# Proving A,B and C are Collinear

1. Apr 1, 2006

### danago

I merged the duplicate thread. This post cannot be deleted!

Integral

Last edited by a moderator: Apr 1, 2006
2. Apr 1, 2006

### Fermat

If A, B and C all lie on the same line, then what can you say about the direction of the lines, AB,AC ?

3. Apr 1, 2006

### danago

Hey. Heres the question:

Ive drawn a diagram:

a=3i-j
b=-i+15j
c=9i-25j

So pretty much, i think i need to prove that $$\overrightarrow{BA}=h\overrightarrow{BC}$$

Ive found that
$$\overrightarrow{BA}=\overrightarrow{BO}+\overrightarrow{OA}$$
$$=-b+a$$
$$=4i-16j$$

$$\overrightarrow{BC}=\overrightarrow{BO}+\overrightarrow{OC}$$
$$=-b+c$$
$$=10i-40j$$

From that, i can see that the i and j components have a set ratio. ie. i:j = 1:4.

For this question, what would i write as my final proof that the three points are collinear? I would use the answers page in my textbook, but it doesnt give answers to questions that are more than 1 line

Dan.

Last edited by a moderator: May 2, 2017
4. Apr 1, 2006

### Fermat

You solution is correct up to the same ratio bit.
Thereafter, sinmply say that, because of the same ratio,
|BC| is a multiple of |AB|
hence BC and AB are parallel.
Since they share a common point, B, then they are collinear.

BTW, your sketch looks like it has OA at (3i + j) rather than (3i - j)

5. Apr 1, 2006

### danago

ok thanks very much for that.

And yea, i made a mistake in my sketch.

6. Apr 1, 2006

### NateTG

Hmm...
Is there an example in the text somewhere?
Since 'use vectors to' is pretty vauge, you can do this a bunch of ways.

For example, you could use the dot product
$$\frac{(\vec{b}-\vec{c}) \cdot (\vec{a}-\vec{b})}{| (\vec{b}-\vec{c})| |(\vec{a}-\vec{b})|}=\pm 1$$
or the cross product
$$(\vec{b}-\vec{c}) \times (\vec{a}-\vec{b}) = \vec{0}$$

Or you could show that all the vectors are on the line
$$y=-4x+11$$