Triangle OAB has OA = a and OB = b. M is the midpoint of OA.
Diagram NOT accurately drawn
(a)Find the vector OM in terms of a.(1)
(b)Find the vector MA in terms of a.(1)
(c)Find the vector MB in terms of a and b.(2)
(Total for Question 1 is 4 marks)
2
Two vectors are given by p = (5, -12) and q = (-10, 24), written as column vectors.
(a)Find |p|.(2)
(b)Show that q is parallel to p.(2)
(c)Find |q|.(1)
(Total for Question 2 is 5 marks)
3
OAB is a triangle with OA = a and OB = b. C is the point on OA such that OC:CA = 1:3. D is the point on OB such that OD:DB = 1:3.
Diagram NOT accurately drawn
(a)Find OC in terms of a.(1)
(b)Find OD in terms of b.(1)
(c)Prove that CD is parallel to AB, and state the ratio CD:AB.(3)
(Total for Question 3 is 5 marks)
4
OABC is a parallelogram, with vertices in the order O, A, B, C. OA = a and OC = c.
Diagram NOT accurately drawn
(a)Write down AB in terms of c.(1)
(b)Find OB in terms of a and c.(1)
(c)F is the midpoint of CB. Find OF in terms of a and c.(2)
(Total for Question 4 is 4 marks)
5
a and b are vectors.
(a)Simplify fully: 3(2a - b) + 2(a + 3b) - 4a(2)
(b)Simplify fully: (1/2)(4a + 6b) - (a - 2b)(2)
(Total for Question 5 is 4 marks)
6
OAB is a triangle with OA = a and OB = b. P is the point on AB such that AP:PB = 3:2.
Diagram NOT accurately drawn
(a)Find AB in terms of a and b.(1)
(b)Find AP in terms of a and b.(1)
(c)Show that OP = (2/5)a + (3/5)b.(2)
(d)Given that OQ = 2a + 3b, show that O, P and Q lie on a straight line, and find the ratio OP:PQ.(3)
(Total for Question 6 is 7 marks)
7
O is the origin. The position vectors of points P, Q and R are OP = 2a - b, OQ = 6a + kb and OR = a + 2b, where a and b are non-parallel vectors and k is a constant. Given that P, Q and R lie on a straight line, find the value of k.
(Total for Question 7 is 4 marks)
8
The position vectors of points A, B, C and D, relative to origin O, are OA = a, OB = a + 2b, OC = 2a + 2b and OD = 2a.
Diagram NOT accurately drawn
(a)Find AB in terms of b.(1)
(b)Find DC in terms of b.(1)
(c)Find AD and BC, and hence prove that ABCD is a parallelogram.(3)
(Total for Question 8 is 5 marks)
9
OABC is a parallelogram with OA = a and OC = c. The diagonals OB and AC intersect at X.
Diagram NOT accurately drawn
(a)Write OB in terms of a and c.(1)
(b)Write AC in terms of a and c.(1)
(c)By expressing OX in two different ways, prove that X is the midpoint of AC.(4)
(Total for Question 9 is 6 marks)
10
OAB is a triangle with OA = a and OB = b. C is the point on OA such that OC = k*a, and D is the point on OB such that OD = k*b, where 0 < k < 1.
Diagram NOT accurately drawn
(a)Find CD in terms of a, b and k.(2)
(b)Show that CD is parallel to AB for all values of k.(2)
(Total for Question 10 is 4 marks)
11
The position vectors of points P, Q and R, relative to the origin O, are OP = (3, 2), OQ = (11, 8) and OR = (-1, -1).
Diagram NOT accurately drawn
(a)Find the vector PQ.(1)
(b)Find the vector PR.(1)
(c)Prove that P, Q and R lie on a straight line.(2)
(Total for Question 11 is 4 marks)
12
X is the point on the line segment AB such that AX:XB = 2:3. O is the origin, with OA = a and OB = b.
