Equation of a Plane with Three Points

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Equation of plane containing points (a,0,0) (0,b,0) (0,0,c)

Vectors
<-a,b,0> <-a,0,c>
Normal vector
<bc,ac,ab>

Plane
Bc(x-a)+ac(y-b)+ab(z-c)=
Bcx+acy+Abz=3abc
Book is showing = abc
 
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hi nameVoid! :smile:
nameVoid said:
Bc(x-a)+ac(y-b)+ab(z-c)=

nooo … bc(x-a)+ac(y-b)+ab(z-c) = -2bc :wink:
 
nameVoid said:
Equation of plane containing points (a,0,0) (0,b,0) (0,0,c)
Normal vector
<bc,ac,ab>
If the point <x, y, z> lies in the plane, what do you have to subtract to get a vector parallel to the plane?
 
nameVoid said:
Equation of plane containing points (a,0,0) (0,b,0) (0,0,c)

Vectors
<-a,b,0> <-a,0,c>
Normal vector
<bc,ac,ab>

Plane
Bc(x-a)+ac(y-b)+ab(z-c)=

(a,b,c) is not a point on the plane, but (a,0,0) is. And what should the right side =?
 
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