MHT CET · Maths · Three Dimensional Geometry
The equation of the plane, passing through the intersection of the planes \(x+y+z=1\) and \(2 x+3 y-z+4=0\) and parallel to \(Y\)-axis is
- A \(x+4 z-1=0\)
- B \(x+4 z-7=0\)
- C \(x-4 z+7=0\)
- D \(x-4 z+1=0\)
Answer & Solution
Correct Answer
(B) \(x+4 z-7=0\)
Step-by-step Solution
Detailed explanation
Equation of plane passing through the intersection of given planes is
\(\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0
\end{aligned}\)
Since the plane is parallel to Y -axis.
\(\begin{aligned}
\therefore \quad & 1+3 \lambda=0 \\
& \Rightarrow \lambda=\frac{-1}{3}
\end{aligned}\)
\(\therefore \quad\) Equation of the required plane is \(x+4 \mathrm{z}-7=0\)
\(\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0
\end{aligned}\)
Since the plane is parallel to Y -axis.
\(\begin{aligned}
\therefore \quad & 1+3 \lambda=0 \\
& \Rightarrow \lambda=\frac{-1}{3}
\end{aligned}\)
\(\therefore \quad\) Equation of the required plane is \(x+4 \mathrm{z}-7=0\)
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