The vector P is parallel to the 2i−j so it has a general form p(2i−j) where p is a constant. The vector Q is parallel to the i+3j so it has a general form q(i+3j) where q is a constant.
Then when these vectors are added together they are equal to the vector F.
Try it from here.
Solution in spoiler.
Spoiler
Thank you! The amount of marks it carried really put me off :/
sometimes i find the distribution of marks to be crazy
Exactly!
I've finished that paper now and I honestly didn't think there was much of a difference between that and an ordinary paper... I'll do June 2015 IAL today and I'll probably do the gold papers tomorrow Hopefully that should be enough
Maths was not changed in 2008 like everything else, so the longer questions that were introduced in most subjects after 2009 when the A* came in don't apply to Maths.
In recent M1 papers, questions from the early 2000s have been recycled, so these papers are worth doing.
Maths was not changed in 2008 like everything else, so the longer questions that were introduced in most subjects after 2009 when the A* came in don't apply to Maths.
In recent M1 papers, questions from the early 2000s have been recycled, so these papers are worth doing.
I won't have time to do loads, so which papers would be worth doing from 2000's - top 3/4??
The vector P is parallel to the 2i−j so it has a general form p(2i−j) where p is a constant. The vector Q is parallel to the i+3j so it has a general form q(i+3j) where q is a constant.
Then when these vectors are added together they are equal to the vector F.
Try it from here.
Solution in spoiler.
Spoiler
Your method makes a lot more sense than what I did. I mean, I got the right answers but I went about it in a different (and probably unnecessarily long) way:
P + Q = F
so I let P = (ai + bj) Q = (ci + dj)
Then I said k(ai + bj) = (2i-j) (aki + bkj) = (2i-j) ak = 2 --> k = 2/a bk = -1 ---> k = -1/b
2/a = -1/b ---> 2b + a = 0
Then I said y(ci + dj) = (i + 3j) cy = 1 ---> y = 1/c dy = 3 ---> y = 3/d 1/c = 3/d ------> 3c - d = 0 (1)
Then...
From P + Q = F,
(ai + bj) + (ci + dj) = (9i + 13j) (a + c)i = 9i -----> a = 9 - c. (b + d)j = 13 ------> b = 13 - d.
Subbing a and b into 2b + a = 0 ---> 2(13-d) + (9-c) = 0 ---> c + 2d = 35 (2)
I then solved 1 & 2 simultaneously to get c = 5, d = 15 ---> Q = (5i + 15j)
Then to work out P I said (ai + bj) = (9i + 13j) - (5i + 15j) (ai + bj) = (4i -2j) --> a = 4, b = -2.
Giving the answers P = (4i - 2j), Q = (5i + 15j).
NOW, I know this method is absolutely stupid and pointlessly long, but would I still get the marks??