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C4 Vectors doubt

Hello, I was getting completely confused on this question and hence thought I would seek ur help . Thanks for all the help in advance!

I am talking about question no .5 on here:
http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/C4/Elmwood%20Papers/Elmwood%20I.pdf

question 5b)

How do we write the vector equation of the line OM? (because this is already the position vector, isn't it ? So I was completely confused. How is this different from the following question:

(NOT FROM THE PAPER) Find the vector equation of the line l passing through A (-2,1,4) and parallel to the vector (2,3,-1). I don't understand how to find out the parallel vector in the case of OM .... HELP ?
If the side lengths are 2 units, would the position vector of OM be (1, 2, 0)? So the vector equation would just be r= (1, 2, 0) I think ?

And parallel vectors are multiples of one another, for example, the vector A (-2, 1, 4) could be parallel to vector B (-4, 2, 8)

I think this is right.. :P
Reply 2
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BabyMaths
Usually you have

r=(the position vector of a point on the line) + (a scalar) x (the direction vector)

In the case of OM the origin is a point on the line so...
Reply 3
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BabyMaths
Original post by Jordi Cornish
If the side lengths are 2 units, would the position vector of OM be (1, 2, 0)? So the vector equation would just be r= (1, 2, 0) I think ?


No, that's just a point.

The line is r=s(i+2j).
Reply 4
Avatar for davros
davros
16
Original post by laurawoods
Hello, I was getting completely confused on this question and hence thought I would seek ur help . Thanks for all the help in advance!

I am talking about question no .5 on here:
http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/C4/Elmwood%20Papers/Elmwood%20I.pdf

question 5b)

How do we write the vector equation of the line OM? (because this is already the position vector, isn't it ? So I was completely confused. How is this different from the following question:

(NOT FROM THE PAPER) Find the vector equation of the line l passing through A (-2,1,4) and parallel to the vector (2,3,-1). I don't understand how to find out the parallel vector in the case of OM .... HELP ?


The question wants you to work in terms of the unit vectors that you are given, i and j . For example, side OA has length 2 units and is in the direction of i. So you can write vector OA as 2i.

Can you see how to get from O to M via C, how to write OC as a vector, and how you might work out CM as vector in terms of the unit vectors shown in the diagram?
Original post by BabyMaths
No, that's just a point.

The line is r=s(i+2j).


But you are given values for the sides so you should use those no?
Reply 6
Avatar for BananaPie
BananaPie
Original post by laurawoods
Hello, I was getting completely confused on this question and hence thought I would seek ur help . Thanks for all the help in advance!

I am talking about question no .5 on here:
http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/C4/Elmwood%20Papers/Elmwood%20I.pdf

question 5b)

How do we write the vector equation of the line OM? (because this is already the position vector, isn't it ? So I was completely confused. How is this different from the following question:

(NOT FROM THE PAPER) Find the vector equation of the line l passing through A (-2,1,4) and parallel to the vector (2,3,-1). I don't understand how to find out the parallel vector in the case of OM .... HELP ?


To answer your last question, a parallel vector will have the same direction vector, albeit a different scale factor. How to represent this properly has been told to you by Babymaths.
Reply 7
Avatar for BabyMaths
BabyMaths
Original post by Jordi Cornish
But you are given values for the sides so you should use those no?
No. :tongue:
Reply 8
Avatar for laurawoods
laurawoods
OP
8
Original post by BabyMaths
No, that's just a point.

The line is r=s(i+2j).


Hello, I am not very sure about this question because this time they have given a POSITION VECTOR (OM) ... so?
Reply 9
Avatar for davros
davros
16
Original post by laurawoods
Hello, I am not very sure about this question because this time they have given a POSITION VECTOR (OM) ... so?


Have you tried using the i and j vectors as I suggested earlier? Can you see what OA, OC, CM etc are in term of these unit vectors?
Reply 10
Avatar for laurawoods
laurawoods
OP
8
Original post by davros
Have you tried using the i and j vectors as I suggested earlier? Can you see what OA, OC, CM etc are in term of these unit vectors?

yes, yes i have tried doing that :smile:
Reply 11
Avatar for davros
davros
16
Original post by laurawoods
yes, yes i have tried doing that :smile:


Are you still stuck on 5b? BabyMaths actually started you off by giving the equation of a general point on OM in the form
r = s(2j+ i)

Now you need to write an equation for a point on the line AB. This line doesn't go through the origin, so you will need to use the more general vector equation of a line that you're used to.
Reply 12
Avatar for laurawoods
laurawoods
OP
8
Original post by davros
Are you still stuck on 5b? BabyMaths actually started you off by giving the equation of a general point on OM in the form
r = s(2j+ i)

Now you need to write an equation for a point on the line AB. This line doesn't go through the origin, so you will need to use the more general vector equation of a line that you're used to.

