AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme]

I bet you there'll be a bitch of a vectors question. And probably one of those "form a differential" questions that leave you wondering if AQA's head office is in hell.

Working through it myself now and we seem to have done the same thing!
Except, shouldn't it be AP, not PA?

As at the right angle, the [-1, 4, 3] would be going away from the angle, so you would want AP (also going away from the angle)

How to you find the Cartesian equation in this form?
I got to the point where y=square root 36(1-cos^2thetacos^2theta)



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Below I've put my complete method (I don't know why I started forming another equation of a line, I'm just really tired so I'm doing some silly things ).

START
I have no idea if there's a concrete method for doing a vector reflection, but this is how I would do it:

The reflected point, R, is of equidistance from the line r as P is, only in the opposite direction and the shortest distance from P to the line occurs when a line through P that intersects with the line is perpendicular to the line.

Let the point of intersection be A.

Therefore, PA = OA - OP. OA is equal to r for some value of U.
=> PA = [-U, 4U-1, 3U]

Doing the scalar product with PA . [-1, 4, 3] = 0 will allow you to solve to find the value for U at which A occurs.

Substituting in the value for U into the equation of the line will give you the position vector of A and, as you know PA = [-U, 4U-1, 3U] you can substitute U in there as well to find PA.

You want to find R, and as R is equidistance from the line in the question you know that PA = AR.
=> PA = OR - OA
=> OR = PA + OA
=> OR = [-U, 4U-1, 3U] + OA (where OA = the equation of the line with the value of U you found substituted in)
=> OR = whatever position vector that came to
=> R is the (x, y, z) form of the position vector you just found
END

NOTE: I'm really tried at the moment so if I've done something stupid please just say .

My predictions for C4 grade boundaries:
A* 6/75
A 5/75
B 4/75
C 3/75

I think these are reasonable.

Argh sorry, how do you know when to do that then?

Considering I was getting 98/99 and 100UMS on all previous mock papers, I'd say something went wrong in the exam considering I must have dropped at least ten marks. It is not necessarily unpreparedness when not doing well in exam, sometimes there can be other factors, such as health, or the ACTUAL EXAM QUESTIONS!!!

Too many people who are not even good at maths are getting A*s.

There was nothing wrong with the exam questions. They were all within the specification. If you thought they were difficult, then grade boundaries should reflect that.

It's good they're making it difficult anyway. Too many people who are not even good at maths are getting A*s.

It's always defined as acute unless they tell you it's not. So if you get a negative cos theta, just make it positive.

This generalisation is absurd and makes you look like a snob.

This seems right, but as I said, shouldn't you be working with AP? Not PA?
Going to give it another shot now. Got it wrong on my first attempt...

Answers:

Below I've put my complete method (I don't know why I started forming another equation of a line, I'm just really tired so I'm doing some silly things ).

START
I have no idea if there's a concrete method for doing a vector reflection, but this is how I would do it:

The reflected point, R, is of equidistance from the line r as P is, only in the opposite direction and the shortest distance from P to the line occurs when a line through P that intersects with the line is perpendicular to the line.

Let the point of intersection be A.

Therefore, PA = OA - OP. OA is equal to r for some value of U.
=> PA = [-U, 4U-1, 3U]

Doing the scalar product with PA . [-1, 4, 3] = 0 will allow you to solve to find the value for U at which A occurs.

Substituting in the value for U into the equation of the line will give you the position vector of A and, as you know PA = [-U, 4U-1, 3U] you can substitute U in there as well to find PA.

You want to find R, and as R is equidistance from the line in the question you know that PA = AR.
=> PA = OR - OA
=> OR = PA + OA
=> OR = [-U, 4U-1, 3U] + OA (where OA = the equation of the line with the value of U you found substituted in)
=> OR = whatever position vector that came to
=> R is the (x, y, z) form of the position vector you just found
END

NOTE: I'm really tried at the moment so if I've done something stupid please just say .

There was nothing wrong with the exam questions. They were all within the specification. If you thought they were difficult, then grade boundaries should reflect that.

It's good they're making it difficult anyway. Too many people who are not even good at maths are getting A*s.