Column vectors are written in the form (x, y), where x is the top (horizontal) number and y is the bottom (vertical) number. u = (4, 1) and v = (-2, 5).
(a)Work out u + v as a column vector.(1)
(b)Work out u - v as a column vector.(1)
(c)Work out 3v as a column vector.(1)
(Total for Question 1 is 3 marks)
2
On a centimetre grid, X is the point (1, 2), Y is the point (5, 6), P is the point (3, 1) and Q is the point (6, 5).
(a)Write the vector XY as a column vector.(1)
(b)Write the vector PQ as a column vector.(1)
(Total for Question 2 is 2 marks)
3
p and q are column vectors.
(a)p = (6, 8). Calculate the magnitude of p, |p|.(2)
(b)q = (-5, 12). Calculate the magnitude of q, |q|.(2)
(Total for Question 3 is 4 marks)
4
r = (3, -1).
(a)s = (-9, 3). Show that s is parallel to r.(1)
(b)w = (12, -4). Determine, showing your working, whether w is parallel to r.(2)
(Total for Question 4 is 3 marks)
5
OABC is a quadrilateral, with vertices in order. OA = a, AB = b and BC = c.
(Total for Question 5 is 2 marks)
6
In triangle OAB, O is the origin. OA = a and OB = b.
Diagram NOT accurately drawn
(a)Find the vector AB in terms of a and b.(1)
(b)Find the vector BA in terms of a and b.(1)
(Total for Question 6 is 2 marks)
7
OABC is a parallelogram, with vertices in order. OA = a and OC = c.
Diagram NOT accurately drawn
(a)Find OB in terms of a and c.(2)
(b)Find CA in terms of a and c.(1)
(Total for Question 7 is 3 marks)
8
In triangle OAB, O is the origin. OA = a and OB = b. M is the midpoint of AB.
Diagram NOT accurately drawn
(Total for Question 8 is 2 marks)
9
v is the column vector (k, 4), where k > 0, and |v| = 5. Find the value of k.
(Total for Question 9 is 3 marks)
10
In triangle OAB, O is the origin. OA = a and OB = b. P is the point on AB such that AP : PB = 1 : 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)Find OP in terms of a and b, giving your answer in its simplest form.(2)
(Total for Question 10 is 4 marks)
11
In triangle OAB, O is the origin. OA = a and OB = b. Q is the point on AB such that AQ : QB = 3 : 2.
Diagram NOT accurately drawn
(Total for Question 11 is 3 marks)
12
A cyclist rides from a cafe C to a viewpoint V. This part of the journey is modelled by the column vector (5, 2) km. She then rides from V to a picnic site P, modelled by the column vector (4, 10) km.
(a)Find the column vector representing the direct journey from C to P.(1)
(b)Calculate the direct distance, in km, from C to P.(2)
(c)Find the column vector representing the return journey directly from P to C.(1)
(Total for Question 12 is 4 marks)
13
In triangle OAB, O is the origin. OA = a and OB = b. R is the point on AB such that OR = (3/4)a + (1/4)b.
Diagram NOT accurately drawn
(Total for Question 13 is 3 marks)
14
ABCDEF is a regular hexagon with centre O; the vertices are labelled in order around the hexagon. OA = a and OB = b.
Diagram NOT accurately drawn
(a)Find AB in terms of a and b.(1)
(b)D is the vertex opposite A. Find OD in terms of a.(1)
(c)E is the vertex opposite B. Find OE in terms of b.(1)
(d)Find EB in terms of b, giving your answer in its simplest form.(1)
(Total for Question 14 is 4 marks)
15
In triangle OAB, O is the origin. OA = a and OB = b. C is the midpoint of OA and D is the midpoint of OB.
Diagram NOT accurately drawn
(a)State OC and OD in terms of a and b.(1)
(b)Find CD in terms of a and b.(2)
(c)Given that AB = b - a, show that CD is parallel to AB.(1)
(Total for Question 15 is 4 marks)
16
OABC is a trapezium, with vertices in order, in which OA is parallel to CB. OA = a, CB = 3a and OC = c. The diagonals OB and AC intersect at E.
