Question

The shortest distance between the lines $\vec{r} = -\hat{i} + t\hat{k}, \; t \in \mathbb{R}$ and $\vec{r} = -\hat{j} + s\hat{i}, \; s \in \mathbb{R}$ is:

• A. $8$
• B. $5$
• C. $3$
• D. $4$
• E. $1$

Hint:

Use vector triple product formula for shortest distance between skew lines.

Answer 1

Answer: (E)

Concept:
• Shortest distance between skew lines: \[ d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \]

Step 1: Identify vectors
\[ \vec{a_1} = (-1,0,0), \quad \vec{d_1} = (0,0,1) \] \[ \vec{a_2} = (0,-1,0), \quad \vec{d_2} = (1,0,0) \]

Step 2: Find cross product
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 0 & 0 & 1 1 & 0 & 0 \end{vmatrix} = (0,1,0) \]

Step 3: Compute numerator
\[ \vec{a_2} - \vec{a_1} = (1,-1,0) \] \[ (1,-1,0)\cdot(0,1,0) = -1 \]

Step 4: Final distance
\[ d = \frac{| -1 |}{1} = 1 \] Final Conclusion:
\[ = 1 \]