The equation of the line of the shortest dist | vedclass

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(B) Let the two lines be $L_1: \frac{x}{1} = \frac{y}{-1} = \frac{z}{1} = r_1$ and $L_2: \frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1} = r_2$. Any

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$\frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z}{-2}$

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(B) Let the two lines be $L_1: \frac{x}{1} = \frac{y}{-1} = \frac{z}{1} = r_1$ and $L_2: \frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1} = r_2$. Any point on $L_1$ is $P(r_1, -r_1, r_1)$ and any point on $L_2$ is $Q(1, -2r_2 - 1, r_2)$. The direction ratios of the line $PQ$ are $(1 - r_1, -2r_2 - 1 + r_1, r_2 - r_1)$. Since $PQ$ is the shortest distance line,it is perpendicular to both $L_1$ (direction $\vec{v_1} = \langle 1, -1, 1 \rangle$) and $L_2$ (direction $\vec{v_2} = \langle 0, -2, 1 \rangle$). $1$) $(1 - r_1)(1) + (-2r_2 - 1 + r_1)(-1) + (r_2 - r_1)(1) = 0 \Rightarrow 1 - r_1 + 2r_2 + 1 - r_1 + r_2 - r_1 = 0 \Rightarrow -3r_1 + 3r_2 + 2 = 0$. $2$) $(1 - r_1)(0) + (-2r_2 - 1 + r_1)(-2) + (r_2 - r_1)(1) = 0 \Rightarrow 4r_2 + 2 - 2r_1 + r_2 - r_1 = 0 \Rightarrow -3r_1 + 5r_2 + 2 = 0$. Subtracting the equations: $2r_2 = 0 \Rightarrow r_2 = 0$. Then $-3r_1 + 2 = 0 \Rightarrow r_1 = 2/3$. The points are $P(2/3, -2/3, 2/3)$ and $Q(1, -1, 0)$. The direction ratios of $PQ$ are $(1 - 2/3, -1 + 2/3, 0 - 2/3) = (1/3, -1/3, -2/3)$,which is proportional to $(1, -1, -2)$. The line passes through $Q(1, -1, 0)$,so the equation is $\frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z}{-2}$.

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

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