MHT CET · Maths · Three Dimensional Geometry
The equation of the plane through the intersection of the planes \(x+y+z=1\) and \(2 x+3 y-z+4=0\) and parallel to \(x-a x i s\) is
- A \(3 y+z-6=0\)
- B \(3 y-z+6=0\)
- C \(y-3 z+6=0\)
- D \(y+3 z-6=0\)
Answer & Solution
Correct Answer
(C) \(y-3 z+6=0\)
Step-by-step Solution
Detailed explanation
Required equation is \((x+y+z-1)+\lambda(2 x+3 y-z+4)=0\)
\(
\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+(4 \lambda-1)=0
\)
But plane is parallel to the \(\mathrm{x}\)-axis
\(
\begin{aligned}
& \Rightarrow \lambda=-\frac{1}{2} \\
& \Rightarrow-\frac{1}{2} y+\frac{3}{2} z-3=0 \\
& \Rightarrow y-3 z+6=0
\end{aligned}
\)
\(
\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+(4 \lambda-1)=0
\)
But plane is parallel to the \(\mathrm{x}\)-axis
\(
\begin{aligned}
& \Rightarrow \lambda=-\frac{1}{2} \\
& \Rightarrow-\frac{1}{2} y+\frac{3}{2} z-3=0 \\
& \Rightarrow y-3 z+6=0
\end{aligned}
\)
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