Q76 / 120 · VITEEE 2010 · Mathematics
If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ intersect, then the value of $\mathrm{k}$ is
- A$\frac{3}{2}$
- Correct answerB$\frac{9}{2}$
- C$-\frac{2}{9}$
- D$-\frac{3}{2}$
Solution
We have the lines
$$
\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}...(i)
$$
and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$...(ii)
Let a point $(2 r+1,3 r-1,4 r+1)$ be on the line Eq. (i). If this is an intersection point of both the lines, then it will lie on Eq. (ii), also
$$
\therefore \frac{2 r+1-3}{1}=\frac{3 r-1-k}{2}=\frac{4 r+1}{1}...(iii)
$$
Taking first and third part of eq. (iii), we get $2 r-2=4 r+1$
$$
\Rightarrow r=-\frac{3}{2}
$$
Taking second and third part of eq. (iii), we get
$$
\begin{array}{l}
3 r-1-k=8 r+2 \\
\Rightarrow 3 r-1-k-8 r-2=0 \Rightarrow k=-5 r-3 \\
\Rightarrow k=-5\left(-\frac{3}{2}\right)-3 \quad\left(\because r=-\frac{3}{2}\right) \\
\Rightarrow k=\frac{15}{2}-3 \Rightarrow k=\frac{9}{2}
\end{array}
$$
Hence, the correct answer is option: 'B'.