Question

Find the shortest distance between the lines: \[ \mathbf{r}_1 = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda (\hat{i} - 2\hat{j} + 3\hat{k}) \] \[ \mathbf{r}_2 = (\hat{i} + 4\hat{k}) + \mu (3\hat{i} - 6\hat{j} + 9\hat{k}) \]

Hint:

Quick Tip: To find the shortest distance between two skew lines, use the formula involving the cross product of direction vectors. The magnitude of the cross product helps in calculating the shortest distance.

Answer 1

The shortest distance \( d \) 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 two lines. From the given equations: - The direction vector of line 1 is \( \mathbf{v}_1 = \hat{i} - 2\hat{j} + 3\hat{k} \)
- The direction vector of line 2 is \( \mathbf{v}_2 = 3\hat{i} - 6\hat{j} + 9\hat{k} \)
- A point on line 1 is \( \mathbf{r}_1 = 2\hat{i} - \hat{j} + 3\hat{k} \)
- A point on line 2 is \( \mathbf{r}_2 = \hat{i} + 4\hat{k} \)
The vector \( \mathbf{r}_2 - \mathbf{r}_1 \) is: \[ \mathbf{r}_2 - \mathbf{r}_1 = (\hat{i} + 4\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) = -\hat{i} + \hat{j} + \hat{k} \] Now, compute the cross product \( \mathbf{v}_1 \times \mathbf{v}_2 \): \[ \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 3 & -6 & 9 \end{vmatrix} \] \[ = \hat{i} \begin{vmatrix} -2 & 3 \\ -6 & 9 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 3 & 9 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 3 & -6 \end{vmatrix} \] \[ = \hat{i} (18 - (-18)) - \hat{j} (9 - 9) + \hat{k} (-6 + 6) \] \[ = 36 \hat{i} + 0 \hat{j} + 0 \hat{k} \] \[ \mathbf{v}_1 \times \mathbf{v}_2 = 36 \hat{i} \] Finally, compute the distance: \[ d = \frac{|(-\hat{i} + \hat{j} + \hat{k}) \cdot (36 \hat{i})|}{|36 \hat{i}|} \] \[ d = \frac{|(-1)(36) + 0 + 0|}{36} = \frac{36}{36} = 1 \] Thus, the shortest distance between the lines is \( \boxed{1} \).