PLEASE does anybody know the answer to Q13? Urgenthttp://www.gcsemathspastpapers.com/gcse-maths-p6j06allqs.htm

Question

Please could anybody Help me and find the answer to Q13 (a) and (b) and show their workings or how they got their answer?

Answer 1

a)
vector PR = vector PS + vector SR = vector PS + vector PQ = m+n
vector QS = vector QP + vector PS = vector PS - vector PQ = n-m
b)
vector PA = vector PS + vector SA
vector PA = vector PQ + vector QA
=> 2 vector PA = vector PQ + vector QA + vector PS + vector SA
= m + n + vector QA - vector QA ( vector QA = - vector SA )
= m + n
=> vector PA = (m+n)/2

or vector PA = 1/2 vector PR (A is the central point of PR)
=> vector PA = (m+n)/2

Answer 2

This is a "oops!" question, the teacher has left out something imporant, like one of the angles of the rhombus. Without that additional information, there is no unique answer. PA is 12 of PR. If you knew the angle PSR, for example, then you could use the law of cosines to find PR, as in the equation below:

PR^2 = PS^2 + SR^2 - 2 PS SR Cos (PSR)

while

QS^2 = PS^2 + SR^2 - 2 PS SR Cos (90 - PSR)

Addendum: With regard to the other answers given (vector geometric answers), there is nothing in the test indicating vectors as part of the curriculum. Vectors, furthermore, are conventionally designated with bold font letters. As a vector "problem", it's a downright no-brainer.

Answer 3

Man, I don't see the complication. Think about vector addition

PR = m+n

QS = n-m