Q74 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
If two lines $L_1: \dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}$ and $L_2: \dfrac{x-3}{1}=\dfrac{y-k}{2}=z$ intersect at a point, then $2 k$ is equal to
- Correct answerA9
- B$\dfrac{9}{2}$
- C1
- D$\dfrac{1}{2}$
Solution
The lines are given by $L_1: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4} = \lambda$ and $L_2: \dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1} = \mu$.
Any point on $L_1$ is $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and any point on $L_2$ is $(\mu + 3, 2\mu + k, \mu)$.
Since the lines intersect, there exist $\lambda$ and $\mu$ such that:
$2\lambda + 1 = \mu + 3 \Rightarrow 2\lambda - \mu = 2$ (1)
$3\lambda - 1 = 2\mu + k \Rightarrow 3\lambda - 2\mu = k + 1$ (2)
$4\lambda + 1 = \mu$ (3)
Substituting (3) into (1):
$2\lambda - (4\lambda + 1) = 2 \Rightarrow -2\lambda = 3 \Rightarrow \lambda = -\dfrac{3}{2}$.
From (3), $\mu = 4(-\dfrac{3}{2}) + 1 = -6 + 1 = -5$.
Substituting $\lambda = -\dfrac{3}{2}$ and $\mu = -5$ into (2):
$3(-\dfrac{3}{2}) - 2(-5) = k + 1$
$-\dfrac{9}{2} + 10 = k + 1$
$\dfrac{11}{2} = k + 1 \Rightarrow k = \dfrac{9}{2}$.
Therefore, $2k = 2 \times \dfrac{9}{2} = 9$.
Answer: 9
Any point on $L_1$ is $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and any point on $L_2$ is $(\mu + 3, 2\mu + k, \mu)$.
Since the lines intersect, there exist $\lambda$ and $\mu$ such that:
$2\lambda + 1 = \mu + 3 \Rightarrow 2\lambda - \mu = 2$ (1)
$3\lambda - 1 = 2\mu + k \Rightarrow 3\lambda - 2\mu = k + 1$ (2)
$4\lambda + 1 = \mu$ (3)
Substituting (3) into (1):
$2\lambda - (4\lambda + 1) = 2 \Rightarrow -2\lambda = 3 \Rightarrow \lambda = -\dfrac{3}{2}$.
From (3), $\mu = 4(-\dfrac{3}{2}) + 1 = -6 + 1 = -5$.
Substituting $\lambda = -\dfrac{3}{2}$ and $\mu = -5$ into (2):
$3(-\dfrac{3}{2}) - 2(-5) = k + 1$
$-\dfrac{9}{2} + 10 = k + 1$
$\dfrac{11}{2} = k + 1 \Rightarrow k = \dfrac{9}{2}$.
Therefore, $2k = 2 \times \dfrac{9}{2} = 9$.
Answer: 9
Hence, the correct answer is option: 'A'.