MHT CET · Maths · Three Dimensional Geometry
The equation of a plane containing the line of intersection of the planes \(2 \mathrm{x}-\mathrm{y}-4=0\) and \(\mathrm{y}+2 \mathrm{z}-4=0\) and passing through the point \((1,1,0)\) is
- A \(x-y-z=0\)
- B \(2 x-z=2\)
- C \(x-3 y-2 z=-2\)
- D \(x+3 y+z=4\)
Answer & Solution
Correct Answer
(A) \(x-y-z=0\)
Step-by-step Solution
Detailed explanation
The equation of plane containing the line of intersection of the planes
\(2 x-y-4=0\) and \(y+2 z-4=0\) is
\((2 x-y-4)+\lambda(y+2 z-4)=0\)
But it passes through \((1,1,0)\)
\(\Rightarrow \lambda=-1\)
\(\Rightarrow\) The required equation of plane is
\(\begin{aligned} & (2 x-y-4)-(y+2 z-4)=0 \\ & \Rightarrow 2 x-2 y-2 z=0 \\ & \Rightarrow x-y-z=0\end{aligned}\)
\(2 x-y-4=0\) and \(y+2 z-4=0\) is
\((2 x-y-4)+\lambda(y+2 z-4)=0\)
But it passes through \((1,1,0)\)
\(\Rightarrow \lambda=-1\)
\(\Rightarrow\) The required equation of plane is
\(\begin{aligned} & (2 x-y-4)-(y+2 z-4)=0 \\ & \Rightarrow 2 x-2 y-2 z=0 \\ & \Rightarrow x-y-z=0\end{aligned}\)
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