The angle between the lines x-3y-4=0, 4y-z+5= | vedclass

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(A) To find the direction ratios of the lines,we express them in symmetric form. For the first line $L_1$: $x-3y-4=0$ and $4y-z+5=0$. Let $y=t$. Then

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$\frac{\pi}{2}$

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(A) To find the direction ratios of the lines,we express them in symmetric form. For the first line $L_1$: $x-3y-4=0$ and $4y-z+5=0$. Let $y=t$. Then $x=3t+4$ and $z=4t+5$. The line is $\frac{x-4}{3} = \frac{y-0}{1} = \frac{z-5}{4}$. Direction vector $\vec{v_1} = (3, 1, 4)$. For the second line $L_2$: $x+3y-11=0$ and $2y-z+6=0$. Let $y=s$. Then $x=-3s+11$ and $z=2s+6$. The line is $\frac{x-11}{-3} = \frac{y-0}{1} = \frac{z-6}{2}$. Direction vector $\vec{v_2} = (-3, 1, 2)$. The cosine of the angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$. $\vec{v_1} \cdot \vec{v_2} = (3)(-3) + (1)(1) + (4)(2) = -9 + 1 + 8 = 0$. Since the dot product is $0$,the lines are perpendicular,so $	heta = \frac{\pi}{2}$.

The angle between the lines $x-3y-4=0, 4y-z+5=0$ and $x+3y-11=0, 2y-z+6=0$ is

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