Question

Find the shortest distance between the lines: \[ \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}), \] \[ \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}). \]

Hint:

For shortest distance, check for skew or parallel lines. Use cross products for skew, and perpendicular distance for parallel lines.

Answer 1

The shortest distance \( d \) between two skew lines is given by: \[ d=\frac{|(\vec{r_1}-\vec{r_2})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}, \] where \( \vec{r_1}=3\hat{i}+3\hat{j}-5\hat{k}, \vec{r_2}=\hat{i}+2\hat{j}-4\hat{k} \), and \( \vec{d_1}=2\hat{i}+3\hat{j}+6\hat{k}, \vec{d_2}=2\hat{i}+3\hat{j}+6\hat{k} \). 1. Compute \( \vec{r_1}-\vec{r_2}=2\hat{i}+\hat{j}-\hat{k} \). 2. Compute \( \vec{d_1}\times\vec{d_2}=\vec{0} \) since \( \vec{d_1} \) and \( \vec{d_2} \) are parallel. Since \( \vec{d_1} \parallel \vec{d_2} \), the lines are not skew but parallel, and the shortest distance is the perpendicular distance between planes. The distance is \( d=0 \).