KCET · Maths · Three Dimensional Geometry
If \((2,3,-1)\) is the foot of the perpendicular from \((4,2,1)\) to a plane, then the equation of the plane is
- A \(2 x+y+2 z-1=0\)
- B \(2 x-y+2 z=0\)
- C \(2 x+y+2 z-5=0\)
- D \(2 x-y+2 z+1=0\)
Answer & Solution
Correct Answer
(D) \(2 x-y+2 z+1=0\)
Step-by-step Solution
Detailed explanation
Since, the line joining the two points is
perpendicular to the plane, it's DR's will given the normal to the plane.
DR's of the normal \(=(4-2,2-3,1+1)=(2,-1,2)\)
Hence, \(a=2, b=-1\) and \(c=2\)
Since, the plane passes through \((2,3,-1)\).
\(\begin{aligned} & d=2(2)+3(-1)-1(2) \\ & d=-1\end{aligned}\)
Hence, the required equation of plane is
\(2 x-y+2 z=-1\)
perpendicular to the plane, it's DR's will given the normal to the plane.
DR's of the normal \(=(4-2,2-3,1+1)=(2,-1,2)\)
Hence, \(a=2, b=-1\) and \(c=2\)
Since, the plane passes through \((2,3,-1)\).
\(\begin{aligned} & d=2(2)+3(-1)-1(2) \\ & d=-1\end{aligned}\)
Hence, the required equation of plane is
\(2 x-y+2 z=-1\)
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