Maths Ext 1 - Got Some Questions (1 Viewer)

[Life'sHard]

Q

Vector Q I've been staring at for ages.

 

[tickboom]

Does this help? It's a similar question and I reckon if you use the same approach, you should be able to work it out ...

 

[Life'sHard]

Does this help? It's a similar question and I reckon if you use the same approach, you should be able to work it out ...

Idk I'm still failing to see the link.

 

[Life'sHard]

https://www.youtube.com /watch?v=iqd9XbUHAwY&feature=emb_title

i put a space so it doesnt automatically made it media file

No I mean the link between the questions lmao. Not the video.

 

[cassicowfan]

idk if its how ur supposed to but u can prove DEC and AEF are similar

so EF/DE=AF/DC

AC=AD+DC
=-DA+DC
=v-u

bcus DE = DA + AE
DE = u + 2/5(AC)
= u + 2/5(v-u)

and EF = EA + AF
= -2/5(AC) + AF
= -2/5(v-u) + AF

so bcus of the similar

(2/5(u-v) + AF)/(u + 2/5(v-u)) = AF/DC

and then u can solve algebraically for AF/DC and u get AF/DC = 2/3

 

[Life'sHard]

idk if its how ur supposed to but u can prove DEC and AEF are similar

so EF/DE=AF/DC

AC=AD+DC
=-DA+DC
=v-u

bcus DE = DA + AE
DE = u + 2/5(AC)
= u + 2/5(v-u)

and EF = EA + AF
= -2/5(AC) + AF
= -2/5(v-u) + AF

so bcus of the similar

(2/5(u-v) + AF)/(u + 2/5(v-u)) = AF/DC

and then u can solve algebraically for AF/DC and u get AF/DC = 2/3

That's actually insane wth. I still want to figure out a vector way but this makes sense tyvm.

 

[CM_Tutor]

To start with, we have and and we seek given that .

Start by expressing and in terms of and , so that we can make use of the given information:

​

Now, we know that and so our goal if to show that , so we need to find a way to get to from what we know about ... and we might note that we haven't yet used the fact that is a parallelogram.

I notice that I want and I have , so we might seek , which we can see is a part of , as is :

​

And, since lies between and , it follows that

​

But, we also know that lies between and , and that (since we have a parallelogram) and so

​

We thus have two forms for :

​

and noting that the vectors and have non-zero magnitude and are not parallel (as they represent sides of a parallelogram), we can conclude that

​

Rearranging the first equation allows us to find :

​

from which we can find :

​

And hence, we have shown that

​

as required.

 

[Life'sHard]

To start with, we have and and we seek given that .

Start by expressing and in terms of and , so that we can make use of the given information:

​

Now, we know that and so our goal if to show that , so we need to find a way to get to from what we know about ... and we might note that we haven't yet used the fact that is a parallelogram.

I notice that I want and I have , so we might seek , which we can see is a part of , as is :

​

And, since lies between and , it follows that

​

But, we also know that lies between and , and that (since we have a parallelogram) and so

​

We thus have two forms for :

​

and noting that the vectors and have non-zero magnitude and are not parallel (as they represent sides of a parallelogram), we can conclude that

​

Rearranging the first equation allows us to find :

​

from which we can find :

​

And hence, we have shown that

​

as required.

Thanks! I’ve also worked it out this morning using vector method like this. This however has nice explanations for future people who want to learn how to solve it.