Question

The shortest distance between the parallel straight lines \[ \overrightarrow{r_1} = \hat{k} + s(\hat{i} + \hat{j}), \quad t, s \in \mathbb{R} \quad {and} \quad \overrightarrow{r_2} = \hat{j} + t(\hat{i} + \hat{j}), \] is:

• A. \( \sqrt{3} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( \frac{1}{\sqrt{2}} \)
• E. \( \frac{1}{\sqrt{3}} \)

Hint:

The shortest distance between two parallel lines can be calculated by finding the difference vector of a point on one line and a point on the other line, taking the dot product with the direction vector, and dividing by the magnitude of the direction vector.

Answer 1

The Correct Option is C

We are given two parallel straight lines:
1. \( \vec{r} = \hat{k} + s(\hat{i} + \hat{j}) \), where \( s \in \mathbb{R} \)
2. \( \vec{r} = \hat{j} + t(\hat{i} + \hat{j}) \), where \( t \in \mathbb{R} \) 
We need to find the shortest distance between these two parallel lines. 
To solve this, we use the formula for the shortest distance \( d \) between two parallel lines given by: \[ d = \frac{| \vec{r_1} - \vec{r_2} \cdot (\vec{v_1} \times \vec{v_2}) |}{|\vec{v_1} \times \vec{v_2}|} \] Here: - \( \vec{r_1} = \hat{k} \) (from the first line)
- \( \vec{r_2} = \hat{j} \) (from the second line)
- \( \vec{v_1} = \hat{i} + \hat{j} \) (direction vector of the first line)
- \( \vec{v_2} = \hat{i} + \hat{j} \) (direction vector of the second line)
Since the lines are parallel, \( \vec{v_1} = \vec{v_2} \), so the cross product \( \vec{v_1} \times \vec{v_2} \) will give a zero vector. 
We now use the formula for the shortest distance between two parallel lines: \[ d = \frac{| (\hat{k} - \hat{j}) \cdot (\hat{i} + \hat{j}) |}{|\hat{i} + \hat{j}|} \] We calculate \( \hat{k} - \hat{j} = (0, -1, 1) \) and \( \hat{i} + \hat{j} = (1, 1, 0) \), and their dot product: \[ (\hat{k} - \hat{j}) \cdot (\hat{i} + \hat{j}) = (0, -1, 1) \cdot (1, 1, 0) = 0 \times 1 + (-1) \times 1 + 1 \times 0 = -1 \] Now, we calculate the magnitude of \( \hat{i} + \hat{j} \): \[ |\hat{i} + \hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \] Thus, the shortest distance is: \[ d = \frac{| -1 |}{\sqrt{2}} = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad d = \frac{\sqrt{3}}{2} \] Thus, the correct answer is option (C), \( \frac{\sqrt{3}}{2} \).