The shortest distance between the lines frac{ | vedclass

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(D) The first line is $L_1: \frac{x-1}{0} = \frac{y+1}{-1} = \frac{z}{1}$. $A$ point on $L_1$ is $A(1, -1, 0)$ and its direction vector is $\vec{c} =

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$\frac{1}{\sqrt{3}}$

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(D) The first line is $L_1: \frac{x-1}{0} = \frac{y+1}{-1} = \frac{z}{1}$. $A$ point on $L_1$ is $A(1, -1, 0)$ and its direction vector is $\vec{c} = (0, -1, 1)$.The second line $L_2$ is the intersection of planes $x+y+z+1=0$ and $2x-y+z+3=0$.Adding the two equations: $(x+y+z+1) + (2x-y+z+3) = 3x + 2z + 4 = 0$,so $x = \frac{-2z-4}{3}$.Substituting $x$ into the first plane: $\frac{-2z-4}{3} + y + z + 1 = 0 \Rightarrow y = -z - 1 + \frac{2z+4}{3} = \frac{-3z-3+2z+4}{3} = \frac{-z+1}{3}$.Thus,$x = \frac{-2z-4}{3}, y = \frac{-z+1}{3}, z = z$. Rewriting in symmetric form: $\frac{x+4/3}{-2/3} = \frac{y-1/3}{-1/3} = \frac{z}{1}$.$A$ point on $L_2$ is $B(-\frac{4}{3}, \frac{1}{3}, 0)$ and its direction vector is $\vec{d} = (-\frac{2}{3}, -\frac{1}{3}, 1)$.Vector $\vec{AB} = B - A = (-\frac{4}{3}-1, \frac{1}{3}-(-1), 0-0) = (-\frac{7}{3}, \frac{4}{3}, 0)$.Cross product $\vec{c} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -1 & 1 \ -2/3 & -1/3 & 1 \end{vmatrix} = \hat{i}(-1 + 1/3) - \hat{j}(0 + 2/3) + \hat{k}(0 - 2/3) = (-\frac{2}{3}, -\frac{2}{3}, -\frac{2}{3})$.Magnitude $|\vec{c} 	imes \vec{d}| = \sqrt{(-\frac{2}{3})^2 + (-\frac{2}{3})^2 + (-\frac{2}{3})^2} = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{12}{9}} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.Shortest distance $d = \frac{|\vec{AB} \cdot (\vec{c} 	imes \vec{d})|}{|\vec{c} 	imes \vec{d}|} = \frac{|(-\frac{7}{3})(-\frac{2}{3}) + (\frac{4}{3})(-\frac{2}{3}) + (0)(-\frac{2}{3})|}{2/\sqrt{3}} = \frac{|\frac{14}{9} - \frac{8}{9}|}{2/\sqrt{3}} = \frac{6/9}{2/\sqrt{3}} = \frac{2/3}{2/\sqrt{3}} = \frac{1}{\sqrt{3}}$.

The shortest distance between the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z+1=0, 2x-y+z+3=0$ is

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