The acute angle theta between the lines 2x = | vedclass

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(D) Step $1$: Rewrite the lines in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line $2x = 3y = -z$, divide by $

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$\frac{\pi}{2}$

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(D) Step $1$: Rewrite the lines in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line $2x = 3y = -z$, divide by $6$: $\frac{x}{3} = \frac{y}{2} = \frac{z}{-6}$. The direction vector is $\vec{v_1} = 3\hat{i} + 2\hat{j} - 6\hat{k}$.Step $2$: For the second line $6x = -y = -4z$, divide by $12$: $\frac{x}{2} = \frac{y}{-12} = \frac{z}{-3}$. The direction vector is $\vec{v_2} = 2\hat{i} - 12\hat{j} - 3\hat{k}$.Step $3$: Use the formula $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.$\vec{v_1} \cdot \vec{v_2} = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$.Step $4$: Since the dot product is $0$, the lines are perpendicular, so $	heta = \frac{\pi}{2}$.

The acute angle $\theta$ between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is:

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