Question

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

Hint:

For skew lines, the shortest distance is given by: \[ d = \frac{| (\mathbf{B} - \mathbf{A}) \cdot (\mathbf{d_1} \times \mathbf{d_2}) |}{|\mathbf{d_1} \times \mathbf{d_2}|}. \]

Answer 1

We are given two skew lines, whose equations are: \[ \mathbf{r}_1 = (4\hat{i} - \hat{j}) + \lambda( \hat{i} + 2\hat{j} - 3\hat{k}), \] \[ \mathbf{r}_2 = ( \hat{i} - \hat{j} - 2\hat{k}) + \mu ( \hat{i} + \hat{j} - 5\hat{k}). \] 
Step 1: Identify points and direction vectors. 
The lines pass through the points: \[ \mathbf{a}_1 = (4\hat{i} - \hat{j}), \quad \mathbf{a}_2 = ( \hat{i} - \hat{j} - 2\hat{k}), \] and are parallel to the direction vectors: \[ \mathbf{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}, \quad \mathbf{b}_2 = \hat{i} + \hat{j} - 5\hat{k}. \] 
Step 2: Find the vector between the points. 
The vector between the two points \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \) is: \[ \mathbf{a}_2 - \mathbf{a}_1 = (\hat{i} - \hat{j} - 2\hat{k}) - (4\hat{i} - \hat{j}) = -3\hat{i} + 2\hat{k}. \] 
Step 3: Calculate the cross product of the direction vectors. 
Now, calculate the cross product of the direction vectors \( \mathbf{b}_1 \) and \( \mathbf{b}_2 \): 

Step 4: Apply the shortest distance formula. 
The formula for the shortest distance \( d \) between two skew lines is given by: \[ d = \frac{| (\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) |}{|\mathbf{b}_1 \times \mathbf{b}_2|} \] Substituting the values: \[ d = \frac{| (-3\hat{i} + 2\hat{k}) \cdot 2\hat{k} |}{| 2\hat{k} |} = \frac{| -6 + 4 |}{2\sqrt{3}} = \frac{2}{2\sqrt{3}}. \] 
Final Answer: Thus, the shortest distance between the two lines is: \[ d = \frac{1}{\sqrt{3}} { units}. \]