Question

Find the shortest distance between parallel lines: \[ \vec{r_1} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \,\,\, \text{and} \,\,\, \vec{r_2} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + \mu (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \]

Hint:

For parallel lines, the shortest distance is calculated using the formula involving the dot product of the difference vector and the direction vector of the lines.

Answer 1

Step 1: Identifying direction vector and point.
For both lines, the direction vector is \( \vec{d} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \).

Step 2: Finding the difference vector.
The difference between a point on line 1 and a point on line 2 is: \[ \vec{a_2} - \vec{a_1} = (3 - 1) \hat{i} + (3 - 2) \hat{j} + (-5 + 4) \hat{k} = 2 \hat{i} + \hat{j} - \hat{k}. \]

Step 3: Using the shortest distance formula.
For parallel lines, the shortest distance is given by: \[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot \vec{d}|}{|\vec{d}|}. \] Here, \( \vec{a_2} - \vec{a_1} = 2 \hat{i} + \hat{j} - \hat{k} \) and \( \vec{d} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \). Now calculate the dot product and magnitude to find the distance.