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:

To find the shortest distance between two skew lines, use the formula involving the cross product of their direction vectors and the difference of a point on each line.

Answer 1

Step 1: Understanding the formula.
The shortest distance between two skew lines is given by the formula: \[ d = \frac{|\vec{b_1} \times \vec{b_2} \cdot (\vec{a_2} - \vec{a_1})|}{|\vec{b_1} \times \vec{b_2}|} \] where: - \( \vec{a_1} \) and \( \vec{a_2} \) are points on lines 1 and 2 respectively, - \( \vec{b_1} \) and \( \vec{b_2} \) are direction vectors of the two lines.

Step 2: Finding the direction vectors.
For line 1: \[ \vec{b_1} = \frac{d}{dt} (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2t) \hat{k} = - \hat{i} + \hat{j} - 2\hat{k}. \] For line 2: \[ \vec{b_2} = \frac{d}{ds} (s + 1) \hat{i} + (2s - 1) \hat{j} - (2s + 1) \hat{k} = \hat{i} + 2 \hat{j} - 2 \hat{k}. \]

Step 3: Finding the difference vector.
The difference vector between a point on line 1 and a point on line 2 is: \[ \vec{a_2} - \vec{a_1} = (s + 1 - 1) \hat{i} + (2s - 1 - t + 2) \hat{j} - (2s + 1 - 3 + 2t) \hat{k}. \]

Step 4: Calculate the cross product.
Now, compute \( \vec{b_1} \times \vec{b_2} \) and substitute the values into the formula for the shortest distance.

Step 5: Conclusion.
Once the cross product and dot product are calculated, substitute the values into the distance formula to get the shortest distance.