Prove 3 vectors are collinear

Question

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.

Answer 1

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.

Answer 2

Two vectors are colinears (and so your the end points) if, and only if there one is multiple of other.

Answer 3

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?

Answer 4

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

$$m_{AB}=m_{AC}=m_{BC}=\frac13$$

Answer 5

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.