The angle between the lines 3x = 2y = -z and | vedclass

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(C) First,we write the equations of the lines in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line $3x = 2y = -z

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$\frac{\pi}{2}$

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(C) First,we write the equations of the lines in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line $3x = 2y = -z$,we divide by $6$ to get $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$. Thus,the direction ratios are $\vec{v_1} = (2, 3, -6)$.For the second line $-x = 6y = -4z$,we divide by $-12$ to get $\frac{x}{12} = \frac{y}{-2} = \frac{z}{3}$. Thus,the direction ratios are $\vec{v_2} = (12, -2, 3)$.The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.$\vec{v_1} \cdot \vec{v_2} = (2)(12) + (3)(-2) + (-6)(3) = 24 - 6 - 18 = 0$.Since the dot product is $0$,the lines are perpendicular,so $	heta = \frac{\pi}{2}$.

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

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