Diagram NOT accurately drawn
(a)Find AX in terms of a and b.(2)
(b)Hence find OX in terms of a and b, giving your answer in the form OX = (1/5)(p*a + q*b), where p and q are integers.(2)
(Total for Question 12 is 4 marks)
13
OAB is a triangle. C is the point on OA such that OC:CA = 2:3. D is the point on OB such that OD:DB = 2:3. OA = a and OB = b.
Diagram NOT accurately drawn
(a)Find CD in terms of a and b.(2)
(b)Given that the area of triangle OAB is 75 cm2, find the area of triangle OCD.(2)
(Total for Question 13 is 4 marks)
14
OAB is a triangle with OA = a and OB = b. T is the point on the line BA extended, beyond A, such that AT = (2/5)AB.
Diagram NOT accurately drawn
(a)Find AB in terms of a and b.(1)
(b)Find OT in terms of a and b.(2)
(Total for Question 14 is 3 marks)
15
OABC is a quadrilateral with OA = a, OB = 2a + 3b and OC = 4a + 5b, where O is the origin. A student claims that OA is parallel to BC. Determine, showing your working, whether the student's claim is correct.
Diagram NOT accurately drawn
(Total for Question 15 is 3 marks)
16
OAB is a triangle with OA = a and OB = b. M is the midpoint of OA. N is the point on OB extended beyond B such that OB = BN. P is the point on AB such that AP:PB = 2:1.
Diagram NOT accurately drawn
(a)Find OM in terms of a.(1)
(b)Find ON in terms of b.(1)
(c)Find OP in terms of a and b.(2)
(d)Prove that M, P and N lie on a straight line.(3)
(e)Find the ratio MP:PN.(1)
(Total for Question 16 is 8 marks)
17
OABC is a trapezium in which OC is parallel to AB, and OC = 3AB. OA = a and OC = c.
Diagram NOT accurately drawn
(a)Find AB in terms of c.(1)
(b)Find OB in terms of a and c.(1)
(c)Find AC in terms of a and c.(1)
(d)The diagonals OB and AC intersect at X. Prove that OX:XB = 3:1.(4)
(Total for Question 17 is 7 marks)
18
O, A, B and C are four points such that OA = a, OB = b and OC = c. P, Q, R and S are the midpoints of OA, AB, BC and CO respectively.
Diagram NOT accurately drawn
(a)Find OP in terms of a.(1)
(b)Find OQ in terms of a and b.(1)
(c)Show that PQ = (1/2)b.(2)
(d)Find OS and OR, and hence show that SR = (1/2)b.(2)
(e)Hence prove that PQRS is a parallelogram.(1)
(Total for Question 18 is 7 marks)
Mark scheme · 8.10 Vectors Proof Questions
Question 1
(a) B1 (1/2)a (oe) (cao)
(a) Answer: OM = (1/2)a
(b) B1 (1/2)a (oe) (cao)
(b) Answer: MA = (1/2)a
(c) M1 correct method, e.g. MB = OB - OM (oe MB = MA + AB)
(c) A1 b - (1/2)a (oe) (cao)
(c) Answer: MB = b - (1/2)a
Question 2
(a) M1√52 + (-12)2 (oe)
(a) A1 13 (cao)
(a) Answer: |p| = 13
(b) M1 compares components, e.g. -10/5 = -2 and 24/-12 = -2 (oe)
(b) A1 q = -2p, so q is parallel to p (cso)
(b) Answer: q = -2p, so q is parallel to p
(c) B1 26 (cao) (ft 2 x their part (a) if q = -2p used)
(c) Answer: |q| = 26
Question 3
(a) B1 (1/4)a (oe) (cao)
