Maths A-Level

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peteryoungy
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#1
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#1
Can I please get some help on this question? Can't answer it.Name:  Screenshot 2022-05-14 at 12.47.06.png
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tonyiptony
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#2
Have you attempted the problem? It would be nice if you could show where you got stumped.
Because I think there is a very obvious first step to this problem.
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peteryoungy
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peteryoungy
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#4
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#4
I was following this template.
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tonyiptony
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(Original post by peteryoungy)
[Attempt]
Ah, I see. Often times it's more important to understand why we take certain methods, rather than "following the template".
I think you got confused at the 4\lambda -3 step.

From the example question, do you know how do we get to the 3(3\lambda -1)=4\lambda +2 step?
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Help_me_pleaseex
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(Original post by peteryoungy)
Can I please get some help on this question? Can't answer it.Name:  Screenshot 2022-05-14 at 12.47.06.png
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Hi after the vector "i" is there anything else?
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peteryoungy
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#7
(Original post by tonyiptony)
Ah, I see. Often times it's more important to understand why we take certain methods, rather than "following the template".
I think you got confused at the 4\lambda -3 step.

From the example question, do you know how do we get to the 3(3\lambda -1)=4\lambda +2 step?
No. I was stuck on that. Can you please explain.
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tonyiptony
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#8
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(Original post by peteryoungy)
No. I was stuck on that. Can you please explain.
Okay, sure.

Let's start with the example question. They want to find some vector that is parallel to i+3j.
So... what vectors are parallel to i+3j? Maybe 2i+6j, 3i+9j,-i-3j, etc...
Do you spot a pattern?
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peteryoungy
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#9
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#9
(Original post by tonyiptony)
Okay, sure.

Let's start with the example question. They want to find some vector that is parallel to i+3j.
So... what vectors are parallel to i+3j? Maybe 2i+6j, 3i+9j,-i-3j, etc...
Do you spot a pattern?
yes I do spot the patter still not sure how it relates to the question though.
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Help_me_pleaseex
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#10
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#10
(Original post by Help_me_pleaseex)
Hi after the vector "i" is there anything else?
Can someone tell me where i went wrong please!!

Key: lambda is represented by K

for this one if a = 2i + 5j and b = 3i - j
if it is parallel to the vector i then does that mean j must be 0?
so then we could do 2i + 5j + 3Ki - Kj = Hi + 0j
compare the j coefficients:
5j - Kj = 0 so Lamda = 5

Thank you
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tonyiptony
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#11
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#11
(Original post by peteryoungy)
yes I do spot the patter still not sure how it relates to the question though.

Let's take it slow. The pattern you're thinking of is probably that whatever the vector is, the ratio the coefficient of i to that of j must be 1:3. In other words, whatever a+\lambda b ends up, the ratio the coefficient of i to that of j must still be 1:3. Thus you'll get to the crucial step.


Note: I actually find \frac{\text{coef. of i}}{\text{coef. of j}}=\frac{1}{3} more readable. But whichever works works.

Can you apply the same logic to solve your original problem?
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peteryoungy
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From the example I did I see that there it is telling me to find vector I. the scale factor from 2i to 3i is 1.5 so will I need to multiply it by 1.5
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tonyiptony
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#13
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#13
(Original post by peteryoungy)
From the example I did I see that there it is telling me to find vector I. the scale factor from 2i to 3i is 1.5 so will I need to multiply it by 1.5
We actually do not care about what the scale factor is. We just want to know what the ratio of coef of i to that of j of the resulting vector should be.
Remember we are not really dealing with either vector a or b, but the resulting vector a+\lambda b
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peteryoungy
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(Original post by tonyiptony)
We actually do not care about what the scale factor is. We just want to know what the ratio of coef of i to that of j of the resulting vector should be.
Remember we are not really dealing with either vector a or b, but the resulting vector a+\lambda b
can you please do this one as an example I honestly don't know the mark scheme isn't that great. Hopefully I will understand. Also if you can do it tiny steps so I can understand it better if you wont mind.
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tonyiptony
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#15
(Original post by peteryoungy)
can you please do this one as an example I honestly don't know the mark scheme isn't that great. Hopefully I will understand. Also if you can do it tiny steps so I can understand it better if you wont mind.
The full answer is provided by Help_me_pleaseex in #10 (with some minor nitpicks, but the structure is there). But if you follow my line of logic of comparing the ratio of the coefficients, it will still end up the same.

When you encounter questions in general, the first step is usually to re-write part the main question into something more workable.
For instance here, a+\lambda b is not really workable for our purpose; rewriting a+\lambda b as "something i + something j" is.

Also I would highly suggest you work on the example question on your own, to see if you actually know what every step you're trying to achieve.
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davros
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(Original post by peteryoungy)
Can I please get some help on this question? Can't answer it.Name:  Screenshot 2022-05-14 at 12.47.06.png
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I don't normally jump in when someone else is helping, but as you've been given a complete solution and seem to be getting a bit bogged down, another way of looking at this is that if vector c is parallel to another vector d, then we must be able to write c = kd where k is some scalar.

In your case, you want a +lb = ki (I'm using 'l' instead of lambda for simplicity) When you expand out the LHS you will end up with :

(something)i + (something else)j = ki

There is no 'j' component on the RHS, so the coefficient of j on the LHS - the "something else" - must be 0. This will give you an equation involving l (or lambda in your case) which you can then solve as normal.


HTH
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Help_me_pleaseex
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#17
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#17
soooo what was the answer for lambda?
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davros
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#18
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#18
(Original post by Help_me_pleaseex)
soooo what was the answer for lambda?
Why not post your working so far and someone can help when you get stuck
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