MHT CET · Maths · Three Dimensional Geometry
The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k} \quad\) and \(\frac{x-1}{\mathrm{k}}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar if
- A \(\mathrm{k}=1\) or \(\mathrm{k}=-1\)
- B \(\mathrm{k}=0\) or \(\mathrm{k}=-3\)
- C \(\mathrm{k}=3\) or \(\mathrm{k}=-3\)
- D \(\mathrm{k}=0\) or \(\mathrm{k}=3\)
Answer & Solution
Correct Answer
(B) \(\mathrm{k}=0\) or \(\mathrm{k}=-3\)
Step-by-step Solution
Detailed explanation
If the given lines are co-planar, we get \(\left|\begin{array}{ccc}2-1 & 3-4 & 4-5 \\ 1 & 1 & -k \\ k & 2 & 1\end{array}\right|=0\)
\(\begin{array}{ll}\therefore & \left|\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1\end{array}\right|=0 \\ \therefore & 1(1+2 k)+1\left(1+k^2\right)-1(2-k)=0 \\ \therefore & 1+2 k+1+k^2-2+k=0 \\ \therefore & k^2+3 k=0 \\ \therefore & k=0 \text { or } k=-3\end{array}\)
\(\begin{array}{ll}\therefore & \left|\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1\end{array}\right|=0 \\ \therefore & 1(1+2 k)+1\left(1+k^2\right)-1(2-k)=0 \\ \therefore & 1+2 k+1+k^2-2+k=0 \\ \therefore & k^2+3 k=0 \\ \therefore & k=0 \text { or } k=-3\end{array}\)
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