Chapter 07 — Vectors (Part 1): Introduction, Notation and Basic Operations

CHAPTER 07: VECTORS

PART 1: INTRODUCTION, VECTOR NOTATION AND BASIC OPERATIONS

Learning objectives
By the end of this lesson, you should be able to:

• Distinguish between scalar and vector quantities.
• Represent vectors using directed line segments, column vectors and unit-vector notation.
• Determine displacement vectors from given points.
• Identify and calculate position vectors.
• Add and subtract vectors correctly.
• Multiply vectors by scalar quantities.
• Calculate the magnitude of two-dimensional and three-dimensional vectors.
• Determine unit vectors in given directions.
• Recognise equal and parallel vectors.

1. INTRODUCTION TO VECTORS

In mathematics and science, quantities can be described by their size alone or by both size and direction. A scalar quantity has magnitude only, while a vector quantity has both magnitude and direction.

1.1 Scalar quantities

Examples of scalar quantities include:

• Mass
• Time
• Temperature
• Distance
• Speed
• Energy

For example, a journey may take 3 hours. Time is a scalar because it has magnitude but no direction.

1.2 Vector quantities

Examples of vector quantities include:

• Displacement
• Velocity
• Acceleration
• Force
• Momentum

For example, a displacement of 5 km east has both magnitude (5 km) and direction (east).

Important distinction:
Distance is a scalar, whereas displacement is a vector. Speed is a scalar, whereas velocity is a vector.

2. REPRESENTATION AND NOTATION OF VECTORS

A vector is commonly represented by a directed line segment. The length represents its magnitude, while the arrow indicates its direction. For example, the vector from point A to point B is written as:

AB⃗

The magnitude of this vector is written as:

|AB⃗|

A vector may also be represented by a bold or ordinary letter with an arrow above it, such as:

a⃗, b⃗, u⃗, v⃗

2.1 Column vector notation in two dimensions

A two-dimensional vector describes movement in the horizontal and vertical directions. A vector with horizontal component x and vertical component y is written as:

a⃗ = ( x
　　 y )

For example:

a⃗ = ( 3
　　 2 )

This represents a movement of 3 units in the positive x-direction and 2 units in the positive y-direction. A negative component indicates movement in the opposite direction. For example:

b⃗ = ( -4
　　 3 )

This represents 4 units to the left and 3 units upwards.

2.2 Unit-vector notation

The standard unit vectors in two dimensions are:

i⃗ = ( 1
　　 0 )　　 j⃗ = ( 0
　　 1 )

Here:

• i⃗ represents one unit in the positive x-direction.
• j⃗ represents one unit in the positive y-direction.

Therefore, a vector can be written in the form:

a⃗ = xi⃗ + yj⃗

For example:

( 5
　-2 ) = 5i⃗ - 2j⃗

2.3 Vectors in three dimensions

In three-dimensional space, a vector has three components: x, y and z.

a⃗ = ( x
　　 y
　　 z )

The three standard unit vectors are:

i⃗ = ( 1
　　 0
　　 0 )　　 j⃗ = ( 0
　　 1
　　 0 )　　 k⃗ = ( 0
　　 0
　　 1 )

Thus:

a⃗ = xi⃗ + yj⃗ + zk⃗

For example:

a⃗ = ( 2
　-3
　 4 ) = 2i⃗ - 3j⃗ + 4k⃗

3. POSITION VECTORS

A position vector describes the position of a point relative to a fixed origin O. Suppose a point P has coordinates (x, y). Its position vector relative to O is:

OP⃗ = ( x
　　 y )

For a point P(x, y, z) in three dimensions:

OP⃗ = ( x
　　 y
　　 z )

Worked Example 1

Find the position vector of point A(4, -3) relative to the origin.

Solution
The position vector is obtained directly from the coordinates of A.

OA⃗ = ( 4
　　 -3 )

Therefore, the position vector of A is (4, -3).

Worked Example 2

Point B has coordinates (-2, 5, 3). Find its position vector in unit-vector notation.

