Question

The lines $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{-1}$, $\frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{4}$ are:

• A. parallel lines
• B. intersecting lines
• C. perpendicular skew lines
• D. Coincident

Answer 1

Answer: (C)

perpendicular skew lines

Answer 2

Let the first line be $L_1$ with direction vector $\vec{d_1} = \langle 1, 2, -1 \rangle$ and a point $P_1 = (-1, 1, 2)$. Let the second line be $L_2$ with direction vector $\vec{d_2} = \langle 2, 1, 4 \rangle$ and a point $P_2 = (1, 0, -1)$.

1. Check for Parallelism: The direction vectors $\vec{d_1}$ and $\vec{d_2}$ are not proportional ($\frac{1}{2} \neq \frac{2}{1} \neq \frac{-1}{4}$), so the lines are not parallel.

2. Check for Perpendicularity: The dot product of the direction vectors is $\vec{d_1} \cdot \vec{d_2} = (1)(2) + (2)(1) + (-1)(4) = 2 + 2 - 4 = 0$. Since the dot product is zero, the lines are perpendicular.

3. Check for Intersection: To check if the lines intersect, we determine if they are coplanar. This is done by checking if the scalar triple product of the vector connecting a point on each line ($\vec{P_1P_2}$) and the direction vectors is zero. $\vec{P_1P_2} = P_2 - P_1 = \langle 1 - (-1), 0 - 1, -1 - 2 \rangle = \langle 2, -1, -3 \rangle$. The scalar triple product is: $\begin{vmatrix} 2 & -1 & -3 \\ 1 & 2 & -1 \\ 2 & 1 & 4 \end{vmatrix} = 2(8 - (-1)) - (-1)(4 - (-2)) + (-3)(1 - 4) = 2(9) + 1(6) - 3(-3) = 18 + 6 + 9 = 33$ Since the scalar triple product is $33 \neq 0$, the lines are not coplanar, and therefore do not intersect.

Since the lines are perpendicular and do not intersect, they are perpendicular skew lines.