Pure: Vectors - Worksheets, Questions and Revision

12 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Pure Mathematics

P10 Pure: Vectors

EDEXCEL 9MA0 · Calculator allowed · about 140 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The vector a = 3i - 4j.

(a)Find |a|.(2)

(b)Find the unit vector in the direction of a.(2)

(c)Given b = -2i + 5j, find a + 2b, and hence find |a + 2b|, giving your answer to 3 significant figures.(4)

(Total for Question 1 is 8 marks)

Two forces acting on a particle are F1 = (2i + 5j) N and F2 = (-6i + j) N.

(a)Find the resultant force F1 + F2, giving your answer in the form π + qj.(2)

(b)Find the magnitude of the resultant force, giving your answer in newtons to 3 significant figures.(2)

(c)Find the angle the resultant force makes with the vector i, measured anticlockwise from the positive i direction, giving your answer in degrees to 1 decimal place.(3)

(Total for Question 2 is 7 marks)

Relative to a fixed origin O, the points A and B have position vectors OA = i + 2j - 3k and OB = 5i - 2j + 9k.

(a)Find the vector AB.(2)

(b)Find |AB|, giving your answer as an exact simplified surd.(2)

(c)M is the midpoint of AB. Find the position vector of M.(2)

(d)The point C has position vector OC = 13i - 10j + 33k. Show that A, B and C are collinear, and find the ratio AB : BC.(4)

(Total for Question 3 is 10 marks)

Relative to a fixed origin O, points P and Q have position vectors p = 2i + 3j and q = 8i - 3j. The point R lies on PQ such that PR : RQ = 2 : 1.

(a)Find the position vector of R.(4)

(b)A further point S has position vector s = 12i + 5j. Find the vector RS, and hence find a unit vector in the direction of RS.(4)

(Total for Question 4 is 8 marks)

Two survey points D and E have position vectors, relative to a fixed origin and measured in metres, OD = 2i - j + 4k and OE = 6i + 3j - 2k.

(a)Find the vector DE, and hence find |DE|, giving your answer in metres to 3 significant figures.(4)

(b)The point F lies on DE such that DF : FE = 3 : 1. Find the position vector of F.(2)

(c)State, with a reason, whether F lies closer to D or to E.(1)

(Total for Question 5 is 7 marks)

The quadrilateral OABC has position vectors, relative to O, of 0, a, a + c and c respectively, where a and c are non-zero, non-parallel vectors.

(a)Show that OABC is a parallelogram.(3)

(b)M is the midpoint of the diagonal OB and N is the midpoint of the diagonal AC. Show that M and N are the same point.(4)

(c)Given that a = 4i - j and c = i + 3j, find the coordinates of the point where the diagonals of OABC intersect.(3)

(Total for Question 6 is 10 marks)

The vector p = 3i + 4j. The vector q = 2n i + (n+5) j, where n is a constant.

(a)Given that p and q are parallel, find the value of n.(3)

(b)When n = 3, state whether p and q act in the same or opposite direction, giving a reason.(2)

(c)Given that |p| = 5, find the possible values of n, other than n = 3, for which |q| = 2|p|.(5)

(Total for Question 7 is 10 marks)

Three forces F1 = (3i + 2j) N, F2 = (-i + 5j) N and F3 = (π + qj) N act on a particle which is in equilibrium.

(a)Find the values of p and q.(3)

(b)Find the magnitude of F3, giving your answer in newtons to 3 significant figures.(2)

(c)A fourth force F4 is now applied so that the resultant of all four forces is 10i N. Find F4.(2)

(Total for Question 8 is 7 marks)

Relative to a fixed origin O, the points A, B and C have position vectors OA = i + 2j - k, OB = 4i - j + 5k and OC = (4t-1)i - 4j + 11k, where t is a constant.

(a)Find the vector AB.(2)

(b)Given that A, B and C are collinear, find the value of t.(5)

(c)Using the value of t found in part (b), show that B is the midpoint of AC.(3)

(Total for Question 9 is 10 marks)

Relative to a fixed origin O, points A and B have position vectors OA = 3i - j + 4k and OB = -i + 7j - 4k. The point X lies on the line segment AB such that AX = t x AB, where 0 ≤ t ≤ 1.

