In my book its given , three points A,B,C with position vectors a,b,c are collinear if and only if there exists scalars x,y,z not all zero simultaneously such that xa + yb + zc = 0, where x + y + z = 0.
Surprisingly, this is also the condition for coplanarity of three vectors. But All COPLANAR vectors are NOT collinear . So where is the the flaw in my argument ?
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As you write, vectors $a,b,c$ are coplanar iff there are scalars $x,y,z$, not all $0$, such that $xa+yb+zc=0$, i.e. iff they are linearly dependent.
The endpoints of $a,b,c$ are collinear iff $c-b$ is parallel to $c-a$, that is, $c-a=t(c-b)$ for some scalar $t$, assuming $b\ne c$.
But then $1a+(-t)b+(t-1)c=0$ and these coefficients sum up to $0$.
Conversely, if $xa+yb+zc=0$ with $x+y+z=0$, then either $x=0$ whence $y=-z$ and $b=c$, or we can divide by $x$ and set $t=-y/x$ to conclude $c-a=t(c-b)$.
Ok i think i've got it now. For coplanarity the condition is that ax + by + cz = 0 for scalars x,y,z where not all x,y,z = 0 , but for collinearity there is an additional condition that x + y + z = 0 so that one point can be shown to to divide the other two points in some ratio .