Question

Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k} \). Let \( L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \lambda \in {R} \) and \( L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \mu \in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec{a} + \vec{b} \), then \( L_3 \) passes through the point:

• A. \( (8, 26, 12) \)
• B. \( (5, 17, 4) \)
• C. \( (2, 8, 5) \)
• D. \( (-1, -1, 1) \)

Hint:

For finding the intersection of two parametric lines, equate their parametric equations and solve for the parameters.

Answer 1

The Correct Option is A

Given Lines:

\[ L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} + 2\hat{j} + \hat{k}) \]

Expanding the equation:

\[ \vec{r} = (\lambda -1)\hat{i} + 2(\lambda +1)\hat{j} + (\lambda +1)\hat{k} \] \[ L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu (2\hat{i} + 7\hat{j} + 3\hat{k}) \]

Expanding the equation:

\[ \vec{r} = 2\mu\hat{i} + (1 + 7\mu)\hat{j} + (1 + 3\mu)\hat{k} \]

Finding the point of intersection: Equating respective components:

From the \(\hat{i}\) component:

\[ \lambda - 1 = 2\mu \quad \text{.....(1)} \]

From the \(\hat{j}\) component:

\[ 2(\lambda + 1) = 1 + 7\mu \quad \text{.....(2)} \]

From the \(\hat{k}\) component:

\[ \lambda + 1 = 1 + 3\mu \quad \text{.....(3)} \]

Solving equations (1), (2), and (3), we get:

\[ \lambda = 3, \quad \mu = 1 \]

Finding \(\vec{a} + \vec{b}\):

\[ \vec{a} + \vec{b} = 3\hat{i} + 9\hat{j} + 4\hat{k} \]

Given the line equation:

\[ L_3 : \vec{r} = 2\hat{i} + 8\hat{j} + 4\hat{k} + \alpha (3\hat{i} + 9\hat{j} + 4\hat{k}) \]

For \(\alpha = 2\):

\[ \vec{r} = 8\hat{i} + 26\hat{j} + 12\hat{k} \]