Question

Find the shortest distance between two lines: \[ \mathbf{r_1} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k}) \] and \[ \mathbf{r_2} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu (2\hat{i} + 3\hat{j} + 6\hat{k}) \]

Hint:

For parallel lines, the shortest distance between them is zero. For skew lines, use the formula involving the cross product to compute the distance.

Answer 1

The shortest distance between two skew lines is given by the formula: \[ d = \frac{|(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{v_1} \times \mathbf{v_2})|}{|\mathbf{v_1} \times \mathbf{v_2}|} \] Where \( \mathbf{r_1} \) and \( \mathbf{r_2} \) are points on the two lines, and \( \mathbf{v_1} \) and \( \mathbf{v_2} \) are the direction vectors of the lines. From the given lines: \[ \mathbf{r_1} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k}), \mathbf{r_2} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu (2\hat{i} + 3\hat{j} + 6\hat{k}) \] The direction vectors are: \[ \mathbf{v_1} = 2\hat{i} + 3\hat{j} + 6\hat{k}, \mathbf{v_2} = 2\hat{i} + 3\hat{j} + 6\hat{k} \] The vector \( \mathbf{r_2} - \mathbf{r_1} \) is: \[ \mathbf{r_2} - \mathbf{r_1} = (3 - 1) \hat{i} + (3 - 2) \hat{j} + (-5 + 4) \hat{k} = 2\hat{i} + \hat{j} - \hat{k} \] Now, compute the cross product \( \mathbf{v_1} \times \mathbf{v_2} \): \[ \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 3 & 6 \end{vmatrix} \] Since the two direction vectors are the same, the cross product is zero: \[ \mathbf{v_1} \times \mathbf{v_2} = 0 \] Thus, the lines are parallel and the shortest distance is zero, as they lie along the same direction.
Final Answer: The shortest distance between the lines is \( \boxed{0} \).