LESSON 01
Vectors and Scalars
Mechanics contains quantities that describe both the size of something and the direction in which it acts. These are vector quantities.
Scalar
A scalar has magnitude only.
- Mass
- Time
- Distance
- Speed
- Energy
Vector
A vector has both magnitude and direction.
- Displacement
- Velocity
- Acceleration
- Force
A physical quantity described by both a magnitude and a direction.
Distance and displacement are not the same. Distance is the total length travelled and is scalar. Displacement describes the directed change in position and is vector.
LESSON 02
Representing Vectors
A vector can be represented geometrically by a directed line segment. The arrowhead indicates direction and the length of the line represents its magnitude.
Magnitude
The size or length of the vector.
Direction
The orientation in which the vector acts.
Two vectors are equal if they have the same magnitude and the same direction, regardless of where they are drawn.
A vector may therefore be translated parallel to itself without changing the vector.
Do not decide whether two vectors are equal merely because they are drawn in different positions. Compare their magnitude and direction.
LESSON 03
Position Vectors
A position vector describes the position of a point relative to a chosen origin.
The vector from the origin O to a point P is called the position vector of P.
Using Cartesian unit vectors, a point with coordinates (x, y) has position vector x i + y j.
Here:
- i is the unit vector in the positive x-direction.
- j is the unit vector in the positive y-direction.
- x and y are the horizontal and vertical components.
WORKED EXAMPLE 01 • Position Vector
Point A has coordinates (4, −3). Find the position vector OA.
The x-coordinate is 4.
The y-coordinate is −3.
LESSON 04
Unit Vectors
A unit vector is a vector with magnitude exactly 1.
The standard Cartesian unit vectors are:
i
Unit vector in the positive x-direction.
j
Unit vector in the positive y-direction.
Therefore a vector such as 3i + 4j has horizontal component 3 and vertical component 4.
Its magnitude is found using Pythagoras.
The coefficient of i or j is a component, not automatically the magnitude of the entire vector.
LESSON 05
Magnitude and Direction
For a vector a = xi + yj, the magnitude is:
The direction can be described using an angle measured from a chosen axis.
The formula above must be interpreted carefully because the signs of x and y determine the quadrant.
WORKED EXAMPLE 02 • Magnitude
Find the magnitude of a = 6i + 8j.
|a| = √(6² + 8²)
|a| = √(36 + 64)
|a| = √100
|a| = 10
WORKED EXAMPLE 03 • Direction
A vector has components 5i + 12j. Find the angle it makes with the positive x-axis.
tan θ = 12 / 5
θ = tan⁻¹(12/5)
θ ≈ 67.4°
If a component is negative, the vector may lie in the second, third or fourth quadrant. Do not simply quote the calculator's inverse-tangent result without checking the direction.
LESSON 06
Vector Addition, Subtraction and Multiplication
Vectors written in component form can be manipulated component by component.
WORKED EXAMPLE 04 • Addition
Let a = 3i + 5j and b = 2i − 7j. Find a + b.
a + b = (3 + 2)i + (5 − 7)j
a + b = 5i − 2j
WORKED EXAMPLE 05 • Scalar Multiplication
Find −3(2i − 4j).
Multiply every component by −3.
−3(2i − 4j) = −6i + 12j
Answer: −6i + 12j
LESSON 07
Displacement Vectors
If a particle moves from point A to point B, its displacement is the vector from A to B.
This formula is one of the most important vector relationships in mechanics.
WORKED EXAMPLE 06 • Displacement
A particle moves from A(2, 3) to B(8, 7). Find its displacement.
OA = 2i + 3j
OB = 8i + 7j
AB = OB − OA
AB = (8 − 2)i + (7 − 3)j
AB = 6i + 4j
Displacement from A to B is B minus A. Reversing the order gives the displacement from B to A.
LESSON 08
Velocity as a Vector
Velocity describes the rate of change of displacement and has both magnitude and direction.
In vector form, if a particle has displacement vector s over a time interval t, its average velocity is:
In component form:
WORKED EXAMPLE 07 • Average Velocity
A particle moves through displacement 12i + 5j metres in 4 seconds. Find its average velocity.
v = (12i + 5j)/4
v = 3i + 1.25j m s⁻¹
Speed is the magnitude of velocity. Velocity itself includes direction.
LESSON 09
Acceleration as a Vector
Acceleration describes the rate of change of velocity.
If a particle changes velocity from u to v over time t:
WORKED EXAMPLE 08 • Acceleration Vector
A particle has initial velocity 2i + 3j m s⁻¹ and final velocity 10i + 7j m s⁻¹ after 4 seconds. Find its acceleration.
Change in velocity:
v − u = (10i + 7j) − (2i + 3j)
= 8i + 4j
Therefore:
a = (8i + 4j)/4
a = 2i + j m s⁻²
A particle can accelerate even if its speed is unchanged, because a change in direction is also a change in velocity.
