Question

Find the shortest distance between the following pairs of lines: \[ \mathbf{r_1} = \hat{i} - 4 \hat{j} + 5 \hat{k} + \mu(5 \hat{i} + 9 \hat{j} + \hat{k}), \quad \mathbf{r_2} = 2 \hat{i} + 8 \hat{j} - 6 \hat{k} + \lambda(3 \hat{i} - 2 \hat{j} + \hat{k}) \]

Hint:

To find the shortest distance between two skew lines, use the formula involving the cross product and the position vectors of points on the lines.

Answer 1

Step 1: Understanding the problem.
We are given two parametric equations of lines. To find the shortest distance between two skew lines, we use the following formula: \[ d = \frac{| (\mathbf{r_2_0} - \mathbf{r_1_0}) \cdot (\mathbf{a_1} \times \mathbf{a_2}) |}{|\mathbf{a_1} \times \mathbf{a_2}|} \] where \( \mathbf{r_1_0} \) and \( \mathbf{r_2_0} \) are position vectors of points on each line, and \( \mathbf{a_1} \) and \( \mathbf{a_2} \) are direction vectors of the lines. Step 2: Extracting vectors.
For the first line \( \mathbf{r_1} \), we have: \[ \mathbf{r_1_0} = \hat{i} - 4 \hat{j} + 5 \hat{k}, \quad \mathbf{a_1} = 5 \hat{i} + 9 \hat{j} + \hat{k} \] For the second line \( \mathbf{r_2} \), we have: \[ \mathbf{r_2_0} = 2 \hat{i} + 8 \hat{j} - 6 \hat{k}, \quad \mathbf{a_2} = 3 \hat{i} - 2 \hat{j} + \hat{k} \] Step 3: Calculating the vector \( \mathbf{r_2_0} - \mathbf{r_1_0} \).
We calculate: \[ \mathbf{r_2_0} - \mathbf{r_1_0} = (2 \hat{i} + 8 \hat{j} - 6 \hat{k}) - (\hat{i} - 4 \hat{j} + 5 \hat{k}) = \hat{i} + 12 \hat{j} - 11 \hat{k} \] Step 4: Finding the cross product \( \mathbf{a_1} \times \mathbf{a_2} \).
Now, we find the cross product \( \mathbf{a_1} \times \mathbf{a_2} \): \[ \mathbf{a_1} \times \mathbf{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
5 & 9 & 1
3 & -2 & 1 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{a_1} \times \mathbf{a_2} = \hat{i} \begin{vmatrix} 9 & 1
-2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 1
3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & 9
3 & -2 \end{vmatrix} \] Simplifying each 2x2 determinant: \[ \mathbf{a_1} \times \mathbf{a_2} = \hat{i}((9)(1) - (1)(-2)) - \hat{j}((5)(1) - (1)(3)) + \hat{k}((5)(-2) - (9)(3)) \] \[ \mathbf{a_1} \times \mathbf{a_2} = \hat{i}(9 + 2) - \hat{j}(5 - 3) + \hat{k}(-10 - 27) \] \[ \mathbf{a_1} \times \mathbf{a_2} = 11 \hat{i} - 2 \hat{j} - 37 \hat{k} \] Step 5: Finding the magnitude of the cross product.
Now, calculate the magnitude of \( \mathbf{a_1} \times \mathbf{a_2} \): \[ |\mathbf{a_1} \times \mathbf{a_2}| = \sqrt{(11)^2 + (-2)^2 + (-37)^2} = \sqrt{121 + 4 + 1369} = \sqrt{1494} \] Thus: \[ |\mathbf{a_1} \times \mathbf{a_2}| = \sqrt{1494} \] Step 6: Finding the shortest distance.
Finally, we can substitute all the values into the formula for the shortest distance: \[ d = \frac{| (\mathbf{r_2_0} - \mathbf{r_1_0}) \cdot (\mathbf{a_1} \times \mathbf{a_2}) |}{|\mathbf{a_1} \times \mathbf{a_2}|} \] Substitute \( \mathbf{r_2_0} - \mathbf{r_1_0} = \hat{i} + 12 \hat{j} - 11 \hat{k} \) and \( \mathbf{a_1} \times \mathbf{a_2} = 11 \hat{i} - 2 \hat{j} - 37 \hat{k} \): \[ d = \frac{|(1)(11) + (12)(-2) + (-11)(-37)|}{\sqrt{1494}} = \frac{|11 - 24 + 407|}{\sqrt{1494}} = \frac{394}{\sqrt{1494}} \] Simplifying: \[ d \approx \frac{394}{38.67} \approx 10.2 \, \text{units} \]