Solution
The position vector is:

OB⃗ = ( -2
　　 5
　　 3 )

In unit-vector notation:

OB⃗ = -2i⃗ + 5j⃗ + 3k⃗

4. DISPLACEMENT VECTORS

A displacement vector describes the change in position when moving from one point to another. If A and B have position vectors:

OA⃗ = a⃗,　　 OB⃗ = b⃗

then the displacement from A to B is:

AB⃗ = OB⃗ - OA⃗ = b⃗ - a⃗

In coordinate form, if A(x₁, y₁) and B(x₂, y₂), then:

AB⃗ = ( x₂ - x₁
　　　 y₂ - y₁ )

In three dimensions:

AB⃗ = ( x₂ - x₁
　　　 y₂ - y₁
　　　 z₂ - z₁ )

Remember: Final point minus starting point.

Worked Example 3

Given A(2, 3) and B(7, 1), find the displacement vector AB⃗.

Solution
Subtract the coordinates of A from those of B.

AB⃗ = ( 7 - 2
　　 1 - 3 )

AB⃗ = ( 5
　　 -2 )

Therefore, the displacement from A to B is (5, -2).

Worked Example 4

Given P(-1, 2, 4) and Q(3, -2, 6), find PQ⃗.

Solution

PQ⃗ = ( 3 - (-1)
　　 -2 - 2
　　 6 - 4 )

PQ⃗ = ( 4
　 -4
　 2 )

Therefore, PQ⃗ = 4i⃗ - 4j⃗ + 2k⃗.

4.1 Reversing the direction of a vector

The vector BA⃗ has the same magnitude as AB⃗ but points in the opposite direction.

BA⃗ = -AB⃗

For example, if:

AB⃗ = ( 3
　　 5 )

then:

BA⃗ = ( -3
　　 -5 )

5. ADDITION OF VECTORS

Vectors can be added by adding their corresponding components. If:

a⃗ = ( a₁
　　 a₂ ),　　 b⃗ = ( b₁
　　 b₂ )

then:

a⃗ + b⃗ = ( a₁ + b₁
　　　　 a₂ + b₂ )

For three-dimensional vectors, add all three corresponding components.

Worked Example 5

Given:

a⃗ = ( 3
　　 4 ),　　 b⃗ = ( 2
　 -1 )

Find a⃗ + b⃗.

Solution

a⃗ + b⃗ = ( 3 + 2
　　　　 4 + (-1) )

a⃗ + b⃗ = ( 5
　　 3 )

Therefore, the sum is (5, 3).

5.1 Geometrical interpretation

Vector addition can be represented using the triangle law or the parallelogram law. In the triangle law, the second vector is placed so that its starting point coincides with the endpoint of the first vector. The resultant is drawn from the beginning of the first vector to the end of the second.

Triangle law:
If a⃗ represents movement from A to B and b⃗ represents movement from B to C, then:

AB⃗ + BC⃗ = AC⃗

6. SUBTRACTION OF VECTORS

Vector subtraction is carried out by subtracting corresponding components.

a⃗ - b⃗ = a⃗ + (-b⃗)

If:

a⃗ = ( a₁
　　 a₂ ),　　 b⃗ = ( b₁
　　 b₂ )

then:

a⃗ - b⃗ = ( a₁ - b₁
　　　　 a₂ - b₂ )

Worked Example 6

Given:

a⃗ = ( 6
　　 2 ),　　 b⃗ = ( 4
　 -3 )

Find a⃗ - b⃗.

Solution

a⃗ - b⃗ = ( 6 - 4
　　　　 2 - (-3) )

a⃗ - b⃗ = ( 2
　　 5 )

Therefore, the answer is (2, 5).

7. MULTIPLICATION OF A VECTOR BY A SCALAR

A scalar is an ordinary number. Multiplying a vector by a scalar changes its magnitude and may reverse its direction. If:

a⃗ = ( x
　　 y )

then:

ka⃗ = ( kx
　　 ky )

where k is a scalar. For example:

a⃗ = ( 2
　　 3 )

3a⃗ = ( 6
　　 9 )

If the scalar is negative, the resulting vector points in the opposite direction.

