C4 Jan 12 Edexcel Post Exam Discussion - Solutions in first post
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsLol exactly what im thinking. Simple vectors as well please(Original post by Sarah456)
Good luck to everyone. High probability i will screw up =/. I hope they gave simple differential equations - nothing too tricky -_-
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Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsA big chance we won't be getting a simple differential equation as the last two years - they were pretty simple. I find vectors comparably alot easier..but yeah, its usually the last part for me, which really kicks me...Damn you C4..-__-and C3...but not as much. =/
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Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tips
Last minute question!
If you have a line, say (1,2,3) + lambda(2,0,1) and you want to find the coordinates of a point on the line, which you know to be perpendicular (from the line to the origin) how would you work out it's coordinates?
I thought you would dot product (x,y,z) and (2,0,1) and then get: 2x+z=0. Then I would sub in random numbers, like x=2, to find z=-4, to get (2,0,-1), but this isn't correct apparently. :-S -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsSorry, I missed out that there's another point, A, on the line, (1,2,3).(Original post by anjelofernando)
do you know the correct answer
But then, say you let the point where the line is perp. to the origin be F, then it has coordinates (x,y,z).
Then, if you work out AF (-1+x,-2+y,-3+z) and OF(x,y,z) and dot product, then you have three unknowns- which is more confusing :-SLast edited by lekha2611; 25-01-2012 at 08:54. -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tips
On vector questions where is asks you to find the coordinates of A (the point of intersection - this tends to be a 1 mark question, I think), I always find lambda and the other symbol first, then substitute in to find A. This is really long though (far too long for one mark!) so how else could I find A?
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Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsPost the question so that we can help you(Original post by alexsasg)
On vector questions where is asks you to find the coordinates of A (the point of intersection - this tends to be a 1 mark question, I think), I always find lambda and the other symbol first, then substitute in to find A. This is really long though (far too long for one mark!) so how else could I find A?
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Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsthis is my opinion im not that sure if im correct though(Original post by lekha2611)
No sorry, but I'm really just looking for methods tbh
the problem is that xyz in this case is positional vector not a directional vector so it cant be used in the dot product , and the data is slightly not enough -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsI know it's a dodgy question, but how else would you find the position vector of a point on the line, given another point and the vector equation of the line itself?(Original post by anjelofernando)
this is my opinion im not that sure if im correct though
the problem is that xyz in this case is positional vector not a directional vector so it cant be used in the dot product , and the data is slightly not enough -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tips
Good luck all

If you see a question that you can't solve immediately, just stay calm and think clearly about what the question is asking you. Remember that all the information you need to answer a question will be written down, or you will have worked it out previously. -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsCan you post an example because the way you're describing it isn't making much sense to me?(Original post by lekha2611)
I know it's a dodgy question, but how else would you find the position vector of a point on the line, given another point and the vector equation of the line itself? -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipshttp://dl.dropbox.com/u/36280449/TSR...20question.bmp(Original post by hassi94)
Can you post an example because the way you're describing it isn't making much sense to me? -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsi got an answer -1,2,2 but in a very long and not sure way(Original post by lekha2611)
Sorry, I missed out that there's another point, A, on the line, (1,2,3).
But then, say you let the point where the line is perp. to the origin be F, then it has coordinates (x,y,z).
Then, if you work out AF (-1+x,-2+y,-3+z) and OF(x,y,z) and dot product, then you have three unknowns- which is more confusing :-S -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tips
I used the dot theorem with OF and the line l1 to find when their product is equal to zero.
OF is given by the equation as (1+2λ,2,3+λ) and the direction of the line as (2,0,1)
Then using the dot product
(2)(1+2λ)+(1)(3+λ)=0
2+4λ+3+λ=0
5λ=-5
so λ = -1Last edited by jhonwds; 25-01-2012 at 09:29. -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsWhy is OF (1+2λ,2,3+λ)?(Original post by jhonwds)
I used the dot theorem with OF and the line l1 to find when their product is equal to zero.
OF is given by the equation as (1+2λ,2,3+λ) and the direction of the line as (2,0,1)
Then using the dot product
(2)(1+2λ)+(1)(3+λ)=0
2+4λ+3+λ=0
5λ=-5
so λ = -1 -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsi got lambda as -1(Original post by jhonwds)
I used the dot theorem with OF and the line l1 to find when their product is equal to zero.
OF is given by the equation as (1+2λ,2,3+λ) and the direction of the line as (2,0,1)
Then using the dot product
(2)(1+2λ)+(1)(3+λ)=0
2+4λ+3+λ=0
5λ=-5
so λ = -1 -
Re: C4 Jan 12 Edexcel - Past Papers, Model answers, tipsThe equation of a line in vector form, gives you the position vector of any point that lies on the line.(Original post by lekha2611)
Why is OF (1+2λ,2,3+λ)?
Taking the line equation we have (1,2,3)+λ(2,0,1) we can say that any point that lies on the line is given by (1+2λ,2+λ(0),3+λ) so (1+2λ,2,3+λ).

