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 $
