MHT CET · Maths · Three Dimensional Geometry
The equation of the plane which passes through \((2,-3,1)\) and is normal to the line joining the points \((3,4,-1)\) and \((2,-1,5)\) is given by
- A \(x+5 y-6 z+19=0\)
- B \(x-5 y+6 z-23=0\)
- C \(x+5 y+6 z+7=0\)
- D \(x-5 y-6 z-11=0\)
Answer & Solution
Correct Answer
(A) \(x+5 y-6 z+19=0\)
Step-by-step Solution
Detailed explanation
The required plane passes through \((2,-3,1)\).
It is normal to the line having d.r.s. \((1,5,-6)\).
\(\therefore \mathrm{x}+5 \mathrm{y}-6 \mathrm{z}=\mathrm{k} \Rightarrow 2+5(-3)-6(1)=1\) i.e. \(\mathrm{k}=-19\)
Hence equation of plane is \(x+5 y-6 z+19=0\)
It is normal to the line having d.r.s. \((1,5,-6)\).
\(\therefore \mathrm{x}+5 \mathrm{y}-6 \mathrm{z}=\mathrm{k} \Rightarrow 2+5(-3)-6(1)=1\) i.e. \(\mathrm{k}=-19\)
Hence equation of plane is \(x+5 y-6 z+19=0\)
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