Vectors in Two Dimensions: Notation, Magnitude and Arithmetic - Worksheets, Questions and Revision

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

Download PDF Jump to mark scheme (page 5)Vectors in Two Dimensions: Notation, Magnitude and Arithmetic revision guide

This topic is chapter 5 of IGCSE Maths Practice Book 2.

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3.4 Vectors in Two Dimensions: Notation, Magnitude and Arithmetic

EDEXCEL 4MA1 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A displacement is 3 units to the right and 5 units up in a plane. Write down the column vector that represents this displacement, using column-vector notation for vectors in two dimensions.

(Total for Question 1 is 1 mark)

Vector u is shown in bold notation as u = 4i - 2j. Write u as a column vector, using column-vector notation for two-dimensional vectors.

(Total for Question 2 is 1 mark)

Given s = [-2; 6], find -2s and give your answer in column-vector form.

(Total for Question 3 is 2 marks)

Write down whether the following pairs of vectors are parallel, perpendicular or neither. (i) [2; 4] and [1; 2] (ii) [3; -6] and [2; 1]

(Total for Question 4 is 2 marks)

The vector v = [6; 8]. Calculate the magnitude, |v|, of v to 3 significant figures.

(Total for Question 5 is 2 marks)

Given vectors a = [2; -3] and b = [-1; 4], find the vector 3a and give your answer in column-vector form.

(Total for Question 6 is 2 marks)

With a = [2; -3] and b = [-1; 4] as above, calculate the vector a + b and give your answer in column form.

(Total for Question 7 is 2 marks)

Vector p = [7; 2] and vector q = [3; -5]. Calculate p - q and give your answer in column-vector form.

(Total for Question 8 is 2 marks)

Let a = [3; 1] and b = [1; 2]. Express the vector 2a - b in column form, and then simplify the components.

(Total for Question 9 is 3 marks)

A translation by vector t = [ -3; 5 ] maps point X with position vector OX = [7; 0] to X'. Find the position vector OX' and give it in column form. Then state the image of the point with position vector [2; -1] under the same translation.

(Total for Question 10 is 3 marks)

A triangle has vertices with position vectors from origin O as OA = [1; 2], OB = [4; 1]. Find the vector AB in column form.

(Total for Question 11 is 3 marks)

A translation T maps point P with position vector OP = [2; 5] to point P' with position vector OP' = [6; 2]. Find the translation vector t that describes T, and state t in column-vector form.

(Total for Question 12 is 3 marks)

The vector r = [5; -2]. Calculate the magnitude |r| and give your answer to 3 significant figures.

(Total for Question 13 is 3 marks)

Given a = [2; -1] and b = [4; 3], show that 3a + 2b = [14; 3] by calculating and giving working.

(Total for Question 14 is 3 marks)

Given vectors m = [6; 15] and n = [ -2; -5 ], show that m = -3 n, and hence state whether m and n are parallel. Give working.

(Total for Question 15 is 4 marks)

Vectors a and b satisfy 2a + 3b = [11; 7] and a - b = [1; -1]. Find a and b by solving the pair of vector equations, giving your answers in column-vector form.

(Total for Question 16 is 4 marks)

Mark scheme · 3.4 Vectors in Two Dimensions: Notation, Magnitude and Arithmetic

Question 1

B1 column vector [3; 5] written in column form, cao
Answer: [3; 5]

Question 2

B1 u = [4; -2] cao
Answer: [4; -2]

Question 3

M1 multiplies each component of s by -2
A1 -2s = [4; -12] cao
Answer: [4; -12]

Question 4

B1 i) parallel, ii) neither; both correct statements
B1 brief justification implicit in identification, e.g. first is a scalar multiple, second is not scalar multiple nor dot product zero
Answer: i) parallel ii) neither

Question 5

M1 uses Pythagoras: |v| = √6² + 8²
A1 correct magnitude 10.0 to 3 s.f., cao
Answer: 10.0

Question 6

M1 multiplies each component of a by 3, e.g. shows 3 x 2 and 3 x (-3)
A1 3a = [6; -9] cao
Answer: [6; -9]

Question 7

M1 adds corresponding components 2 + (-1) and -3 + 4
A1 a + b = [1; 1] cao
Answer: [1; 1]

Question 8

M1 subtracts corresponding components: 7 - 3 and 2 - (-5)
A1 p - q = [4; 7] cao
Answer: [4; 7]

Question 9

M1 shows method: forms 2a and subtracts b, e.g. 2[3;1] - [1;2]
M1 performs component arithmetic correctly, e.g. [6;2] - [1;2]
A1 gives final simplified vector [5; 0] cao
Answer: [5; 0]

Question 10

M1 adds t to OX correctly to find OX', and adds t to [2; -1] for the second image
M1 computes components correctly for both images
A1 OX' = [4; 5] and image of [2; -1] is [-1; 4], cao
Answer: [4; 5] and [-1; 4]

Question 11

M1 uses AB = OB - OA or shows subtraction of position vectors
M1 performs components subtraction: [4 - 1; 1 - 2]
A1 AB = [3; -1] cao
Answer: [3; -1]

Question 12

M1 uses t = OP' - OP or equivalent
M1 computes components 6 - 2 and 2 - 5 correctly
A1 t = [4; -3] cao
Answer: [4; -3]

Question 13

M1 uses Pythagoras: |r| = √5² + (-2)²
M1 evaluates the square root correctly, e.g. √25+4 = √29
A1 gives |r| = 5.39 to 3 s.f., cao
Answer: 5.39

Question 14

M1 forms 3a and 2b correctly, e.g. 3[2; -1] and 2[4;3]
M1 adds the two resulting vectors componentwise
A1 correct result 3a + 2b = [14; 3] cao

Question 15

M1 compares components and shows factor -3: -3[-2; -5] = [6;15]
M1 states the conclusion m = -3 n
B1 states that since one is a scalar multiple of the other they are parallel
A1 final statement is correct and clear, cao

Question 16

M1 writes the system of equations for components, or uses vector elimination: from a - b = [1; -1] get a = b + [1; -1]
M1 substitutes into 2a + 3b = [11;7] and simplifies to an equation in b, for both components
M1 solves for b correctly, giving b = [2; 3]
A1 finds a = b + [1; -1] = [3; 2] cao

Vectors in Two Dimensions: Notation, Magnitude and Arithmetic - Worksheets, Questions and Revision

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