Vectors & Section Formula: Proving Collinearity

[andrew.c]

Help! Vectors!

Homework Statement

In the triangle ABC, D divides AB in the ratio 3:2, E divides DC in the ratio 1:5 and F divides AC in the ratio 1:2. Show (using position vectors and the section formula) that B, E and F are collinear and find BE:EF

Homework Equations

Section formula = $$\frac{m \textbf{a} = n\textbf{b}}{m+n}$$

The Attempt at a Solution

Other than drawing this out, and spotting that they look like they're kind of in a line.
This isn't due for homework or anything, I'm just revising, so it would be really useful if someone cold explain this to me.

Any ideas?

 

[tiny-tim]

Hi andrew.c!

Use the section formula to find the vectors D E and F …

for example, F = … ?

 

[andrew.c]

F = $$\frac{a+2c}{3}$$ ?

 

[tiny-tim]

F divides AC in the ratio 1:2.

F = $$\frac{a+2c}{3}$$ ?

Almost … but it's closer to A,

so f = (2a + c)/3 ​

Next, what are D and E?

 

[andrew.c]

AD/DB = 3/2

sec. formula...

$$\frac{3a+2b}{5}$$

---

DE/EC = 1/5

sec. formula...

$$\frac{d+5c}{6}$$
---
I still don't really understand which value in the ratio is m and which is n, I am just reading it as AD/DB = m/n. Is this right?

 

[tiny-tim]

I still don't really understand which value in the ratio is m and which is n …

Forget the formula …

I never remember which way round it is …

I just ask myself each time "which one is it nearer?"

so if DE:EC = 1:5, then it's nearer D, so … ?

 

[andrew.c]

ok, i get that logic :)

so it would be...

$$D = \frac{2a+3b}{5}$$

and

$$E = \frac{5d+6}{c}$$

 

[tiny-tim]

Yup!

(except for the obvious mistake … quick! edit it before anyone else notices! )

ok, now you have the vectors for B E and F …

how can you prove they're collinear?

 

[andrew.c]

OK, so I have...

$$D = \frac{2a+3b}{5}$$ $$E=\frac{5d+c}{6}$$ and $$F = \frac{2a+C}{3}$$

For collinearity, I need to prove that BE is parallel to EF, with a common point at E?

I tried this, but not sure if its right - i got the bottom lines to be the same, but my notes indicate that they should be a scalar multiple of each other!

------

Sub. D into E to get...

$$E = \frac{2a+3b+c}{6}$$

$$BE = e-b =\frac{2a+3b+c}{6} - b =\frac{2a+3b+c}{6}-\frac{6b}{6} =\frac{2a-3b+c}{6}$$

$$EF = f-e =\frac{2a+c}{3}-\frac{2a+3b+c}{6} =\frac{4a-2c}{6} -\frac{2a+3b+c}{6} =\frac{2a-3b+c}{6}$$

Since they are equal, and E was a common point, they are collinear?

----------------------

Is BE:EF just 1:1 ?

Thanks for your help btw!

 

[tiny-tim]

… Since they are equal, and E was a common point, they are collinear?

----------------------

Is BE:EF just 1:1 ?

Thanks for your help btw!

Yes, that's very good

(except you typed a minus for a plus in the last line )

the vectors BE and EF are exactly the same, in other words BE = EF, so yes, obviously it's 1:1

 

[andrew.c]

Ta muchly for your help!

Now i need to tackle vectors intercepting planes. Oh joy of joys!