Key Concepts and Formulas

• Vector Equation of a Line: A line passing through a point with position vector $\vec{p}$ and parallel to a direction vector $\vec{d}$ is given by $\vec{r} = \vec{p} + t\vec{d}$, where $t$ is a scalar parameter.
• Intersection of Two Lines: To find the intersection of two lines $\vec{r}_1 = \vec{p}_1 + \lambda\vec{d}_1$ and $\vec{r}_2 = \vec{p}_2 + \mu\vec{d}_2$, we set $\vec{p}_1 + \lambda\vec{d}_1 = \vec{p}_2 + \mu\vec{d}_2$ and solve the resulting system of linear equations for $\lambda$ and $\mu$.
• Point on a Line: A point with position vector $\vec{P}$ lies on the line $\vec{r} = \vec{p} + t\vec{d}$ if $\vec{P} - \vec{p}$ is parallel to $\vec{d}$, meaning $\vec{P} - \vec{p} = t\vec{d}$ for some scalar $t$.

Step-by-Step Solution

We are given two lines $L_1$ and $L_2$, defined by: $L_1 : \overrightarrow{r}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda \vec{a}$, where $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ $L_2: \overrightarrow{r}=(\hat{j}+\hat{k})+\mu \vec{b}$, where $\vec{b}=2 \hat{i}+7 \hat{j}+3 \hat{k}$

Line $L_3$ passes through the intersection of $L_1$ and $L_2$ and is parallel to $\vec{a}+\vec{b}$.

Step 1: Find the point of intersection of $L_1$ and $L_2$.

To find the intersection point, we equate the vector equations of $L_1$ and $L_2$: $(-\hat{i}+2 \hat{j}+\hat{k})+\lambda (\hat{i}+2 \hat{j}+\hat{k}) = (\hat{j}+\hat{k})+\mu (2 \hat{i}+7 \hat{j}+3 \hat{k})$ Group the components for $\hat{i}$, $\hat{j}$, and $\hat{k}$: $(-1+\lambda)\hat{i} + (2+2\lambda)\hat{j} + (1+\lambda)\hat{k} = (2\mu)\hat{i} + (1+7\mu)\hat{j} + (1+3\mu)\hat{k}$ Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ gives a system of linear equations:

1. $-1+\lambda = 2\mu \quad \Rightarrow \quad \lambda - 2\mu = 1$ (Equation 1)
2. $2+2\lambda = 1+7\mu \quad \Rightarrow \quad 2\lambda - 7\mu = -1$ (Equation 2)
3. $1+\lambda = 1+3\mu \quad \Rightarrow \quad \lambda - 3\mu = 0$ (Equation 3)

From Equation 3, we get $\lambda = 3\mu$. Substitute this into Equation 1: $(3\mu) - 2\mu = 1 \quad \Rightarrow \quad \mu = 1$ Now substitute $\mu=1$ back into $\lambda = 3\mu$: $\lambda = 3(1) = 3$ We verify these values in Equation 2: $2(3) - 7(1) = 6 - 7 = -1$. The values are consistent.

Now, we find the position vector of the intersection point by substituting $\lambda=3$ into the equation for $L_1$: $\vec{r}_{int} = (-\hat{i}+2 \hat{j}+\hat{k}) + 3(\hat{i}+2 \hat{j}+\hat{k})$ $\vec{r}_{int} = (-\hat{i}+3\hat{i}) + (2\hat{j}+6\hat{j}) + (\hat{k}+3\hat{k})$ $\vec{r}_{int} = 2\hat{i} + 8\hat{j} + 4\hat{k}$ The point of intersection is $(2, 8, 4)$.

Step 2: Determine the direction vector of $L_3$.

Line $L_3$ is parallel to $\vec{a}+\vec{b}$. $\vec{d}_{L_3} = \vec{a}+\vec{b} = (\hat{i}+2 \hat{j}+\hat{k}) + (2 \hat{i}+7 \hat{j}+3 \hat{k})$ $\vec{d}_{L_3} = (1+2)\hat{i} + (2+7)\hat{j} + (1+3)\hat{k}$ $\vec{d}_{L_3} = 3\hat{i} + 9\hat{j} + 4\hat{k}$

Step 3: Write the vector equation of $L_3$.

