Q86 / 180 · COMEDK 2023 (Memory Based) · Mathematics

If two lines $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point, then $2 k$ is equal to

  • Correct answer
    A
    9
  • B
    $\frac{1}{2}$
  • C
    $\frac{9}{2}$
  • D
    1

Solution

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ Now, any point $P$ that lies on the lines $L_1$ has the form $(1+2 \lambda,-1+3 \lambda, 1+4 \lambda)$. Now, on putting $x=1+2 \lambda, y=-1+3 \lambda$ and $z=1+4 \lambda$ into the equation of lines $L_2$, we get $\begin{aligned} & \frac{1+2 \lambda-3}{1}=\frac{-1+3 \lambda-k}{2}=1+4 \lambda \\ & \Rightarrow \quad \frac{1+2 \lambda-3}{1}=1+4 \lambda \\ & -2 \lambda=3 \Rightarrow \lambda=\frac{-3}{2} \\ & \frac{-1+3 \lambda-k}{2}=1+4 \lambda \\ & -1+3 \lambda-k=2+8 \lambda \\ & -5 \lambda=3+k \\ & -5\left(-\frac{3}{2}\right)=3+k \quad[\because \lambda=-3 / 2] \\ & k=\frac{15}{2}-3 \\ & k=\frac{9}{2} \\ & 2 k=9 \\ & \end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions