-1}{2} = \frac{z+1}{2}$. The direction vector is $\vec{v_2} = 0\hat{i} + 2\hat{j} + 2\hat{k}$.Step $3$: The cosine of the angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.Step $4$: Calculate the dot product: $\vec{v_1} \cdot \vec{v_2} = (2)(0) + (2)(2) + (0)(2) = 4$.Step $5$: Calculate the magnitudes: $|\vec{v_1}| = \sqrt{2^2 + 2^2 + 0^2} = \sqrt{8} = 2\sqrt{2}$ and $|\vec{v_2}| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.Step $6$: $\cos 	heta = \frac{4}{(2\sqrt{2})(2\sqrt{2})} = \frac{4}{8} = \frac{1}{2}$.Step $7$: Therefore, $	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

If $\theta$ is the angle between the lines $\frac{x - 1}{2} = \frac{2y + 3}{4}; z = -2$ and $x = 1; \frac{y - 1}{2} = \frac{z + 1}{2}$, then

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