Question

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

Hint:

Memorize the formula for the shortest distance between skew lines. The numerator is the scalar triple product of \( (\vec{a}_2 - \vec{a}_1) \), \( \vec{b}_1 \), and \( \vec{b}_2 \), which represents the volume of the parallelepiped formed by these vectors. The denominator is the magnitude of the cross product, representing the area of the base parallelogram.

Answer 1

Step 1: Understanding the Concept:
The two given lines are skew (not parallel and not intersecting). The shortest distance between two skew lines is the length of the perpendicular line segment between them. We can find this using a standard vector formula.
Step 2: Key Formula or Approach:
The vector equations of the lines are of the form \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b}_2 \).
The shortest distance (d) between them 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, we identify the vectors:
\( \vec{a}_1 = \hat{i} + \hat{j} + 0\hat{k} \)
\( \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k} \)
\( \vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k} \)
\( \vec{b}_2 = 3\hat{i} + \hat{j} + 2\hat{k} \)
1. Calculate \( \vec{a}_2 - \vec{a}_1 \):
\[ \vec{a}_2 - \vec{a}_1 = (2\hat{i} + \hat{j} - \hat{k}) - (\hat{i} + \hat{j}) = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} - \hat{k} \] 2. Calculate \( \vec{b}_1 \times \vec{b}_2 \):
\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -1 & 1
3 & 1 & 2 \end{vmatrix} \] \[ = \hat{i}((-1)(2) - (1)(1)) - \hat{j}((2)(2) - (1)(3)) + \hat{k}((2)(1) - (-1)(3)) \] \[ = \hat{i}(-2 - 1) - \hat{j}(4 - 3) + \hat{k}(2 + 3) = -3\hat{i} - \hat{j} + 5\hat{k} \] 3. Calculate the dot product \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \):
\[ (\hat{i} + 0\hat{j} - \hat{k}) \cdot (-3\hat{i} - \hat{j} + 5\hat{k}) = (1)(-3) + (0)(-1) + (-1)(5) = -3 - 5 = -8 \] 4. Calculate the magnitude \( |\vec{b}_1 \times \vec{b}_2| \):
\[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + (-1)^2 + 5^2} = \sqrt{9 + 1 + 25} = \sqrt{35} \] 5. Calculate the shortest distance d:
\[ d = \left| \frac{-8}{\sqrt{35}} \right| = \frac{8}{\sqrt{35}} \] Step 4: Final Answer:
The shortest distance between the two lines is \( \frac{8}{\sqrt{35}} \) units.