(a) Answer: OC = (1/4)a
(b) B1 (1/4)b (oe) (cao)
(b) Answer: OD = (1/4)b
(c) M1 CD = OD - OC = (1/4)b - (1/4)a (oe)
(c) M1 AB = OB - OA = b - a (oe)
(c) A1 CD = (1/4)AB, so CD is parallel to AB; CD:AB = 1:4 (cso)
(c) Answer: CD is parallel to AB; CD:AB = 1:4
Question 4
(a) B1 c (oe) (cao)
(a) Answer: AB = c
(b) B1 a + c (oe) (ft their part (a))
(b) Answer: OB = a + c
(c) M1 correct method, e.g. CB = OA = a, so OF = OC + (1/2)CB
(c) M1 OP = OA + AP = a + (3/5)(b - a) (ft their part (b))
(c) A1 correctly simplifies to (2/5)a + (3/5)b (cso)
(c) Answer: OP = (2/5)a + (3/5)b
(d) M1 recognises OP = (1/5)(2a + 3b) (ft their part (c))
(d) A1 OP = (1/5)OQ, and since O, P and Q share the common point O with OP a scalar multiple of OQ, O, P and Q are collinear (cso)
(d) A1 OP:PQ = 1:4 (cao)
(d) Answer: O, P, Q are collinear; OP:PQ = 1:4
Question 7
M1 finds PQ = OQ - OP = 4a + (k+1)b (oe)
M1 finds PR = OR - OP = -a + 3b (oe)
M1 sets PQ = m x PR and equates coefficients of a: 4 = -m, so m = -4
A1 k = -13 (cao)
Answer: k = -13
Question 8
(a) B1 2b (cao)
(a) Answer: AB = 2b
(b) B1 2b (cao)
(b) Answer: DC = 2b
(c) M1 AD = OD - OA = a (oe)
(c) M1 BC = OC - OB = a (oe)
(c) A1 AD = BC and AB = DC, so both pairs of opposite sides are equal and parallel, hence ABCD is a parallelogram (cso)
(c) Answer: AD = BC = a and AB = DC = 2b, so ABCD is a parallelogram
Question 9
(a) B1 a + c (oe) (cao)
(a) Answer: OB = a + c
(b) B1 c - a (oe) (cao)
(b) Answer: AC = c - a
(c) M1 X lies on OB, so OX = m(a + c) for some scalar m (ft their part (a))
(c) M1 X lies on AC, so OX = OA + h.AC = (1-h)a + hc for some scalar h (ft their part (b))
(c) dM1 equates coefficients of a and c, since a and c are non-parallel: m = 1-h and m = h, giving h = m = 1/2
(c) A1 OX = (1/2)(a+c), so AX = (1/2)(c-a) = (1/2)AC, so X is the midpoint of AC (cso)
(c) Answer: X is the midpoint of AC (h = 1/2)
Question 10
(a) M1 CD = OD - OC = k*b - k*a
(a) A1 k*(b - a) (oe) (cao)
(a) Answer: CD = k*(b - a)
(b) M1 AB = OB - OA = b - a (ft their part (a))
(b) A1 CD = k x AB, and since k is a scalar for any value of k with 0 < k < 1, CD is a scalar multiple of AB, so CD is parallel to AB for all such k (cso)
(b) Answer: CD = k*AB, parallel to AB for all values of k
Question 11
(a) B1 (8, 6) (cao)
(a) Answer: PQ = (8, 6)
(b) B1 (-4, -3) (cao)
(b) Answer: PR = (-4, -3)
(c) M1 compares PR with PQ, e.g. -(1/2) x (8, 6) = (-4, -3) (ft their parts (a) and (b))
(c) A1 PR = -(1/2)PQ, so PR is parallel to PQ, and since P is a common point, P, Q and R are collinear (cso)
(c) Answer: P, Q, R are collinear (PR = -(1/2)PQ)
Question 12
(a) M1 AB = OB - OA = b - a, and AX = (2/5)AB
(a) A1 (2/5)(b - a) (oe) (cao)
(a) Answer: AX = (2/5)(b - a)
(b) M1 OX = OA + AX = a + (2/5)(b - a) (ft their part (a))
