1. Vector Help

Sorry to bother you guys twice on the same day but hopefully, you'll be able to help me out.

I have drawn a diagram (a lovely diagram if I say so myself...) and here is what I need to do:

Find the ratio in which M divides AC and BD

Also, AB is parallel to DC and HALF ITS LENGTH.

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First, I tried to find AC (which turns out to be y+2x) and then AM would be k(y+2x) where k is a scalar but apart from that, I truly am stuck. I have a feeling I need to somehow find AM and BM but how?

Thanks.

2. Hello,

Originally Posted by sqleung
Sorry to bother you guys twice on the same day but hopefully, you'll be able to help me out.
It's not a problem ^-^

I have drawn a diagram (a lovely diagram if I say so myself...) and here is what I need to do:

Find the ratio in which M divides AC and BD

Also, AB is parallel to DC and HALF ITS LENGTH.

-----

First, I tried to find AC (which turns out to be y+2x) and then AM would be k(y+2x) where k is a scalar but apart from that, I truly am stuck. I have a feeling I need to somehow find AM and BM but how?

Thanks.
I don't know how you calculated AC...

Nevermind, you don't need it !

Do you know Thales theorem ?
AB is parallel to CD.
Therefore, there is this equality :
$\frac{AB}{CD}=\frac{AM}{AC}=\frac{BM}{BD}$

3. Sorry, that was meant to say DC, not AC

Anyway, I am not familiar with the Thale's Theorem you speak of. Is there perhaps another way around this problem?

4. Originally Posted by sqleung
Sorry, that was meant to say DC, not AC

Anyway, I am not familiar with the Thale's Theorem you speak of. Is there perhaps another way around this problem?
DC=2x, I don't see why you talk about y o.O

For Thales Theorem... I see it's called "intercept theorem" in English : Intercept theorem - Wikipedia, the free encyclopedia

5. Whoops. It actually IS AC I calculated, not DC.

With vector addition, I found that it's y + 2x

Maybe my diagram's a little TOO small. There are vectors drawn on there (probably not very clear) - vector x and vector y

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I've never used the intercept theorem before either. I was thinking about trying to use the properties of vectors or something to figure it out because I'm pretty sure that's what the teacher is after because he'd be pretty amazed and confused about me using a theorem that we haven't come across before.

Sorry for being a pain and not explaining it properly before.

6. Oi !
I've just understood that you were talking about vectors !!!
I thought it was the distance AC you were talking about because for me, x and y are lengths, sorry ~

Use this : $$\vec{...}$$

7. Originally Posted by Moo
Oi !
I've just understood that you were talking about vectors !!!
I thought it was the distance AC you were talking about because for me, x and y are lengths, sorry ~

Use this : $$\vec{...}$$
Probably was my fault for making such a tiny picture!

Anyway, now that we've got that cleared up, I think you may have made a little error in your Math tags somewhere.

8. Originally Posted by sqleung
...

I have drawn a diagram (a lovely diagram if I say so myself...) and here is what I need to do:

Find the ratio in which M divides AC and BD
...
According to your drawing you get:

$\overrightarrow{AC}=\vec y + 2 \vec x$ ......... Since M is placed on this vector $\overrightarrow{AM}=k(\vec y + 2 \vec x)$

The direction of

$\overrightarrow{BD}=-\vec x + \vec y$

If you go from A to M you have 2 different ways:

$\overrightarrow{AM}=k(\vec y + 2 \vec x)$ ........ [1]

$\overrightarrow{AM}=\vec x + m(-\vec x+ \vec y)$ ........ [2]

Both way must be equal because they both end at M:

$k(\vec y + 2 \vec x) = \vec x + m(-\vec x+ \vec y)$

$k \vec y + 2k \vec x = (1-m)\vec x + m \vec y$

${\color{red}2k} \vec x + {\color{green} k} \vec y = {\color{red}(1-m)}\vec x + {\color{green}m }\vec y$

That means you get a system of simultaneous equations:

$\left| \begin{array}{lcr}2k&=&1-m \\k&=&m\end{array}\right.$

Solve for k and m. I've got $k=m=\frac13$

Since $\overline{AM}=\frac13 \cdot \overline{AC}$ the ratio is 1 : 2.

9. Thankyou both very much for your help