Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Let $\vec{a} = \hat{i} - \hat{j}, \vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}, \vec{c} = -\hat{i} - \hat{j} + \hat{k}$ and $\vec{d} = 2\hat{i} - \hat{j} + \hat{k}$, then the shortest distance between the lines $\vec{r} = \vec{a} + t\vec{b}$ and $\vec{r} = \vec{c} + p\vec{d}$ is $k$, then the value of $\frac{1}{k^2}$ is ______.
Step-by-Step Solution
Key Concept: The shortest distance between skew lines equals the magnitude of the projection of a vector joining any two points on the lines onto the common perpendicular direction.
The shortest distance between two skew lines is given by $d = \left|\frac{(\vec{a} - \vec{c}) \cdot (\vec{a} \times \vec{d})}{|\vec{a} \times \vec{d}|}\right|$ where $\vec{a}, \vec{b}$ are direction vectors and $\vec{c}, \vec{d}$ are points on the respective lines.
Correct Answer: I need to find the shortest distance between two lines and then calculate 1/k².
Given:
- Line 1: $\vec{r} = \vec{a} + t\vec{b}$ where $\vec{a} = \hat{i} - \hat{j}$, $\vec{b} = \hat{i}