Q65 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
Consider the lines $L_1$ and $L_2$ given by the following vector equations:
$L_1: \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} + t\hat{j})$ $L_2: \vec{r} = (4\hat{i} + a\hat{j} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$
If $a = -2$ and the lines intersect, then the value of 't' is:
$L_1: \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} + t\hat{j})$ $L_2: \vec{r} = (4\hat{i} + a\hat{j} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$
If $a = -2$ and the lines intersect, then the value of 't' is:
- A$1$
- B$-1$
- C$0$
- Correct answerD$-3$
Solution
The general point on line $L_1$ is given by $(1 + 3\lambda)\hat{i} + (1 + \lambda t)\hat{j} - \hat{k}$.
Given $a = -2$, the general point on line $L_2$ is given by $(4 + 2\mu)\hat{i} - 2\hat{j} + (-1 + 3\mu)\hat{k}$.
Since the lines intersect, their corresponding components must be equal for some values of $\lambda$ and $\mu$. Equating the $\hat{i}$, $\hat{j}$, and $\hat{k}$ components, we get:
$1 + 3\lambda = 4 + 2\mu$
$1 + \lambda t = -2$
$-1 = -1 + 3\mu$
From the third equation:
$3\mu = 0 \Rightarrow \mu = 0$
Substituting $\mu = 0$ into the first equation:
$1 + 3\lambda = 4 + 0 \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1$
Substituting $\lambda = 1$ into the second equation:
$1 + (1)t = -2 \Rightarrow t = -3$
Answer: $-3$
Given $a = -2$, the general point on line $L_2$ is given by $(4 + 2\mu)\hat{i} - 2\hat{j} + (-1 + 3\mu)\hat{k}$.
Since the lines intersect, their corresponding components must be equal for some values of $\lambda$ and $\mu$. Equating the $\hat{i}$, $\hat{j}$, and $\hat{k}$ components, we get:
$1 + 3\lambda = 4 + 2\mu$
$1 + \lambda t = -2$
$-1 = -1 + 3\mu$
From the third equation:
$3\mu = 0 \Rightarrow \mu = 0$
Substituting $\mu = 0$ into the first equation:
$1 + 3\lambda = 4 + 0 \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1$
Substituting $\lambda = 1$ into the second equation:
$1 + (1)t = -2 \Rightarrow t = -3$
Answer: $-3$
Hence, the correct answer is option: 'D'.