Q60 / 120 · VITEEE 2014 · Mathematics

Two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point, if $k$ is equal to

  • A
    $\frac{2}{9}$
  • B
    $\frac{1}{2}$
  • Correct answer
    C
    $\frac{9}{2}$
  • D
    $\frac{1}{6}$

Solution

$\begin{array}{l} \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=r(\text { say }) \\ \Rightarrow x=2 r+1, y=3 r-1, z=4 r+1 \end{array}$ Since, the two lines intersect. So, putting above values in second line, we get $\begin{array}{l} \frac{2 r+1-3}{1}=\frac{3 r-1-k}{2}=\frac{4 r+1}{1} \\ 2 r-2=4 r+1 \end{array}$ $\begin{aligned} \Rightarrow & \mathrm{r}=-3 / 2 \\ & \text { Also } 3 \mathrm{r}-1-\mathrm{k}=8 \mathrm{r}+2 \\ \Rightarrow & \mathrm{k}=-5 \mathrm{r}-3=\frac{15}{2}-3=\frac{9}{2} \end{aligned}$

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

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