Showing that three vectors with the same origin have their endpoints on the same line I just started taking a course on Linear Algebra at university, and there is one problem that has kept me stumped for hours now:


*

*there are three vectors a, b, c in 2d space

*they all have the same origin

*they satisfy the equation 2a - 3b + c = 0


I'm supposed to show that the endpoints of those vectors are all on the same line.
Now, taking any two vectors with the same origin, you can always find a line that goes through their endpoints.
I got to the point that they must lie on the same line if (-c + b) = (-c + a) + (-a + b), but this is far from constituting a proof, and I don't see where the equation mentioned above comes into play.
Any help is greatly appreciated.
Thanks!
 A: Say, $\vec{a}=\vec{OA}$, $\vec{b}=\vec{OB}$, $\vec{c}=\vec{OC}$.
Note that, $2\vec{a}-3\vec{b}+\vec{c}=0\implies3(\vec{a}-\vec{b})=\vec{a}-\vec{c}$
Then $\vec{BA}=\vec{a}-\vec{b}$ and $\vec{CA}=\vec{a}-\vec{c}$ must be co-linear. But then $A,B,C$ are co-linear.
A: You have $2(\vec{a}-\vec{b}) = (\vec{b}-\vec{c})$ or $3(\vec{a}-\vec{b}) = (\vec{a}-\vec{c})$.  
So the three points must lie on a line with the point $b$ a third of the way along the line from $a$ to $c$. 
A: $$2a-3b+c=0$$ then 
$$3a-3b+c=a $$ so 
$$c+3(a-b)=a$$
$$c=-3(a-b)+a$$
Notice that $(a-b)$ is direction vector of   the line passing through the points $a$ and $b$. Thus, $c$ is also on that line as above equality shows.we are done.
A: Try to convince yourself that if all of your vectors have the same origin and do not lie on the same line then the only $c_{1},c_{2},c_{3}\in \mathbb{R}$ such that $c_{1}x+c_{2}y+c_{3}z=0$ is $c_{1}=c_{2}=c_{3}=0$. Hint: If they have the same origin then the only time the cross is if they lie on the same line. 
Now, if $a,b$ lie on the same line then it must be the case that we can write $a=c_{1}b$ with $c_{1}>0$. If $b,c$ lie on the same line then it must be the case that we can write $b=c_{2}c$. If $c,a$ lie on the same line it must be the case that we can write $c=c_{3}a$. Subtract and we get three equations of the form $x_{i}-y_{i}=0$ for $i=1,2,3$. Now sum these equations and you should be able to figure it out from there.
