Vectors question: OACB is a trapezium whose diagonals, OC and AB, intersect at P.

[shuphie]

MY WORKING:

Why I am confused: (2 things, both shown by a yellow line)
1) At the first yellow highlighted area, a 2a is included in working out AP and OP in the mark scheme, specifically the 2a is meant to be after 5by-2ay
This was my thinking (I did not include this 2a in my working, but it was added after because the mark scheme had it in):
Why is 2a in the mark scheme?
AP = y(AB)
AB = 5b-2a
AP = 5by - 2ay
OP = x(OC)
OC = 3b+2a
(This would all work the other way, right? If I chose to do OC = 5b+2a instead?)
OP = 3bx+2ax

2) Normally in vector questions we would find AP and OP then set their factors equal immediately, but in this example we set the actual equations as equal:
AP = 5by - 2ay
OP = 3bx + 2ax
5by-2ay(+2a in the ms) = 3bx+2ax
Do we do this because there is a point of intersection? I don't actually understand what setting them equal *means* compared to when we just say
(5y)b + (2-2y)a
(3x)b + (2x)a
5y=3x and 2-2y=2x
Without the 2a also we would never get 2-2y=2x so the answers would change

Thank you

 

[Dr.Peterson]

(This would all work the other way, right? If I chose to do OC = 5b+2a instead?)

No. Why did you think that BC = 2a? That is clearly not parallel to OA.

Why is 2a in the mark scheme?

I have no idea. Please show the mark scheme. (It's very hard to read your handwriting to see what you are referring to, and even that doesn't make clear what they actually say.)

Normally in vector questions we would find AP and OP then set their factors equal immediately

You appear to be saying that AP and OP are parallel. Why?? This is probably the location of your error.

 

[Dr.Peterson]

I found that I could make your writing (mostly) readable by replacing black with white:

​

To answer your question about setting things equal, what you are setting equal are two different expressions for the same vector, OP -- namely, 2xa + 3xb and 2a - 2ya + 5yb.

And the 2a you ask about comes from expressing OP as OA + AP. You can't set OP equal to AP, which you wanted to do. They are not the same vector. Very likely, you are misunderstanding what was done in other problems, not seeing that vectors, not just points, have to be equal to do this.

 

[blamocur]

Posting a solution after 1 week has lapsed.
[imath]\gdef\ve#1{\mathbf #1}[/imath]
Using notations [imath]\ve A, \ve C, \ve B\, \ve P[/imath] for vectors OA, OC, OB and OP respectively. Since P lies on OC we know that for some scalar [imath]x[/imath]
[math]\ve P = x\ve C = 2x\ve a + 3x\ve b[/math]Since P also lies on AB we know that for same scalar [imath]y[/imath]
[math]\ve P = y\ve A + (1-y) \ve B = 2y\ve a + 5\ve b - 5y \ve b[/math]By subtracting one from another we get one vector equation:
[math](2x-2y)\ve a + (3x-5+5y)\ve b = 0[/math]But since [imath]\ve a[/imath] and [imath]\ve b[/imath] are linearly independent vectors we know that both coefficients must be 0:
[math]\begin{aligned} x - y &=& 0\\ 3x + 5y &=& 5 \end{aligned}[/math]Solving the equations gives us [imath]x = \frac{5}{8}[/imath] and
[math]\ve P = \frac{5}{4}\ve a + \frac{15}{8}\ve b[/math]