If the two lines l_{1}: frac{x-2}{3}=frac{y+1 | vedclass

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(B) The line $l_{1}$ is given by $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}$. The direction vector is $\vec{v_{1}} = (3, -2, 0)$.The line $l_{2}$ is

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$\sec^{-1}\left(\frac{29}{4}\right)$

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(B) The line $l_{1}$ is given by $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}$. The direction vector is $\vec{v_{1}} = (3, -2, 0)$.The line $l_{2}$ is given by $\frac{x-1}{1}=\frac{y+3/2}{\alpha/2}=\frac{z+5}{2}$. The direction vector is $\vec{v_{2}} = (1, \alpha/2, 2)$.Since $l_{1} \perp l_{2}$,their dot product is zero: $(3)(1) + (-2)(\alpha/2) + (0)(2) = 0$.$3 - \alpha = 0 \Rightarrow \alpha = 3$.The line $l_{3}$ is given by $\frac{x-1}{-3}=\frac{y-1/2}{-2}=\frac{z-0}{4}$. The direction vector is $\vec{v_{3}} = (-3, -2, 4)$.For $\alpha = 3$,the direction vector of $l_{2}$ is $\vec{v_{2}} = (1, 3/2, 2)$.The angle $	heta$ between $l_{2}$ and $l_{3}$ is given by $\cos 	heta = \frac{|\vec{v_{2}} \cdot \vec{v_{3}}|}{||\vec{v_{2}}|| ||\vec{v_{3}}||}$.$\vec{v_{2}} \cdot \vec{v_{3}} = (1)(-3) + (3/2)(-2) + (2)(4) = -3 - 3 + 8 = 2$.$||\vec{v_{2}}|| = \sqrt{1^{2} + (3/2)^{2} + 2^{2}} = \sqrt{1 + 9/4 + 4} = \sqrt{29/4} = \frac{\sqrt{29}}{2}$.$||\vec{v_{3}}|| = \sqrt{(-3)^{2} + (-2)^{2} + 4^{2}} = \sqrt{9 + 4 + 16} = \sqrt{29}$.$\cos 	heta = \frac{2}{(\sqrt{29}/2) 	imes \sqrt{29}} = \frac{2}{29/2} = \frac{4}{29}$.Therefore,$	heta = \cos^{-1}\left(\frac{4}{29}ight) = \sec^{-1}\left(\frac{29}{4}ight)$.

If the two lines $l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular,then the angle between the lines $l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is

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