Line $L_3$ passes through $(2, 8, 4)$ and has direction vector $3\hat{i} + 9\hat{j} + 4\hat{k}$. The vector equation of $L_3$ is: $\overrightarrow{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + \nu (3\hat{i} + 9\hat{j} + 4\hat{k})$ where $\nu$ is a scalar parameter.

This can be written in parametric form as: $x = 2 + 3\nu$ $y = 8 + 9\nu$ $z = 4 + 4\nu$

Step 4: Check which option point lies on $L_3$.

We test each option by substituting its coordinates into the parametric equations and checking for a consistent value of $\nu$.

(A) $(-1, -1, 1)$

• For $x$: $-1 = 2 + 3\nu \Rightarrow 3\nu = -3 \Rightarrow \nu = -1$
• For $y$: $-1 = 8 + 9\nu \Rightarrow 9\nu = -9 \Rightarrow \nu = -1$
• For $z$: $1 = 4 + 4\nu \Rightarrow 4\nu = -3 \Rightarrow \nu = -3/4$ The values of $\nu$ are not consistent.

(B) $(2, 8, 5)$

• For $x$: $2 = 2 + 3\nu \Rightarrow 3\nu = 0 \Rightarrow \nu = 0$
• For $y$: $8 = 8 + 9\nu \Rightarrow 9\nu = 0 \Rightarrow \nu = 0$
• For $z$: $5 = 4 + 4\nu \Rightarrow 4\nu = 1 \Rightarrow \nu = 1/4$ The values of $\nu$ are not consistent.

(C) $(8, 26, 12)$

• For $x$: $8 = 2 + 3\nu \Rightarrow 3\nu = 6 \Rightarrow \nu = 2$
• For $y$: $26 = 8 + 9\nu \Rightarrow 9\nu = 18 \Rightarrow \nu = 2$
• For $z$: $12 = 4 + 4\nu \Rightarrow 4\nu = 8 \Rightarrow \nu = 2$ The values of $\nu$ are consistent ($\nu=2$). This point lies on $L_3$.

(D) $(5, 17, 4)$

• For $x$: $5 = 2 + 3\nu \Rightarrow 3\nu = 3 \Rightarrow \nu = 1$
• For $y$: $17 = 8 + 9\nu \Rightarrow 9\nu = 9 \Rightarrow \nu = 1$
• For $z$: $4 = 4 + 4\nu \Rightarrow 4\nu = 0 \Rightarrow \nu = 0$ The values of $\nu$ are not consistent.

The line $L_3$ passes through the point $(8, 26, 12)$, which corresponds to option (C).

Common Mistakes & Tips

• Algebraic Errors: Be meticulous when solving the system of linear equations for $\lambda$ and $\mu$. A small arithmetic mistake can lead to an incorrect intersection point.
• Consistency Check: Always verify the found parameters $\lambda$ and $\mu$ in all three equations. Similarly, when checking a point on $L_3$, ensure the parameter $\nu$ is consistent across all coordinates.
• Direction Vector Calculation: Double-check the addition of vectors $\vec{a}$ and $\vec{b}$ to ensure the direction vector of $L_3$ is correctly computed.

Summary

The problem required us to find a line $L_3$ that passes through the intersection of two given lines, $L_1$ and $L_2$, and is parallel to the sum of their direction vectors ($\vec{a}+\vec{b}$). We first found the point of intersection of $L_1$ and $L_2$ by solving a system of linear equations derived from their vector forms. Next, we computed the direction vector of $L_3$ by adding $\vec{a}$ and $\vec{b}$. Using the intersection point and the direction vector, we formulated the vector equation of $L_3$. Finally, we tested each given option point by substituting its coordinates into the parametric form of $L_3$ to see which point yielded a consistent parameter value. The point $(8, 26, 12)$ consistently satisfied the equation of $L_3$, indicating it lies on the line.

The final answer is \boxed{(8, 26, 12)} which corresponds to option (C).