LESSON 10
Relative Position and Displacement
Vector subtraction can be used to determine the position of one particle relative to another.
If particles A and B have position vectors a and b respectively, then:
This is simply the position of B minus the position of A.
WORKED EXAMPLE 09 • Relative Position
At a particular instant, A has position vector 4i + 2j and B has position vector 9i − 4j. Find AB.
AB = OB − OA
AB = (9i − 4j) − (4i + 2j)
AB = 5i − 6j
AB = 5i − 6j
AB and BA have the same magnitude but opposite directions.
LESSON 11
Vector Equations of Motion
Mechanics equations can be written using vectors by treating each component consistently.
This can be applied directly to vector quantities when the acceleration is constant.
When using vector equations, each term must have compatible vector units and the vector directions must be handled correctly.
WORKED EXAMPLE 10 • Constant Acceleration
A particle has initial velocity 3i + 2j m s⁻¹ and constant acceleration i − 2j m s⁻². Find its velocity after 5 seconds.
v = u + at
v = (3i + 2j) + (i − 2j)(5)
v = 3i + 2j + 5i − 10j
v = 8i − 8j m s⁻¹
If you find a vector equation difficult, separate it into its i-component and j-component and solve each independently.
LESSON 12
Exam Method: A Reliable Vector Workflow
Most vector errors in mechanics are not difficult mathematics; they are caused by incorrect interpretation or sign errors.
- Identify the points and directions.
- Write position vectors clearly.
- Use final position minus initial position for displacement.
- Keep the i and j components separate.
- Check signs carefully.
- Use Pythagoras for magnitude.
- Use trigonometry for direction.
- Include correct units where appropriate.
LESSON 13
Advanced Worked Examples
EXAMPLE 11 • Distance Between Two Points
A has position vector 2i − 5j and B has position vector 8i + 3j. Find the distance AB.
AB = (8 − 2)i + (3 − (−5))j
AB = 6i + 8j
|AB| = √(6² + 8²)
|AB| = 10
EXAMPLE 12 • Finding an Unknown Component
A vector has magnitude 13 and is 5i + kj. Find the possible values of k.
√(5² + k²) = 13
25 + k² = 169
k² = 144
k = ±12
EXAMPLE 13 • Equal Vectors
If a = (2p + 1)i + 7j and b = 9i + 7j are equal, find p.
Equal vectors have equal corresponding components.
2p + 1 = 9
2p = 8
p = 4
EXAMPLE 14 • Finding Time from Displacement
A particle travels with constant velocity 4i + 3j m s⁻¹. Find its displacement after 6 seconds.
s = vt
s = (4i + 3j)(6)
s = 24i + 18j m
EXAMPLE 15 • Position After a Given Time
A particle starts at position 2i − j metres. It moves with constant velocity 3i + 4j m s⁻¹. Find its position after 5 seconds.
Final position = initial position + displacement.
Displacement = vt
= (3i + 4j)(5)
= 15i + 20j
Final position:
(2i − j) + (15i + 20j)
17i + 19j m
EXAMPLE 16 • Finding Acceleration
A particle's velocity changes from 7i − 2j to 3i + 10j in 4 seconds. Find the acceleration.
Δv = (3i + 10j) − (7i − 2j)
Δv = −4i + 12j
a = Δv / 4
a = −i + 3j m s⁻²
LESSON 14
Common Exam Traps
AB = B − A, not A − B.
Distance is scalar. Displacement is vector.
A negative i or j component represents direction opposite to the positive axis.
For xi + yj, magnitude is √(x² + y²), not x + y.
Always consider the signs of both components before reporting a direction.
Position/displacement should normally be in metres, velocity in m s⁻¹ and acceleration in m s⁻².
The magnitude of velocity gives speed, but velocity itself includes direction.
Write the vector relationship first. Substitute numbers second. This dramatically reduces sign and direction errors.
LESSON 15
Vector Formula Sheet
Position → displacement → velocity → acceleration. Each step involves comparing the appropriate vector quantity over time.
LESSON 16
Final Exam Checklist
- Can you distinguish vectors from scalars?
- Can you write a position vector using i and j?
- Can you find displacement from two position vectors?
- Can you add and subtract vectors?
- Can you multiply a vector by a scalar?
- Can you calculate vector magnitude?
- Can you determine the direction of a vector?
- Can you calculate average velocity from displacement?
- Can you calculate acceleration from two velocity vectors?
- Can you use vector equations of motion?
- Can you interpret negative components correctly?
- Can you distinguish AB from BA?
The 100-question assessment below mixes definitions, calculations, interpretation and exam-style vector problems.
ASSESSMENT SYSTEM
100-Question Vectors Exam
Test the complete topic. Each question has four possible answers. You can move freely between questions and return to previous questions before submitting.
VECTOR CORE EXAM
100 MCQs • Mechanics 1 • No timer
Select one answer for each question. Use the numbered palette to jump between questions. Submit when you are finished to receive your score and full explanations.