Question

Find the shortest distance between the lines \[ \vec{r_1} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2t) \hat{k} \] \text{and} \[ \vec{r_2} = (s + 1) \hat{i} + (2s - 1) \hat{j} - (2s + 1) \hat{k}. \]

Hint:

For finding the shortest distance between two skew lines, use the formula involving the cross product of the direction vectors and the vector joining points on the lines.

Answer 1

The shortest distance between two skew lines is given by the formula: \[ d = \frac{|(\vec{r_2_0} - \vec{r_1_0}) \cdot (\vec{a_1} \times \vec{a_2})|}{|\vec{a_1} \times \vec{a_2}|} \] where \( \vec{r_1_0} \) and \( \vec{r_2_0} \) are position vectors of points on the lines, and \( \vec{a_1} \) and \( \vec{a_2} \) are direction vectors of the lines. From the given vectors, \( \vec{r_1} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2t) \hat{k} \), \( \vec{r_2} = (s + 1) \hat{i} + (2s - 1) \hat{j} - (2s + 1) \hat{k} \). So, the direction vectors are: \[ \vec{a_1} = -\hat{i} + \hat{j} - 2\hat{k}, \vec{a_2} = \hat{i} + 2\hat{j} - 2\hat{k}. \] The point on line 1 can be taken as \( \vec{r_1_0} = \hat{i} - 2\hat{j} + 3\hat{k} \), and the point on line 2 can be taken as \( \vec{r_2_0} = \hat{i} - \hat{j} - \hat{k} \). Now, calculate the cross product \( \vec{a_1} \times \vec{a_2} \): \[ \vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} \] \[ = \hat{i}(1 \times -2 - 2 \times -2) - \hat{j}(-1 \times -2 - 1 \times -2) + \hat{k}(-1 \times 2 - 1 \times 1). \] \[ = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) \] \[ = 2\hat{i} - 4\hat{j} - 3\hat{k}. \] Now calculate the numerator: \[ \vec{r_2_0} - \vec{r_1_0} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \hat{j} - 4\hat{k}. \] Now, calculate the dot product: \[ (\vec{r_2_0} - \vec{r_1_0}) \cdot (\vec{a_1} \times \vec{a_2}) = ( \hat{j} - 4 \hat{k}) \cdot (2 \hat{i} - 4 \hat{j} - 3 \hat{k}) \] \[ = (0) + (-4 \times -4) + (-4 \times -3) = 16 + 12 = 28. \] Now, calculate the magnitude of \( \vec{a_1} \times \vec{a_2} \): \[ |\vec{a_1} \times \vec{a_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}. \] Finally, the shortest distance is: \[ d = \frac{|28|}{\sqrt{29}} = \frac{28}{\sqrt{29}}. \]
Conclusion: The shortest distance between the two lines is \[ \boxed{\frac{28}{\sqrt{29}}}. \]