Discussion
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
Answer: E
The shortest distance is 1.
Line 1 passes through \(A(-1, 0, 0)\) with direction \(\vec{d_1} = \hat{k}\); line 2 passes through \(B(0, -1, 0)\) with direction \(\vec{d_2} = \hat{i}\).
\[ \begin{aligned} \vec{d_1} \times \vec{d_2} &= \hat{k} \times \hat{i} = \hat{j} \\ \overrightarrow{AB} &= \hat{i} - \hat{j} \\ d &= \frac{|\overrightarrow{AB} \cdot \hat{j}|}{|\hat{j}|} = |-1| = 1 \end{aligned} \]
Geometrically, line 1 is vertical through \(x = -1, y = 0\) and line 2 lies along \(y = -1\) in the plane \(z = 0\); they are skew and 1 unit apart in the \(y\)-direction.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development