Q50 / 90 · JEE Main 2012 (Offline) · Mathematics

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k$ is equal to

  • A
    $-1$
  • B
    $\frac{2}{9}$
  • Correct answer
    C
    $\frac{9}{2}$
  • D
    $0$

Solution

Any point on $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=t$ is $(2 t+1,3 t-1,4 t+1)$ And any point on $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mathrm{s}$ is $(s+3,2 s+k, s)$ Given lines are intersecting $\Rightarrow \mathrm{t}=-\frac{3}{2} \text { and } \mathrm{s}=-5 \quad \therefore \mathrm{k}=\frac{9}{2}$

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

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