Question

A straight line passes through the point whose position vector is $\hat{k}$. The straight line also passes through the point of intersection of the lines $\vec{r} = \hat{j} + \lambda \hat{i}, \lambda \in \mathbb{R}$ and $\vec{r} = \hat{i} + s\hat{j}, s \in \mathbb{R}$. Then the equation of the straight line is:

• A. $\vec{r} = \hat{k} + t(\hat{i} + \hat{j} - \hat{k}), \; t \in \mathbb{R}$
• B. $\vec{r} = \hat{k} + t(\hat{i} - \hat{j} - \hat{k}), \; t \in \mathbb{R}$
• C. $\vec{r} = \hat{k} + t(\hat{i} - \hat{j} + \hat{k}), \; t \in \mathbb{R}$
• D. $\vec{r} = \hat{k} + t(-\hat{i} + \hat{j} + 2\hat{k}), \; t \in \mathbb{R}$
• E. $\vec{r} = \hat{k} + t(-\hat{i} + 2\hat{j} - \hat{k}), \; t \in \mathbb{R}$

Hint:

Direction vector of line through two points is their difference.

Answer 1

The Correct Option is A

Concept:
• Intersection of two lines gives a point
• Direction vector = difference of two points

Step 1: Find intersection point
From $\vec{r} = \hat{j} + \lambda \hat{i} \Rightarrow (\lambda,1,0)$
From $\vec{r} = \hat{i} + s\hat{j} \Rightarrow (1,s,0)$ Equating: \[ \lambda = 1,\quad s = 1 \] Intersection point = $(1,1,0)$

Step 2: Given point
\[ (0,0,1) \]

Step 3: Find direction vector
\[ (1,1,0) - (0,0,1) = (1,1,-1) \]

Step 4: Write equation
\[ \vec{r} = \hat{k} + t(\hat{i} + \hat{j} - \hat{k}) \] Final Conclusion:
Option (A)