Question

Find the shortest distance between two lines \( l_1 \) and \( l_2 \) whose vector equations are: \[ \mathbf{r} = \hat{i} + \hat{j} + \lambda (2\hat{i} - \hat{j} + \hat{k}) \] and \[ \mathbf{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu (3\hat{i} - 5\hat{j} + 2\hat{k}). \]

Hint:

For the shortest distance between two skew lines, use the formula: \[ d = \frac{| (\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1}) |}{|\mathbf{b_1} \times \mathbf{b_2}|}. \]

Answer 1

To find the shortest distance between two skew lines, we use the formula:

\[ d = \frac{| (\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1}) |}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

where:

• \(\mathbf{a_1}\) and \(\mathbf{a_2}\) are position vectors of any points on the two lines.
• \(\mathbf{b_1}\) and \(\mathbf{b_2}\) are the direction vectors of the two lines.

From the given equations:

\[ \mathbf{a_1} = (1,1,0), \quad \mathbf{a_2} = (2,1,-1) \] \[ \mathbf{b_1} = (2,-1,1), \quad \mathbf{b_2} = (3,-5,2). \]

First, we compute \( \mathbf{b_1} \times \mathbf{b_2} \):

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} \] \[ = \hat{i} ( (-1)(2) - (1)(-5) ) - \hat{j} ( (2)(2) - (1)(3) ) + \hat{k} ( (2)(-5) - (-1)(3) ). \] \[ = \hat{i} (-2 + 5) - \hat{j} (4 - 3) + \hat{k} (-10 + 3). \] \[ = 3\hat{i} - \hat{j} - 7\hat{k}. \]

Now, compute the magnitude:

\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59}. \]

Next, compute \( (\mathbf{a_2} - \mathbf{a_1}) \):

\[ \mathbf{a_2} - \mathbf{a_1} = (2 - 1, 1 - 1, -1 - 0) = (1,0,-1). \]

Now, compute the dot product:

\[ (\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1}) \] \[ = (3\hat{i} - \hat{j} - 7\hat{k}) \cdot (1,0,-1). \] \[ = (3 \times 1) + (-1 \times 0) + (-7 \times -1). \] \[ = 3 + 0 + 7 = 10. \]

Finally, compute the shortest distance:

\[ d = \frac{|10|}{\sqrt{59}} = \frac{10}{\sqrt{59}}. \]