Q86 / 180 · KCET 2014 · Mathematics
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 co-planar if
- Correct answerA\( \mathrm{K}=0 \)
- B\( K=-1 \)
- C\( \mathrm{K}=2 \)
- D\( \mathrm{K}=3 \)
Solution
Given equation,
\[ \begin{array}{l} \frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K} \rightarrow(1) \\ \frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1} \rightarrow(2) \end{array} \]
For co-planarity, we have
\[ \begin{array}{l} \left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{2} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|=0 \\ \Rightarrow\left|\begin{array}{cccc} 1-2 & 4-3 & 5-4 \\ 1 & 1 & -K \\ K & 2 & 1 \end{array}\right|=0 \\ \Rightarrow\left|\begin{array}{cccc} 1 & 1 & 1 & -K \\ K & 2 & 1 & =0 \end{array}\right|=0 \end{array} \]
\[ \begin{array}{l} \frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K} \rightarrow(1) \\ \frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1} \rightarrow(2) \end{array} \]
For co-planarity, we have
\[ \begin{array}{l} \left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{2} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|=0 \\ \Rightarrow\left|\begin{array}{cccc} 1-2 & 4-3 & 5-4 \\ 1 & 1 & -K \\ K & 2 & 1 \end{array}\right|=0 \\ \Rightarrow\left|\begin{array}{cccc} 1 & 1 & 1 & -K \\ K & 2 & 1 & =0 \end{array}\right|=0 \end{array} \]
Hence, the correct answer is option: 'A'.