Question

Find the shortest distance between the lines \[ \mathbf{r}_1 = (i + 2j + k) + \lambda(i - j + k) \quad \text{and} \quad \mathbf{r}_2 = (2i - j - k) + \mu(2i + j + 2k). \] 

Hint:

When finding the shortest distance between two skew lines, use the cross product and the formula involving the direction vectors and the vector between the lines.

Answer 1

Step 1: The shortest distance between two skew lines is given by the formula: \[ d = \frac{|(\mathbf{r}_2 - \mathbf{r}_1) \cdot (\mathbf{a}_1 \times \mathbf{a}_2)|}{|\mathbf{a}_1 \times \mathbf{a}_2|}. \] Here, \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \) are the direction vectors of the lines, and \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) are points on the lines. 

Step 2: The direction vectors are: \[ \mathbf{a}_1 = i - j + k, \quad \mathbf{a}_2 = 2i + j + 2k. \] 

Step 3: The vector from \( \mathbf{r}_1 \) to \( \mathbf{r}_2 \) is: \[ \mathbf{r}_2 - \mathbf{r}_1 = (2i - j - k) - (i + 2j + k) = i - 3j - 2k. \] 

Step 4: Find the cross product of \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \): \[ \mathbf{a}_1 \times \mathbf{a}_2 = \begin{vmatrix} i & j & k 
1 & -1 & 1 
2 & 1 & 2 \end{vmatrix} = (1 \cdot 2 - 1 \cdot 1) \hat{i} - (1 \cdot 2 - 1 \cdot 2) \hat{j} + (1 \cdot 1 - (-1) \cdot 2) \hat{k} = \hat{i} + 0 \hat{j} + 3 \hat{k}. \] 

Step 5: Now, calculate the distance: \[ d = \frac{|(i - 3j - 2k) \cdot (i + 3k)|}{|\hat{i} + 3\hat{k}|}. \] Compute the dot product and magnitude to find the shortest distance. \bigskip