Maths Ext 1 - Got Some Questions (1 Viewer)

Life'sHard

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Vector Q I've been staring at for ages.
 

tickboom

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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 ...

 

cassicowfan

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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

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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

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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

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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.
 

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