The lines frac{x - 1}{-1} = frac{y + 2}{1} = | vedclass

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(D) Step $1$: Identify the direction vectors of the two lines. The direction vector of the first line is $\vec{b_1} = -1\hat{i} + 1\hat{j} - 2\hat{k}$

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skew lines

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(D) Step $1$: Identify the direction vectors of the two lines. The direction vector of the first line is $\vec{b_1} = -1\hat{i} + 1\hat{j} - 2\hat{k}$ and the direction vector of the second line is $\vec{b_2} = 1\hat{i} + 1\hat{j} - 2\hat{k}$.Step $2$: Check for perpendicularity by calculating the dot product $\vec{b_1} \cdot \vec{b_2} = (-1)(1) + (1)(1) + (-2)(-2) = -1 + 1 + 4 = 4$. Since the dot product is not $0$, the lines are not perpendicular.Step $3$: Check for intersection. Both lines pass through the point $(1, -2, z)$. For the first line, at $x=1, y=-2$, we have $z=3$. For the second line, at $x=1, y=-2$, we have $z=-1$. Since the lines pass through the same point $(1, -2, z)$ only if $z$ values match, and they do not, we check if they intersect at any other point. Setting the parametric equations equal: $1-t_1 = 1+t_2 \implies t_1+t_2=0$ and $-2+t_1 = -2+t_2 \implies t_1=t_2$. Thus $t_1=t_2=0$. At $t_1=0, t_2=0$, the points are $(1, -2, 3)$ and $(1, -2, -1)$. These are distinct points. Since the lines share a common point $(1, -2)$ in the $xy$-plane but have different $z$-coordinates, they are skew lines.

The lines $\frac{x - 1}{-1} = \frac{y + 2}{1} = \frac{z - 3}{-2}$ and $\frac{x - 1}{1} = \frac{y + 2}{1} = \frac{z + 1}{-2}$ are

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