Diagram NOT accurately drawn
(a)Show that OB = 3a + c.(1)
(b)Show that AC = c - a.(1)
(c)E lies on OB, so OE = h(3a + c) for some scalar h. E also lies on AC, so AE = k(c - a) for some scalar k. By writing OE two ways and comparing the coefficients of a and c, find h and hence find OE in terms of a and c.(3)
(Total for Question 16 is 5 marks)
17
In triangle OAB, O is the origin. OA = a and OB = b. F is the point on OA such that OF = (2/5)a. G is the point on AB such that AG : GB = 3 : 1.
Diagram NOT accurately drawn
(Total for Question 17 is 4 marks)
18
In triangle OAB, O is the origin. OA = a and OB = b. M is the midpoint of AB. G is the point on OM such that OG : GM = 2 : 1. N is the midpoint of OB.
Diagram NOT accurately drawn
(a)Find OM in terms of a and b.(1)
(b)Find OG in terms of a and b.(2)
(c)Show that A, G and N lie on a straight line.(3)
(Total for Question 18 is 6 marks)
19
OABC is a parallelogram, with vertices in order. OA = a and OC = c. The diagonals OB and AC intersect at X.
Diagram NOT accurately drawn
(a)Show that OB = a + c.(1)
(b)Show that AC = c - a.(1)
(c)X lies on OB, so OX = h(a + c) for some scalar h. X also lies on AC, so AX = k(c - a) for some scalar k. Show that h = k = 1/2, and state what this proves about the diagonals of the parallelogram.(4)
(Total for Question 19 is 6 marks)
Mark scheme · 5.21D Vectors: Fluency and Exam Drill
Question 1
(a) B1 (2, 6) cao
(a) Answer: (2, 6)
(b) B1 (6, -4) cao
(b) Answer: (6, -4)
(c) B1 (-6, 15) cao
(c) Answer: (-6, 15)
Question 2
(a) B1 (4, 4) cao
(a) Answer: (4, 4)
(b) B1 (3, 4) cao
(b) Answer: (3, 4)
Question 3
(a) M1√62 + 82 oe
(a) A1 10 cao
(a) Answer: 10
(b) M1√(-5)2 + 122 oe
(b) A1 13 cao
(b) Answer: 13
Question 4
(a) B1 cso: s = -3r (or r = -(1/3)s), a scalar multiple of r, so s is parallel to r
(a) Answer: s = -3r, so s is parallel to r
(b) M1 attempts to write w as a scalar multiple of r, e.g. compares 12/3 and -4/-1
(b) A1 cso: w = 4r, so w is parallel to r (and in the same direction, since the scalar is positive)
(b) Answer: w = 4r, so w is parallel to r
Question 5
M1 uses OA + AB + BC + CO = 0 (or an equivalent complete path back to O)
A1 -a - b - c oe, e.g. -(a + b + c)
Answer: CO = -a - b - c
Question 6
(a) B1 b - a cao
(a) Answer: AB = b - a
(b) B1 a - b ft from (a), the negative of their AB
(b) Answer: BA = a - b
Question 7
(a) M1 recognises AB = OC = c (opposite sides of a parallelogram) and OB = OA + AB
(a) A1 a + c cao
(a) Answer: OB = a + c
(b) B1 a - c cao
(b) Answer: CA = a - c
Question 8
M1 OM = OA + (1/2)(b - a), or OM = (1/2)(a + b), correct method
A1 (1/2)(a + b) oe, e.g. (a+b)/2
Answer: OM = (a + b)/2
Question 9
M1 k2 + 42 = 52 oe
M1 dep: k2 = 9
A1 k = 3 cao (rejects k = -3 since k > 0)
Answer: k = 3
Question 10
(a) B1 b - a cao
(a) Answer: AB = b - a
(b) B1 (1/3)(b - a) oe, ft their AB
(b) Answer: AP = (1/3)(b - a)
(c) M1 OP = OA + AP = a + (1/3)(b - a), ft their AP
