Question

Find the shortest distance between the lines: \[ \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}, \quad \frac{x + 2}{1} = \frac{y - 3}{-2} = \frac{z + 1}{2} \]

Hint:

To find the shortest distance between skew lines, use: \[ D = \frac{|\vec{PQ} \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|} \] It uses cross and dot product of direction vectors and vector between lines.

Answer 1

Let lines be: \[ L_1: \vec{r}_1 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + \lambda \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}, \quad L_2: \vec{r}_2 = \begin{bmatrix} -2 \\ 3 \\ -1 \end{bmatrix} + \mu \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \] Direction vectors: \[ \vec{d}_1 = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}, \quad \vec{d}_2 = \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \] Vector between points: \[ \vec{PQ} = \vec{r}_2 - \vec{r}_1 = \begin{bmatrix} -3 \\ 4 \\ -1 \end{bmatrix} \] Shortest distance \( D = \frac{|\vec{PQ} \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|} \) Compute cross product: \[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -2 & 2 \end{vmatrix} \\ = \hat{i}(3 \cdot 2 - 4 \cdot (-2)) - \hat{j}(2 \cdot 2 - 4 \cdot 1) + \hat{k}(2 \cdot (-2) - 3 \cdot 1) \\ = \hat{i}(6 + 8) - \hat{j}(4 - 4) + \hat{k}(-4 - 3) \\ = \begin{bmatrix} 14 \\ 0 \\ -7 \end{bmatrix} \] \[ |\vec{d}_1 \times \vec{d}_2| = \sqrt{14^2 + 0 + 49} = \sqrt{245} \] Dot product: \[ \vec{PQ} \cdot (\vec{d}_1 \times \vec{d}_2) = (-3)(14) + (4)(0) + (-1)(-7) = -42 + 0 + 7 = -35 \] \[ D = \frac{| -35 |}{\sqrt{245}} = \frac{35}{\sqrt{245}} = \boxed{\frac{7}{\sqrt{5}}} \] Final Answer: \[ \boxed{\frac{7}{\sqrt{5}}} \]