(a)Find the vector AB.(2)

(b)Show that OX = (3-4t)i + (8t-1)j + (4-8t)k.(2)

(c)Given that |OX| = √10, show that 18t² - 13t + 2 = 0, and hence find the possible values of t.(7)

(Total for Question 10 is 11 marks)

In triangle OAB, OA = a and OB = b. The point P lies on OA such that OP = (2/5)a, and the point Q lies on AB such that AQ = (3/4)AB.

(a)Find AB in terms of a and b.(1)

(b)Show that PQ = (3/20)(5b - a), fully simplified.(4)

(c)A point R lies on OB such that OR = (1/4)b. Determine, showing full working, whether P, Q and R are collinear.(4)

(Total for Question 11 is 9 marks)

A harbourmaster models three navigation buoys A, B and C using position vectors relative to a fixed origin O (the harbour), with distances in metres: OA = 4i + 2j, OB = 16i + 8j, OC = 22i - 4j.

(a)Show that O, A and B are collinear, and find the ratio OA : AB.(3)

(b)Find the vectors BC and AC, and show that |AB|^2 = |BC|^2 = 180 and |AC|^2 = 360.(5)

(c)Hence show that triangle ABC is a right-angled isosceles triangle, stating clearly where the right angle is located.(3)

(d)Find the exact perimeter of triangle ABC, in metres.(2)

(Total for Question 12 is 13 marks)

Mark scheme · P10 Pure: Vectors

Question 1

(a) M1 √3² + (-4)²
(a) A1 = 5 cao
(a) Answer: |a| = 5
(b) M1 divide a by |a| found in (a)
(b) A1 = (3/5)i - (4/5)j oe
(b) Answer: (3/5)i - (4/5)j
(c) M1 attempt 2b and add to a
(c) A1 a + 2b = -i + 6j cao
(c) M1 √(-1)² + 6²
(c) A1 awrt 6.08
(c) Answer: a + 2b = -i + 6j, |a + 2b| = awrt 6.08

Question 2

(a) M1 add corresponding components
(a) A1 = -4i + 6j (N) cao
(a) Answer: -4i + 6j (N)
(b) M1 √(-4)² + 6²
(b) A1 awrt 7.21 N
(b) Answer: awrt 7.21 N
(c) M1 attempt reference angle = arctan(6/4)
(c) dM1 adjust for 2nd quadrant: 180 - reference angle
(c) A1 awrt 123.7 degrees
(c) Answer: awrt 123.7 degrees

Question 3

(a) M1 OB - OA
(a) A1 = 4i - 4j + 12k cao
(a) Answer: AB = 4i - 4j + 12k
(b) M1 √4² + (-4)² + 12²
(b) A1 = 4sqrt(11) cao (exact)
(b) Answer: 4sqrt(11)
(c) M1 (OA + OB)/2
(c) A1 = 3i + 0j + 3k oe (3,0,3)
(c) Answer: OM = 3i + 3k
(d) M1 find BC = OC - OB
(d) A1 BC = 8i - 8j + 24k
(d) M1 recognise BC = 2 x AB (scalar multiple, so BC parallel to AB)
(d) A1 cso: common point B, so A, B, C collinear; ratio AB : BC = 1 : 2
(d) Answer: A, B, C collinear; AB : BC = 1 : 2

Question 4

(a) M1 find PQ = q - p
(a) A1 PQ = 6i - 6j
(a) M1 r = p + (2/3)PQ
(a) A1 r = 6i - j cao
(a) Answer: r = 6i - j
(b) M1 RS = s - r
(b) A1 RS = 6i + 6j
(b) M1 find |RS| = 6sqrt(2) and divide RS by |RS|
(b) A1 unit vector = (√2/2)i + (√2/2)j oe (awrt 0.707i + 0.707j)
(b) Answer: RS = 6i + 6j; unit vector = (√2/2)i + (√2/2)j

Question 5

(a) M1 OE - OD
(a) A1 DE = 4i + 4j - 6k
(a) M1 √4²+4²+(-6)²
(a) A1 awrt 8.25 m
(a) Answer: DE = 4i+4j-6k; |DE| = awrt 8.25 m
(b) M1 F = D + (3/4)DE (ft from (a))
(b) A1 F = 5i + 2j - 0.5k oe
(b) Answer: OF = 5i + 2j - 0.5k
(c) B1 F is closer to E, since DF:FE = 3:1 means FE is only 1 part of the 4 compared to DF's 3 parts
(c) Answer: F is closer to E