Worked Example 7

Given:

p⃗ = ( -2
　　 5 )

Find:

1. 4p⃗
2. -2p⃗

Solution (i)

4p⃗ = ( -8
　　 20 )

(ii)

-2p⃗ = ( 4
　 -10 )

8. MAGNITUDE OF A VECTOR

The magnitude of a vector is its length or size. It is always non-negative.

8.1 Magnitude in two dimensions

For:

a⃗ = ( x
　　 y )

the magnitude is:

|a⃗| = √(x² + y²)

This follows from Pythagoras' theorem.

Worked Example 8

Find the magnitude of:

a⃗ = ( 6
　　 8 )

Solution

|a⃗| = √(6² + 8²)

= √(36 + 64)

= √100 = 10

Therefore, |a⃗| = 10 units.

8.2 Magnitude in three dimensions

For:

a⃗ = ( x
　　 y
　　 z )

the magnitude is:

|a⃗| = √(x² + y² + z²)

Worked Example 9

Find the magnitude of:

b⃗ = ( 2
　-3
　 6 )

Solution

|b⃗| = √(2² + (-3)² + 6²)

= √(4 + 9 + 36)

= √49 = 7

Therefore, |b⃗| = 7 units.

8.3 Magnitude of a displacement vector

The magnitude of AB⃗ gives the straight-line distance between points A and B. For A(x₁, y₁) and B(x₂, y₂):

|AB⃗| = √[(x₂ - x₁)² + (y₂ - y₁)²]

Worked Example 10

Find the distance between A(1, 2) and B(4, 6).

Solution First determine the displacement:

AB⃗ = ( 4 - 1
　　 6 - 2 ) = ( 3
　　　　　　　　 4 )

Then calculate its magnitude:

|AB⃗| = √(3² + 4²)

= √25 = 5

The distance AB is 5 units.

9. UNIT VECTORS

A unit vector is a vector whose magnitude is exactly 1. The unit vector in the direction of a non-zero vector a⃗ is obtained by dividing the vector by its magnitude.

â = a⃗ / |a⃗|

This is also written as:

Unit vector = Original vector ÷ Magnitude

Worked Example 11

Find the unit vector in the direction of:

a⃗ = ( 3
　　 4 )

Solution First find the magnitude:

|a⃗| = √(3² + 4²) = 5

Divide each component by 5:

â = ( 3/5
　　 4/5 )

Therefore, the unit vector is:

â = ⅗i⃗ + ⅘j⃗

Check:

|â| = √[(3/5)² + (4/5)²] = √(9/25 + 16/25) = 1

Worked Example 12

Find the unit vector in the direction of:

b⃗ = ( 2
　-3
　 6 )

Solution The magnitude is:

|b⃗| = √(4 + 9 + 36) = 7

Therefore:

b̂ = ( 2/7
　-3/7
　 6/7 )

In unit-vector notation:

b̂ = 2/7 i⃗ - 3/7 j⃗ + 6/7 k⃗

Important: A zero vector has magnitude zero and has no defined direction. Therefore, a unit vector cannot be obtained by dividing the zero vector by its magnitude.

10. EQUAL VECTORS

Two vectors are equal if they have:

• The same magnitude.
• The same direction.

For column vectors, equality means that their corresponding components are equal. For example:

a⃗ = ( 3
　　 5 ),　　 b⃗ = ( 3
　　 5 )

Therefore:

a⃗ = b⃗

Two directed line segments can be in different positions and still represent equal vectors, provided they have the same length and direction.

Worked Example 13

Find x and y if:

( 2x - 1
　　 y + 3 ) = ( 7
　　 8 )

Solution Equate corresponding components. First component:

2x - 1 = 7

2x = 8 ⇒ x = 4

Second component:

y + 3 = 8

y = 5

Therefore:

x = 4,　　 y = 5

11. PARALLEL VECTORS

Two non-zero vectors are parallel if one is a scalar multiple of the other. If:

a⃗ = kb⃗

where k is a non-zero scalar, then a⃗ and b⃗ are parallel. For example:

a⃗ = ( 2
　　 4 ),　　 b⃗ = ( 1
　　 2 )

Since:

a⃗ = 2b⃗

the vectors are parallel. If k is positive, the vectors point in the same direction. If k is negative, they point in opposite directions.

