3D Geometry 2 Question 5
####5. If the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, then $k$ can have
(a) any value
(b) exactly one value
(c) exactly two values
(d) exactly three values
(2012 Main)
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Answer:
Correct Answer: 5. (c)
Solution:
- Condition for two lines are coplanar.
$ \left|\begin{array}{ccc} x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right|=0 $
where, $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ are the points lie on lines (i) and (ii) respectively and $<l_{1}, m_{1}, n_{1}>$ and $<l_{2}, m_{2}, n_{2}>$ are the direction cosines of the lines (i) and (ii), respectively.
$ \begin{aligned} & \therefore \quad\left|\begin{array}{ccc} 2-1 & 3-4 & 4-5 \\ 1 & 1 & -k \\ k & 2 & 1 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{rrr} 1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{array}\right|=0 \\ & \Rightarrow \quad 1(1+2 k)+\left(1+k^{2}\right)-(2-k)=0 \\ & \Rightarrow \quad k^{2}+2 k+k=0 \\ & \Rightarrow \quad k^{2}+3 k=0 \\ & \Rightarrow \quad k=0,-3 \end{aligned} $
If 0 appears in the denominator, then the correct way of representing the equation of straight line is
$ \frac{x-2}{1}=\frac{y-3}{1} ; z=4 $
