Question

Find the shortest distance between the lines \(\vec{r} = (2\hat{i} + \hat{j} – \hat{k}) + \mu(3\hat{i} – 5\hat{j} + 2\hat{k})\) and \(\vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k})\).

Hint:

Before applying the skew lines formula, quickly check if the lines are parallel by seeing if \(\vec{b_1}\) is a scalar multiple of \(\vec{b_2}\). If they are, a different formula is needed. If the scalar triple product \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\) equals zero, the lines are coplanar and intersect, so the shortest distance is 0.

Answer 1

Step 1: Understanding the Concept:
The two given lines are skew lines, meaning they are non-parallel and do not intersect.
The shortest distance between two skew lines is the length of the perpendicular line segment connecting them.
Step 2: Key Formula or Approach:
The shortest distance (d) between two skew lines \(\vec{r} = \vec{a_1} + \lambda\vec{b_1}\) and \(\vec{r} = \vec{a_2} + \mu\vec{b_2}\) is given by the formula:
\[ d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \] Step 3: Detailed Explanation or Calculation:
From the given equations of the lines, we identify the vectors:
Line 1: \(\vec{a_1} = 2\hat{i} + \hat{j} - \hat{k}\), \(\vec{b_1} = 3\hat{i} - 5\hat{j} + 2\hat{k}\)
Line 2: \(\vec{a_2} = \hat{i} + \hat{j}\), \(\vec{b_2} = 2\hat{i} - \hat{j} + \hat{k}\)
First, calculate \((\vec{a_2} - \vec{a_1})\):
\[ \vec{a_2} - \vec{a_1} = (\hat{i} + \hat{j}) - (2\hat{i} + \hat{j} - \hat{k}) = -\hat{i} + 0\hat{j} + \hat{k} \] Next, calculate the cross product \((\vec{b_1} \times \vec{b_2})\):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -5 & 2
2 & -1 & 1 \end{vmatrix} \] \[ = \hat{i}((-5)(1) - (2)(-1)) - \hat{j}((3)(1) - (2)(2)) + \hat{k}((3)(-1) - (-5)(2)) \] \[ = \hat{i}(-5 + 2) - \hat{j}(3 - 4) + \hat{k}(-3 + 10) \] \[ = -3\hat{i} + \hat{j} + 7\hat{k} \] Now, calculate the magnitude of the cross product, \(|\vec{b_1} \times \vec{b_2}|\):
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + 1^2 + 7^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \] Next, calculate the dot product \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\):
\[ (-\hat{i} + \hat{k}) \cdot (-3\hat{i} + \hat{j} + 7\hat{k}) = (-1)(-3) + (0)(1) + (1)(7) = 3 + 7 = 10 \] Finally, substitute these values into the shortest distance formula:
\[ d = \left| \frac{10}{\sqrt{59}} \right| = \frac{10}{\sqrt{59}} \] Step 4: Final Answer
The shortest distance between the two lines is \(\frac{10}{\sqrt{59}}\) units.