Help with A-Level Maths Question

[GreenRugxoxo]

DS is struggling with Maths homework. I have attached the question he can't do.
Anyone know how to do it?
Thanks in advance.

Sorry forgot to attach picture!

First, as we know OA = 3AP, that tells us that AP = OA / 3, and so OP = OA + AP = (4/3)OA. This is useful because we want to relate things to vector a which is the vector OA. (vectors in bold, scalars not in bold).

So vector OP = (4/3) OA (because OP and OA are in the same direction), so OP = (4/3) a or 4 a / 3.

I can't read it well enough to help (and might not be able to anyway these days).

Has he done the first bit and is stuck on the 'hence or otherwise', or is he stuck on the first part?

For something to be a parallelogram the vectors for opposite sides need to be the same (in both direction and magnitude).

Well if DadDadDad is on the case you don't need me anyway.

OY is trickier, but from the information given, it's clear that OY is in the same direction as OX but half the distance, so OY = (1/2) OX.

Now, we gradually need to relate OX to a and b.

First, it should be obvious that OX = OA + AX, and we know OA is a. So what about AX ? We know AX = (1/2) AB.
AB = AO + OB = -a + b (this is a common trick in vectors, so DS should get familiar with it).
So AX = (1/2) (-a + b) = -(1/2) a + (1/2) b.

Now we can put it all together:
OY = (1/2) OX
= (1/2) ( OA + AX )
= (1/2) ( a - (1/2) a + (1/2) b )
= (1/2) ( (1/2) a + (1/2) b )
= (1/4) a + (1/4) b.

I'll hit post but I have a feeling I've made a mistake.

Yes, that looks right. If you complete the parallelogram with sides OA and OB, then OX is half the diagonal, and OY is quarter of the diagonal.

Thanks, Teen, but I've got to go. I pasted the image into Word, rotated and enlarged to make it easier to read!

For the next bit, we have two ways to get from Y to Q.

Either we go [1]: YO + OP + PQ
or we go [2]: YQ

These two must be equal, [1] = [2] - write that down and express everything in terms of a and b. (You either know these already or you've just been given it). Let me know how far you get with that.

OY = 1/3a + 1/3b
YQ = OP + PQ -OY
By substituting the values of the vectors, you should get m=1 and n=1/3.

You can then write AQ = AP + PQ which should equal OY.

Similarly by showing OA = YQ OAQY is a parallelogram.

@DadDadDad @xsquared
Sorry for the delay in replying. We have been having tea. Thank you so much for these helpful posts . I will show them to DS!