Prove 3 vectors are collinear I am asked to prove A(2,4), B(8,6), C(11,7) are collinear using vectors.
I can work AB by subtracting A from B and BC by subtracting B from C in vector form.
I can say that BC = 2AB.
But I don't understand why this proves they are collinear.
 A: If $A=(2,4), \; B=(8,6), \; C=(11,7)$, then
$$
\vec{AB} = B - A = (6,2) \quad ; \quad
\vec{BC} = C - B = (3,1)
$$
So $\vec{AB} = 2 \vec{BC}$, which is different from your conclusion.
Anyway, this says that $\vec{AB}$ and $\vec{BC}$ are parallel, since one is just a multiple of the other. So, to get from $B$ to $C$, you go in the same direction you went to get from $A$ to $B$ (no turn is required). This means that $A$, $B$, $C$ must all lie on the same line.
A: Because now you can parametrice:
$$A=B-BA$$
$$B=B$$
$$C=B-BC=B+\frac12 BA$$
Now can you see that they are on one line?
A: I think the easiest way to check this is by slopes: three points on a plane are collinear if and only if the slopes between any two of them are equal, and here you have
$$m_{AB}=m_{AC}=m_{BC}=\frac13$$
A: They're indeed colinear.
To see this, use:

Proposition 0. Given some points $A_0,\cdots,A_{n-1}$ in $\mathbb{R}^n$, these points are colinear iff the span of $\{A_1-A_0,\cdots,A_{n-1}-A_0\}$ has dimension at most $1$.

So begin by computing $B-A$ and $C-A$:
$$B-A = (6,2), \qquad C-A = (9,3)$$
Now you can use linear algebra to find out the dimension of the span of the above two vectors; in fact, it is obviously just $1$, since they're scalar multiples of each other. Hence the original triple is indeed colinear.
There's higher dimension versions, of course. For example:

Proposition 1. Given some points $A_0,\cdots,A_{n-1}$ in $\mathbb{R}^n$, these points are coplanar iff the span of $\{A_1-A_0,\cdots,A_{n-1}-A_0\}$ has dimension at most $2$.

The take-home message is that we can often answer questions about affine geometry by first subtracting by an appropriate vector so that now we're just doing linear algebra.
A: Two vectors are colinears (and so your the end points) if, and only if there one is multiple of other.