Worked Example 14

Determine whether the vectors:

p⃗ = ( 6
　-9 ),　　 q⃗ = ( -2
　　 3 )

are parallel.

Solution Compare their components:

6/(-2) = -3

(-9)/3 = -3

Both ratios are equal. Hence:

p⃗ = -3q⃗

Therefore, the vectors are parallel and point in opposite directions.

12. IMPORTANT LAWS OF VECTOR ALGEBRA

For vectors a⃗ and b⃗ and scalars m and n, the following laws apply:

Law	Statement
Commutative law	a⃗ + b⃗ = b⃗ + a⃗
Associative law	(a⃗ + b⃗) + c⃗ = a⃗ + (b⃗ + c⃗)
Zero vector	a⃗ + 0⃗ = a⃗
Additive inverse	a⃗ + (-a⃗) = 0⃗
Scalar distribution	m(a⃗ + b⃗) = ma⃗ + mb⃗
Scalar associativity	m(na⃗) = (mn)a⃗

13. EXAMINATION TIPS AND COMMON ERRORS

Remember these points:

1. For displacement, subtract the starting point from the final point.
2. When adding or subtracting column vectors, work with corresponding components.
3. When multiplying by a scalar, multiply every component.
4. Magnitude is always non-negative.
5. For a unit vector, divide every component by the magnitude.
6. Equal vectors must have equal corresponding components.
7. Parallel vectors must be scalar multiples of one another.
8. Do not confuse a position vector with a displacement vector.

14. EXERCISES

Attempt all questions before opening the solutions.

Exercise A: Basic understanding and notation

1. Distinguish between a scalar and a vector quantity, giving three examples of each.
2. Write the vector (4, -7) in unit-vector notation.
3. Express 3i⃗ - 5j⃗ + 2k⃗ as a column vector.
4. Write down the three standard unit vectors in three dimensions.
5. Explain why displacement is a vector but distance is a scalar.

Exercise B: Position and displacement vectors

1. Find the position vector of A(5, -2).
2. Find the position vector of P(-3, 4, 2).
3. Given A(2, -1) and B(8, 5), find AB⃗.
4. Given P(1, 2, -3) and Q(4, -2, 5), find PQ⃗.
5. If OA⃗ = (3, 4) and OB⃗ = (-2, 6), find BA⃗.

Exercise C: Vector operations

1. Given a⃗ = (2, 5) and b⃗ = (4, -3), find a⃗ + b⃗.
2. Given p⃗ = (7, -2) and q⃗ = (3, 6), find p⃗ - q⃗.
3. If a⃗ = (2, -1, 3) and b⃗ = (4, 5, -2), find a⃗ + b⃗.
4. Given u⃗ = (-3, 2), find 5u⃗.
5. Given v⃗ = (4, -2, 1), find -3v⃗.

Exercise D: Magnitude and unit vectors

1. Find the magnitude of (5, 12).
2. Find the magnitude of (-3, 4).
3. Find the magnitude of (2, -3, 6).
4. Find the unit vector in the direction of (8, 6).
5. Find the unit vector in the direction of (1, 2, 2).

Exercise E: Equal and parallel vectors

1. Find x and y if (x + 2, 2y - 1) = (7, 9).
2. Determine whether (4, 6) and (2, 3) are parallel.
3. Determine whether (3, -5) and (-6, 10) are parallel.
4. Find k if (6, 12) = k(2, 4).
5. Find the scalar k if (3, -6, 9) = k(1, -2, 3).

