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. \( (5, 17, 4) \)
• B. \( (8, 26, 12) \)
• 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

To solve the problem, we need to determine the point of intersection of the lines \( L_1 \) and \( L_2 \), and then find the equation of the line \( L_3 \) that passes through this intersection point and is parallel to \( \vec{a} + \vec{b} \). Finally, we verify which given point lies on \( L_3 \).

1. Finding the Point of Intersection:
The position vectors of the lines \( L_1 \) and \( L_2 \) are given as:

\( (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} + 2\hat{j} + \hat{k}) = (\hat{j} + \hat{k}) + \mu (2\hat{i} + 7\hat{j} + 3\hat{k}) \)

Equating the coefficients of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \), we obtain the following system of equations:

\begin{align*} -1 + \lambda &= 2\mu \\ 2 + 2\lambda &= 1 + 7\mu \\ 1 + \lambda &= 1 + 3\mu \end{align*}

From the third equation, \( 1 + \lambda = 1 + 3\mu \), we get \( \lambda = 3\mu \).
Substituting \( \lambda = 3\mu \) into the first equation:

\( -1 + 3\mu = 2\mu \implies \mu = 1 \)
Substituting \( \mu = 1 \) back into \( \lambda = 3\mu \):

\( \lambda = 3(1) = 3 \)

Verifying these values in the second equation:

\( 2 + 2(3) = 1 + 7(1) \implies 8 = 8 \)
Since all three equations are satisfied, the lines intersect. Substituting \( \lambda = 3 \) into the equation for \( L_1 \):

\( \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + 3(\hat{i} + 2\hat{j} + \hat{k}) \)

\( \vec{r} = -\hat{i} + 2\hat{j} + \hat{k} + 3\hat{i} + 6\hat{j} + 3\hat{k} = 2\hat{i} + 8\hat{j} + 4\hat{k} \)

Thus, the point of intersection is \( (2, 8, 4) \).

2. Equation of Line \( L_3 \):
The line \( L_3 \) passes through the point \( (2, 8, 4) \) and is parallel to \( \vec{a} + \vec{b} \). Calculating \( \vec{a} + \vec{b} \):

\( \vec{a} + \vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 7\hat{j} + 3\hat{k}) = 3\hat{i} + 9\hat{j} + 4\hat{k} \)

The equation of \( L_3 \) is:

\( \vec{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + t(3\hat{i} + 9\hat{j} + 4\hat{k}), \quad t \in \mathbb{R} \)

3. Verifying Which Point Lies on \( L_3 \):
A point \( (x, y, z) \) lies on \( L_3 \) if:

\( x = 2 + 3t, \quad y = 8 + 9t, \quad z = 4 + 4t \)

We check the option \( (5, 17, 8) \):

\( 5 = 2 + 3t \implies 3t = 3 \implies t = 1 \)
\( 17 = 8 + 9t \implies 9t = 9 \implies t = 1 \)
\( 8 = 4 + 4t \implies 4t = 4 \implies t = 1 \)

Since \( t = 1 \) satisfies all three coordinates, the point \( (5, 17, 8) \) lies on \( L_3 \).

Final Answer:
The final answer is \( \boxed{(5, 17, 8)} \).