(c) A1 (2/3)a + (1/3)b oe, e.g. (2a + b)/3
(c) Answer: OP = (2a + b)/3
Question 11
M1 AB = b - a
M1 OQ = a + (3/5)(b - a), correct method for the ratio
A1 (2/5)a + (3/5)b oe, e.g. (2a + 3b)/5
Answer: OQ = (2a + 3b)/5
Question 12
(a) B1 (9, 12) cao
(a) Answer: (9, 12)
(b) M1√92 + 122 ft from (a)
(b) A1 15 (km) cao
(b) Answer: 15 km
(c) B1 (-9, -12) ft from (a), the negative of their CP
(c) Answer: (-9, -12)
Question 13
M1 sets up OR = OA + AR = a + t(b - a) = (1-t)a + tb for AR:RB = t : (1-t), or equivalent
M1 compares coefficients with OR = (3/4)a + (1/4)b to find t = 1/4
A1 AR : RB = 1 : 3 oe
Answer: AR : RB = 1 : 3
Question 14
(a) B1 b - a cao
(a) Answer: AB = b - a
(b) B1 -a cao, using that O is the midpoint of the diagonal AD in a regular hexagon
(b) Answer: OD = -a
(c) B1 -b cao, using that O is the midpoint of the diagonal BE
(c) Answer: OE = -b
(d) B1 2b oe, ft their OE
(d) Answer: EB = 2b
Question 15
(a) B1 OC = (1/2)a and OD = (1/2)b, both correct
(a) Answer: OC = (1/2)a, OD = (1/2)b
(b) M1 CD = OD - OC = (1/2)b - (1/2)a
(b) A1 (1/2)(b - a) oe, e.g. (b-a)/2
(b) Answer: CD = (1/2)(b - a)
(c) B1 cso: CD = (1/2)AB, so CD is a scalar multiple of AB, so CD is parallel to AB
(c) Answer: CD = (1/2)AB, so CD is parallel to AB
Question 16
(a) B1 cso: OB = OC + CB = c + 3a = 3a + c
(a) Answer: OB = 3a + c
(b) B1 cso: AC = OC - OA = c - a
(b) Answer: AC = c - a
(c) M1 OE = OA + AE = a + k(c - a), set equal to h(3a + c): h(3a+c) = a + k(c-a)
(c) M1 compares coefficients of a and c (independent vectors) to get 3h = 1 - k and h = k, then solves simultaneously
(c) A1 h = 1/4 and OE = (3/4)a + (1/4)c oe
(c) Answer: h = 1/4, OE = (3/4)a + (1/4)c
Question 17
M1 OF = (2/5)a
M1 AG = (3/4)(b - a), leading to OG = OA + AG = (1/4)a + (3/4)b
M1 FG = OG - OF
A1 -(3/20)a + (3/4)b oe, e.g. (3/4)b - (3/20)a
Answer: FG = (3/4)b - (3/20)a
Question 18
(a) B1 (1/2)(a + b) oe
(a) Answer: OM = (1/2)(a + b)
(b) M1 OG = (2/3)OM, ft their OM
(b) A1 (1/3)(a + b) oe
(b) Answer: OG = (1/3)(a + b)
(c) M1 finds AG = OG - OA = (1/3)(a+b) - a = (1/3)b - (2/3)a, ft their OG
(c) M1 finds AN = ON - OA = (1/2)b - a, using ON = (1/2)b
(c) A1 cso: AG = (2/3)AN, so AG is a scalar multiple of AN; since A is a common point, A, G and N lie on a straight line (with AG : GN = 2 : 1)
(c) Answer: AG = (2/3)AN, so A, G and N are collinear
Question 19
(a) B1 cso: OB = OA + AB = a + c, since AB = OC = c
(a) Answer: OB = a + c
(b) B1 cso: AC = OC - OA = c - a
(b) Answer: AC = c - a
(c) M1 writes OX = OA + AX = a + k(c - a), sets equal to h(a + c): h(a+c) = a + k(c-a)
(c) M1 compares coefficients of a and c (independent vectors) to get h = 1 - k and h = k
(c) M1 solves simultaneously to get h = k = 1/2
(c) A1 cso: OX = (1/2)(a+c) is the midpoint of OB, and AX = (1/2)(c-a) is the midpoint of AC, so the diagonals of the parallelogram bisect each other
(c) Answer: h = k = 1/2; the diagonals of the parallelogram bisect each other