3D Geometry 2 Question 6
6. 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 the value of $k$ is
(a) $\frac{3}{2}$
(b) $\frac{9}{2}$
(c) $-\frac{2}{9}$
(d) $-\frac{3}{2}$
(2004, 1M)
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Answer:
Correct Answer: 6. (b)
Solution:
- Since, the lines intersect, therefore they must have a point in common, i.e.
and
$ \begin{aligned} & \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \\ & \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu \\ & x=2 \lambda+1, y=3 \lambda-1 \\ & z=4 \lambda+1 \end{aligned} $
$ \Rightarrow \quad x=2 \lambda+1, y=3 \lambda-1 $
and $\quad x=\mu+3, y=2 \mu+k, z=\mu$ are same.
$ \Rightarrow \quad \begin{aligned} 2 \lambda+1 & =\mu+3 \\ 3 \lambda-1 & =2 \mu+k \\ 4 \lambda+1 & =\mu \end{aligned} $
On solving Ist and IIIrd terms, we get,
$ \begin{aligned} & \lambda & =-\frac{3}{2} \text { and } \mu=-5 \\ \therefore & k & =3 \lambda-2 \mu-1 \end{aligned} $
$ \begin{array}{lll} \Rightarrow & & k=3-\frac{3}{2}-2(-5)-1=\frac{9}{2} \\ \therefore & k & =\frac{9}{2} \end{array} $
