MHT CET · Maths · Three Dimensional Geometry
The equation of a plane, containing the line of intersection of the planes \(2 x-y-4=0\) and \(y+2 z-4=0\) and passing through the point \((2,1,0)\), is
- A \(3 x-2 y+z=4\)
- B \(3 x+2 y+z=4\)
- C \(3 x-2 y-z=4\)
- D \(3 x+2 y-z=-4\)
Answer & Solution
Correct Answer
(C) \(3 x-2 y-z=4\)
Step-by-step Solution
Detailed explanation
Equation of plane passing through the intersection of given planes is
\(2 x-y-4+\lambda(y+2 z-4)=0\)
Since, the plane passes through \((2,1,0)\)
\(\begin{aligned}
& 2(2)-1-4+\lambda(1+2(0)-4)=0 \\
& 4-1-4-3 \lambda=0 \\
& -1-3 \lambda=0 \\
& \lambda=\frac{-1}{3}
\end{aligned}\)
Substituting \(\lambda=\frac{-1}{3}\) in equation (i), we get
\(\begin{aligned}
& 2 x-y-4-\frac{1}{3}(y+2 z-4)=0 \\
& 6 x-3 y-12-y-2 z+4=0 \\
& 6 x-4 y-2 z-8=0 \\
& 3 x-2 y-z=4
\end{aligned}\)
\(2 x-y-4+\lambda(y+2 z-4)=0\)
Since, the plane passes through \((2,1,0)\)
\(\begin{aligned}
& 2(2)-1-4+\lambda(1+2(0)-4)=0 \\
& 4-1-4-3 \lambda=0 \\
& -1-3 \lambda=0 \\
& \lambda=\frac{-1}{3}
\end{aligned}\)
Substituting \(\lambda=\frac{-1}{3}\) in equation (i), we get
\(\begin{aligned}
& 2 x-y-4-\frac{1}{3}(y+2 z-4)=0 \\
& 6 x-3 y-12-y-2 z+4=0 \\
& 6 x-4 y-2 z-8=0 \\
& 3 x-2 y-z=4
\end{aligned}\)
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