Proving A,B and C are Collinear

[danago]

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


Integral

 

[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 ?

 

[danago]

Hey. Heres the question:

Points A, B and C have position vectors 3i-j, -i+15j and 9i-25j respectively. Use vectors to prove that A, B and C are collinear.

Ive drawn a diagram:
http://img2.freeimagehosting.net/uploads/b74251caf2.gif

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 doesn't give answers to questions that are more than 1 line

Thanks in advance,
Dan.

 

[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)

 

[danago]

ok thanks very much for that.

And yea, i made a mistake in my sketch.

 

[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

Or your drawing works