Hello , I understand it now ! thanks!

Just another question that I was stuck on...pls can u help me with this one ?

http://www.carlgauss.co.uk/PDFs/A-Level%20Papers/C3/C3%20-%20Specimen%20Paper.pdf

Question no 7 part ii)

I don't know how to prove it ?

thanks for all the help in advance! :smile:
Reply 13
Avatar for davros
davros
16
Original post by laurawoods
Hello , I understand it now ! thanks!

Just another question that I was stuck on...pls can u help me with this one ?

http://www.carlgauss.co.uk/PDFs/A-Level%20Papers/C3/C3%20-%20Specimen%20Paper.pdf

Question no 7 part ii)

I don't know how to prove it ?

thanks for all the help in advance! :smile:


Not sure what techniques you have learned but I would work out dx/dy and then take the reciprocal (because dy/dx = 1/(dx/dy)).

Hint: when you have differentiated tan(y/2) you'll want to express the derivative in terms of x, and you should know a trig identity that will help you.

(Actually, if you're not comfortable with that method you can actually differentiate both sides with respect to x as it stands, and use the chain rule on the RHS. It comes to the same thing as my first suggested method!)
Reply 14
Avatar for laurawoods
laurawoods
OP
8
Original post by davros
Not sure what techniques you have learned but I would work out dx/dy and then take the reciprocal (because dy/dx = 1/(dx/dy)).

Hint: when you have differentiated tan(y/2) you'll want to express the derivative in terms of x, and you should know a trig identity that will help you.

(Actually, if you're not comfortable with that method you can actually differentiate both sides with respect to x as it stands, and use the chain rule on the RHS. It comes to the same thing as my first suggested method!)


Hello thanks ! I was a bit confused about this concept of the long term value given an exponential function:

If there is something like P= 20 + 60e^-0.1t (where P is population of a town in millions and the t is time in years)

and then something like theta = 70e^-0.1t + 2


When we are finding out the long term values in both of these cases, how is it different? Because for the first one, it would always be above 20 , but for the second one it is 2 exactly. How are they different, and how do we spot if it is exactly or round about there (if you can understand what I mean) ?

Also, if it was something like P= 20+3e^t then what would be the long term value for the population? I was a bit confused with this one because unlike the other examples, this one doesn't have a minus sign.

Thanks for all ur help! :smile: If you could clear up my misunderstandings that would be great! Thanks once again ! :smile:
(edited 11 years ago)
Reply 15
Avatar for BabyMaths
BabyMaths
Original post by laurawoods


When we are finding out the long term values in both of these cases, how is it different? Because for the first one, it would always be above 20 , but for the second one it is 2 exactly. How are they different, and how do we spot if it is exactly or round about there (if you can understand what I mean) ?

Also, if it was something like P= 20+3e^t then what would be the long term value for the population?


It's not different. 70e−0.1t+2>270e^{-0.1t}+2>2. What made you think it is ever 2?

As for your last question, well, it just keeps getting bigger, doesn't it.
Reply 16
Avatar for davros
davros
16
Original post by laurawoods
Hello thanks ! I was a bit confused about this concept of the long term value given an exponential function:

If there is something like P= 20 + 60e^-0.1t (where P is population of a town in millions and the t is time in years)

and then something like theta = 70e^-0.1t + 2


When we are finding out the long term values in both of these cases, how is it different? Because for the first one, it would always be above 20 , but for the second one it is 2 exactly. How are they different, and how do we spot if it is exactly or round about there (if you can understand what I mean) ?

Also, if it was something like P= 20+3e^t then what would be the long term value for the population? I was a bit confused with this one because unlike the other examples, this one doesn't have a minus sign.

Thanks for all ur help! :smile: If you could clear up my misunderstandings that would be great! Thanks once again ! :smile:


BabyMaths has already explained this to you but there's no difference! the 2 expressions are always bigger than 20 and 2 respectively for any finite number you put in, but as t gets bigger and bigger, the negative exponential gets smaller and smaller, so the long term values are 20 and 2.

In your final example, P just keeps getting bigger and bigger as t increases, so you can either say that P tends to infinity, or that there is no finite long-term value.

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