Exercise F: Examination-style problems

1. The position vectors of points A and B relative to the origin are:
   OA⃗ = ( 2
   　　 3 ),　　 OB⃗ = ( 8
   　　 7 )
   Find:
   1. AB⃗.
   2. BA⃗.
   3. The magnitude of AB⃗.
2. Given:
   a⃗ = ( 3
   　　 4 ),　　 b⃗ = ( -1
   　　 2 )
   Find:
   1. 2a⃗ - b⃗.
   2. |2a⃗ - b⃗|.
3. A point P has position vector:
   OP⃗ = ( 2
   　-1
   　 2 )
   Find:
   1. The magnitude of OP⃗.
   2. The unit vector in the direction of OP⃗.
4. The vectors a⃗ and b⃗ are given by:
   a⃗ = ( x
   　　 6 ),　　 b⃗ = ( 4
   　　 y )
   If a⃗ = b⃗, find x and y.
5. A displacement vector is given by:
   d⃗ = ( 5
   　-12 )
   Find its magnitude and the unit vector in its direction.

15. EXPANDABLE ANSWERS AND SOLUTIONS

Exercise A — Answers

1. A scalar has magnitude only; a vector has magnitude and direction. Scalars: mass, time, temperature. Vectors: displacement, velocity, force. 2. 4i⃗ - 7j⃗. 3. (3, -5, 2). 4. i⃗ = (1,0,0), j⃗ = (0,1,0), k⃗ = (0,0,1). 5. Displacement includes direction; distance measures total length only.

Exercise B — Worked answers

1. OA⃗ = (5, -2). 2. OP⃗ = (-3, 4, 2). 3. AB⃗ = (8-2, 5-(-1)) = (6, 6). 4. PQ⃗ = (4-1, -2-2, 5-(-3)) = (3, -4, 8). 5. BA⃗ = OA⃗ - OB⃗ = (3-(-2), 4-6) = (5, -2).

Exercise C — Worked answers

1. (2+4, 5-3) = (6, 2). 2. (7-3, -2-6) = (4, -8). 3. (2+4, -1+5, 3-2) = (6, 4, 1). 4. 5(-3, 2) = (-15, 10). 5. -3(4, -2, 1) = (-12, 6, -3).

Exercise D — Worked answers

1. √(25+144) = √169 = 13. 2. √(9+16) = 5. 3. √(4+9+36) = 7. 4. Magnitude = √(64+36) = 10. Unit vector = (8/10, 6/10) = (4/5, 3/5). 5. Magnitude = √(1+4+4) = 3. Unit vector = (1/3, 2/3, 2/3).

Exercise E — Worked answers

1. x+2=7 gives x=5. Also 2y-1=9 gives y=5. 2. Yes. (4,6) = 2(2,3). 3. Yes. (3,-5) = -½(-6,10). 4. 6=2k gives k=3. The second component confirms the result. 5. 3=k gives k=3. The other components also confirm k=3.

Exercise F — Full solutions

Question 1 (a) AB⃗ = OB⃗ - OA⃗ = (8-2, 7-3) = (6,4). (b) BA⃗ = (-6,-4). (c) |AB⃗| = √(6²+4²) = √52 = 2√13.

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Question 2 (a) First calculate 2a⃗: 2a⃗ = (6,8). Therefore: 2a⃗ - b⃗ = (6-(-1), 8-2) = (7,6). (b) Magnitude = √(7²+6²) = √85.

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Question 3 (a) |OP⃗| = √(2²+(-1)²+2²) = √9 = 3. (b) Unit vector = (2/3, -1/3, 2/3).

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Question 4 Equating corresponding components: x=4 and y=6.

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Question 5 Magnitude = √(5²+(-12)²) = √169 = 13. Unit vector = (5/13, -12/13).

16. PART 1 SUMMARY

The central formulae from this lesson are:

Displacement: AB⃗ = OB⃗ - OA⃗

Addition: (a₁,a₂) + (b₁,b₂) = (a₁+b₁,a₂+b₂)

Scalar multiplication: k(a₁,a₂) = (ka₁,ka₂)

2D magnitude: |a⃗| = √(x²+y²)

3D magnitude: |a⃗| = √(x²+y²+z²)

Unit vector: â = a⃗/|a⃗|

Parallel vectors: a⃗ = kb⃗

End of Chapter 07 — Part 1
Make sure you can calculate displacement, magnitude and unit vectors confidently before moving to the next part. These skills will be used repeatedly in more advanced vector problems.

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