Question 6

(a) M1 find AB = OB - OA
(a) A1 AB = c, so AB = OC
(a) A1 cso: one pair of opposite sides (AB, OC) equal and parallel, so OABC is a parallelogram
(a) Answer: AB = OC = c, so OABC is a parallelogram
(b) M1 OM = (1/2)(a+c)
(b) A1 correct OM
(b) M1 ON = OA + (1/2)AC = a + (1/2)(c-a)
(b) A1 cso: ON simplifies to (1/2)(a+c) = OM, so M and N coincide
(b) Answer: OM = ON = (1/2)(a+c), so M and N are the same point
(c) M1 substitute a, c into (1/2)(a+c)
(c) M1 (1/2)(5i+2j)
(c) A1 coordinates (2.5, 1) cao
(c) Answer: (2.5, 1)

Question 7

(a) M1 set up (2n)/3 = (n+5)/4 oe from parallel condition
(a) M1 solve: 8n = 3n + 15
(a) A1 n = 3 cao
(a) Answer: n = 3
(b) B1 same direction
(b) B1 reason: q = 2p, a positive scalar multiple of p
(b) Answer: Same direction, since q = 2p
(c) M1 form |q|^2 = (2n)² + (n+5)²
(c) A1 = 5n² + 10n + 25
(c) M1 set equal to (2x5)² = 100 and simplify to n² + 2n - 15 = 0
(c) dM1 solve the quadratic, e.g. (n+5)(n-3) = 0
(c) A1 n = -5 (rejecting the given n = 3) cao
(c) Answer: n = -5

Question 8

(a) M1 set sum of i- and j-components equal to zero (equilibrium)
(a) A1 p = -2
(a) A1 q = -7
(a) Answer: p = -2, q = -7
(b) M1 √p²+q² using values from (a)
(b) A1 awrt 7.28 N
(b) Answer: awrt 7.28 N
(c) M1 recognise F1+F2+F3 = 0 from part (a), so F4 alone equals the new resultant
(c) A1 F4 = 10i N (+0j) cao ft
(c) Answer: F4 = 10i N

Question 9

(a) M1 OB - OA
(a) A1 = 3i - 3j + 6k cao
(a) Answer: AB = 3i - 3j + 6k
(b) M1 find AC = OC - OA
(b) A1 AC = (4t-2)i - 6j + 12k
(b) M1 use j- or k-component ratio to find AC = m x AB with m = 2
(b) dM1 solve i-component: 4t - 2 = 3m
(b) A1 t = 2 cao
(b) Answer: t = 2
(c) M1 find BC = OC - OB using t = 2
(c) A1 BC = 3i - 3j + 6k
(c) A1 cso: BC = AB, so B is the midpoint of AC
(c) Answer: BC = AB, so B is the midpoint of AC

Question 10

(a) M1 OB - OA
(a) A1 = -4i + 8j - 8k cao
(a) Answer: AB = -4i + 8j - 8k
(b) M1 OX = OA + t x AB
(b) A1 cso: correctly simplified to given expression
(b) Answer: OX = (3-4t)i + (8t-1)j + (4-8t)k
(c) M1 form |OX|^2 = (3-4t)² + (8t-1)² + (4-8t)²
(c) A1 = 144t² - 104t + 26 (unsimplified sum correct)
(c) M1 set equal to 10 and rearrange
(c) A1 cso: 18t² - 13t + 2 = 0 shown
(c) dM1 solve the quadratic (factorise or formula)
(c) A1 t = 1/2 or t = 2/9 cao (both values)
(c) B1 ft: both values lie in 0 ≤ t ≤ 1, so both give valid positions of X
(c) Answer: t = 1/2 or t = 2/9

Question 11

(a) B1 AB = b - a oe
(a) Answer: AB = b - a
(b) M1 find OQ = a + (3/4)(b-a)
(b) A1 OQ = (1/4)a + (3/4)b
(b) M1 PQ = OQ - OP
(b) A1 cso: PQ = (3/20)(5b - a) fully simplified as printed
(b) Answer: PQ = (3/20)(5b - a)
(c) M1 find PR = OR - OP
(c) A1 PR = (1/4)b - (2/5)a
(c) M1 attempt to compare PR with PQ as a scalar multiple (compare coefficients of a and b)
(c) A1 cso: coefficient ratios 1/3 (for b) and 8/3 (for a) are unequal, so PR is not a scalar multiple of PQ, hence P, Q, R are not collinear
(c) Answer: P, Q and R are not collinear