Question

Find the shortest distance between the lines whose vector equations are:
\(\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}\) and \(\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}\).

Hint:

Before using the skew lines formula, quickly check if the lines are parallel by seeing if their direction vectors (\(\vec{b_1}\) and \(\vec{b_2}\)) are scalar multiples of each other. If they are, a different formula for parallel lines must be used. In this case, \(\vec{b_1}\) and \(\vec{b_2}\) are clearly not parallel.

Answer 1

Step 1: Understanding the Concept:
The problem is to find the shortest distance between two skew lines given in vector form. The standard method involves using a formula that uses the position vectors of a point on each line and the direction vectors of the lines.
Step 2: Key Formula or Approach:
First, rewrite the line equations in the standard form \(\vec{r} = \vec{a} + \lambda \vec{b}\).
Line 1: \(\vec{r} = \vec{a_1} + t\vec{b_1}\)
Line 2: \(\vec{r} = \vec{a_2} + s\vec{b_2}\)
The formula for the shortest distance (SD) between two skew lines is:
\[ \text{SD} = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \]
Step 3: Detailed Explanation:
1. Identify vectors from the line equations:
Line 1: \(\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - 2\hat{k})\).
So, \(\vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k}\) and \(\vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k}\).
Line 2: \(\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + 2\hat{j} - 2\hat{k})\).
So, \(\vec{a_2} = \hat{i} - \hat{j} - \hat{k}\) and \(\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}\).
2. Calculate \(\vec{a_2} - \vec{a_1}\):
\[ \vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = 0\hat{i} + \hat{j} - 4\hat{k} \]
3. Calculate \(\vec{b_1} \times \vec{b_2}\):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 - (-4)) - \hat{j}(2 - (-2)) + \hat{k}(-2 - 1) = 2\hat{i} - 4\hat{j} - 3\hat{k} \]
4. Calculate \(|\vec{b_1} \times \vec{b_2}|\):
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \]
5. Calculate \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\):
\[ (0\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k}) = (0)(2) + (1)(-4) + (-4)(-3) = 0 - 4 + 12 = 8 \]
6. Calculate the Shortest Distance:
\[ \text{SD} = \left| \frac{8}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}} \]
Step 4: Final Answer:
The shortest distance between the two lines is \(\frac{8}{\sqrt{29}}\) units.