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Question
A plane passes through a fixed point (a,b,c). Then the locus of the foot of the perpendicular to it from the origin is
A plane passes through a fixed point (a,b,c). Then the locus of the foot of the perpendicular to it from the origin is
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Solution
The correct option is D x2+y2+z2−ax−by−cz=0
Let (α,β,γ) be the foot of the perpendicular from the origin to the plane.
Now this plane passes through (α,β,γ) and the direction ratios of its normal is α,β,γ.
Therefore, the equation of this plane is given by
α(x−α)+β(y−β)+γ(z−γ)=0
This plane pass through the fixed point (a,b,c).
⇒ α(a−α)+β(b−β)+γ(c−γ)=0⇒aα−α2+bβ−β2+cγ−γ2=0⇒α2+β2+γ2−aα−bβ−cγ=0
Hence, the locus of the foot of perpendicular to the variable plane from the origin is
x2+y2+z2−ax−by−cz=0
x2+y2+z2−ax−by−cz=0
Let (α,β,γ) be the foot of the perpendicular from the origin to the plane.
Now this plane passes through (α,β,γ) and the direction ratios of its normal is α,β,γ.
Therefore, the equation of this plane is given by
α(x−α)+β(y−β)+γ(z−γ)=0
This plane pass through the fixed point (a,b,c).
⇒ α(a−α)+β(b−β)+γ(c−γ)=0⇒aα−α2+bβ−β2+cγ−γ2=0⇒α2+β2+γ2−aα−bβ−cγ=0
Hence, the locus of the foot of perpendicular to the variable plane from the origin is
x2+y2+z2−ax−by−cz=0
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