(b) A1 OX = (1/5)(3a + 2b); p = 3, q = 2 (cao)
(b) Answer: OX = (1/5)(3a + 2b); p = 3, q = 2
Question 13
(a) M1 OC = (2/5)a and OD = (2/5)b, so CD = OD - OC
(a) A1 (2/5)(b - a) (oe) (cao)
(a) Answer: CD = (2/5)(b - a)
(b) M1 recognises triangles OCD and OAB are similar (common angle O, CD parallel to AB) with linear scale factor 2/5, so uses area scale factor (2/5)2 = 4/25
(b) A1 12 cm2 (cao)
(b) Answer: Area of triangle OCD = 12 cm2
Question 14
(a) B1 b - a (oe) (cao)
(a) Answer: AB = b - a
(b) M1 recognises T is beyond A away from B, so AT = -(2/5)(b - a), and OT = OA + AT (ft their part (a))
(b) A1 (7/5)a - (2/5)b (oe) (cao)
(b) Answer: OT = (7/5)a - (2/5)b
Question 15
M1 finds BC = OC - OB = 2a + 2b (oe)
A1 compares with OA = a and states that BC cannot equal k*a for any scalar k, since a and b are non-parallel vectors and BC has a non-zero b-component
A1 correctly concludes the student's claim is incorrect: OA is not parallel to BC (cso)
Answer: The student is incorrect; OA is not parallel to BC
Question 16
(a) B1 (1/2)a (oe) (cao)
(a) Answer: OM = (1/2)a
(b) B1 2b (oe) (cao)
(b) Answer: ON = 2b
(c) M1 AP = (2/3)(b - a), and OP = OA + AP
(c) A1 (1/3)a + (2/3)b (oe) (cao)
(c) Answer: OP = (1/3)a + (2/3)b
(d) M1 finds MP = OP - OM = -(1/6)a + (2/3)b (ft their parts (a) and (c))
(d) M1 finds MN = ON - OM = -(1/2)a + 2b (ft their parts (a) and (b))
(d) A1 MN = 3 x MP (since -(1/2) = 3 x -(1/6) and 2 = 3 x (2/3)), so MN is a scalar multiple of MP, and since M is a common point, M, P and N are collinear (cso)
(d) Answer: M, P and N are collinear (MN = 3MP)
(e) B1 1:2 (cao) (ft their part (d))
(e) Answer: MP:PN = 1:2
Question 17
(a) B1 (1/3)c (oe) (cao)
(a) Answer: AB = (1/3)c
(b) B1 a + (1/3)c (oe) (ft their part (a))
(b) Answer: OB = a + (1/3)c
(c) B1 c - a (oe) (cao)
(c) Answer: AC = c - a
(d) M1 X lies on OB, so OX = m(a + (1/3)c) for some scalar m (ft their part (b))
(d) M1 X lies on AC, so OX = OA + h.AC = (1-h)a + hc for some scalar h (ft their part (c))
(d) dM1 equates coefficients of a and c: m = 1-h and m/3 = h; substitutes to get m = 1 - m/3, leading to m = 3/4 and h = 1/4
(d) A1 OX = (3/4)OB, so OX:XB = 3:1 (cso)
(d) Answer: OX:XB = 3:1
Question 18
(a) B1 (1/2)a (oe) (cao)
(a) Answer: OP = (1/2)a
(b) B1 (1/2)(a + b) (oe) (cao)
(b) Answer: OQ = (1/2)(a + b)
(c) M1 PQ = OQ - OP = (1/2)(a+b) - (1/2)a (ft their parts (a) and (b))
(c) A1 correctly simplifies to (1/2)b (cso)
(c) Answer: PQ = (1/2)b
(d) M1 OS = (1/2)c (S is the midpoint of CO) and OR = (1/2)(b+c) (R is the midpoint of BC)
(d) A1 SR = OR - OS = (1/2)b (cso)
(d) Answer: OS = (1/2)c, OR = (1/2)(b+c), SR = (1/2)b
(e) B1 PQ = SR = (1/2)b, so PQ is parallel and equal to SR, therefore PQRS is a parallelogram (cso, ft their parts (c) and (d))
(e) Answer: PQRS is a parallelogram, since PQ = SR = (1/2)b