JEE Advanced · Mathematics · 31. 3D Geometry
If the straight lines \(\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}\) and \(\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}\) are coplanar, then the plane (s) containing these two lines is (are)
- A \(y+2 z=-1\)
- B \(y+z=-1\)
- C \(y-z=-1\)
- D \(y-2 z=-1\)
Answer & Solution
Correct Answer
(C) \(y-z=-1\)
Step-by-step Solution
Detailed explanation
Given that lines are coplanar.
\(\therefore\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|=\left|\begin{array}{ccc}
2 & 0 & 0 \\
2 & k & 2 \\
5 & 2 & k
\end{array}\right|=0 \Rightarrow k=\pm 2\)
For \(k=2\), equation of the plane is given by \(\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2\end{array}\right|=0 \Rightarrow y-z+1=0\)
For \(k=-2\), equation of the plane is given by \(\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & -2 & 2 \\ 5 & 2 & -2\end{array}\right|=0 \Rightarrow y+z+1=0\)
\(\therefore\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|=\left|\begin{array}{ccc}
2 & 0 & 0 \\
2 & k & 2 \\
5 & 2 & k
\end{array}\right|=0 \Rightarrow k=\pm 2\)
For \(k=2\), equation of the plane is given by \(\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2\end{array}\right|=0 \Rightarrow y-z+1=0\)
For \(k=-2\), equation of the plane is given by \(\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & -2 & 2 \\ 5 & 2 & -2\end{array}\right|=0 \Rightarrow y+z+1=0\)
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