MHT CET · Maths · Three Dimensional Geometry
The equation of a plane passing through the intersection of two planes \(x+2 y-3 z+2=0\) and \(6 x+y+z+1=0\) and parallel to the line \(x-1=y+2=7-z\) is
- A \(5 x-y+4 z+1=0\)
- B \(5 x+y+4 z+1=0\)
- C \(5 x-y+4 z=1\)
- D \(5 x+y+4 z=1\)
Answer & Solution
Correct Answer
(C) \(5 x-y+4 z=1\)
Step-by-step Solution
Detailed explanation
Equation of plane passing through the line of intersection of given planes is \((x+2 y-3 z+2)+\lambda(6 x+y+z+1)=0\)
\((1+6 \lambda) x+(2+\lambda) y+(-3+\lambda) z+(2+\lambda)=0\)
This is parallel to the line \(\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-7}{-1}\)
\(\therefore(1+6 \lambda)(1)+(2+\lambda)(1)+(-3+\lambda)(-1)=\) \(0 \Rightarrow \lambda=-1\)
Hence required equation of plane is
\(-5 x+y-4 z+1=0 \Rightarrow 5 x-y+4 z=1\)
\((1+6 \lambda) x+(2+\lambda) y+(-3+\lambda) z+(2+\lambda)=0\)
This is parallel to the line \(\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-7}{-1}\)
\(\therefore(1+6 \lambda)(1)+(2+\lambda)(1)+(-3+\lambda)(-1)=\) \(0 \Rightarrow \lambda=-1\)
Hence required equation of plane is
\(-5 x+y-4 z+1=0 \Rightarrow 5 x-y+4 z=1\)
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