Free Vectors in two dimensions questions for IGCSE Mathematics (0580), part of Transformations & vectors. Each question has the answer and the marking points an examiner would look for.
Generate a full Vectors in Two Dimensions worksheet →All Transformations & vectors questionsOABC is a parallelogram. O is the origin, OA = a and OC = c. M is the midpoint of AB.
Find AB in terms of c.
Find OM in terms of a and c.
N is the point on OB such that ON : NB = 2 : 1. Find MN in terms of a and c.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | c c B1 | 1 |
| (b) | a + ½c OA + AM M1 a + ½c A1 | 2 |
| (c) | MN = −⅓a + ⅙c ON = ⅔(a + c) M1 MN = ON − OM M1 −⅓a + ⅙c A1 | 3 |
O is the origin. The position vector of A is a and the position vector of B is b.
P is the point on AB such that AP : PB = 3 : 1.
M is the midpoint of OA.
R is the point with position vector 32b.
Find, in terms of a and/or b, in its simplest form, the vector
from A to B,
from O to P,
from M to P.
Show that M, P and R lie on a straight line, and find the ratio MP : PR.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a)(i) | b − a b − a B1 | 1 |
| (ii) | 14a + 34b a + 34(b − a) M1 14a + 34b A1 | 2 |
| (iii) | −14a + 34b (vector OP) − ½a M1 −14a + 34b A1 | 2 |
| (b) | vector MR = −½a + 32b = 2 × (vector MP); parallel with common point M, so collinear. MP : PR = 1 : 1 vector MR = −½a + 32b B1 vector MR = 2 × vector MP M1 parallel and share point M, so collinear A1 MP : PR = 1 : 1 B1 | 4 |
a = 11 b = 25
Find 2a − b.
Find |a + b|.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0−3 One component correct B1 0−3 B1 | 2 |
| (b) | 6.71 √(32 + 62) M1 6.71 oe A1 | 2 |
OAB is a triangle with OA = a and OB = b.
P is the point on AB such that AP : PB = 1 : 3.
M is the midpoint of OA. The point R is such that OR = 3OP.
Find, in terms of a and b, in their simplest form, the vector OP.
Find MR in terms of a and b.
The line BR meets the line OA, extended, at the point X, where OX = ta. Find the value of t.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 34a + 14b OP = a + 14(b − a) M1 34a + 14b A1 | 2 |
| (b) | 74a + 34b MR = MO + OR M1 74a + 34b A1 | 2 |
| (c) | t = 9 BR = −b + OR, and OX = OB + μBR M1 coefficient of b equal to zero: μ = 4 M1 t = 9 A1 | 3 |
Use vectors in this question.
In triangle OAB, M is the midpoint of OA and N is the midpoint of OB. Prove that MN is parallel to AB and half its length.
PQRS is any quadrilateral. Prove that the midpoints of its four sides are the vertices of a parallelogram.
Prove that this parallelogram is a rhombus if and only if the diagonals PR and QS of the original quadrilateral are equal in length.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | MN = ½b − ½a = ½(b − a) = ½AB OA = a, OB = b, AB = b − a B1 MN = ½b − ½a M1 = ½AB A1 so parallel and half the length A1 | 4 |
| (b) | each pair of opposite sides of the midpoint quadrilateral equals ½PR or ½QS position vectors p, q, r, s and midpoints ½(p + q) etc. M1 side joining midpoints of PQ and QR = ½(r − p) A1 side joining midpoints of RS and SP = ½(p − r) A1 opposite sides equal and parallel (same for the other pair) M1 so a parallelogram A1 | 5 |
| (c) | adjacent sides are ½PR and ½QS adjacent sides of the parallelogram are ½(r − p) and ½(s − q) M1 rhombus ⇔ these lengths are equal ⇔ PR = QS A1 both directions argued A1 | 3 |
OAB is a triangle with OA = a and OB = b.
P is the point on AB such that AP : PB = 1 : 3.
Q is the point on OB such that OQ = 13b.
The lines OP and AQ intersect at X.
Find OP in terms of a and b.
By writing OX = λOP and AX = μAQ, find the values of λ and μ.
Find the ratio OX : XP in its simplest form.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 34a + 14b a + 14(b − a) M1 34a + 14b A1 | 2 |
| (b) | λ = 23, μ = 12 OX = a + μ(−a + 13b) M1 equate coefficients of a and of b M1 two correct equations A1 λ = 23 A1 μ = 12 A1 | 5 |
| (c) | 2 : 1 2 : 1 B1 | 1 |
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