Vectors : Detailed Notes, Solved Examples & Exam Questions | Grade 11 Mathematics Unit 5

Introduction to Vectors

Hello dear student! Welcome to Unit 5 of your Grade 11 Mathematics. In this unit, we are going to learn about Vectors. Have you ever wondered how a pilot knows the exact direction and speed to fly an airplane? Or how a football player decides the exact force and direction to kick a ball? All of these involve something called vectors. Let us start from the very beginning.

In your earlier classes, you studied many quantities like mass, time, temperature, speed, and distance. These quantities only tell you “how much” — they have magnitude (size) only. But some quantities need more information — they tell you “how much” AND “in which direction.” That is the key difference!

Scalars and Vectors

Scalar Quantities

A scalar is a quantity that has only magnitude (size or numerical value). It does not have any direction.

Examples: Mass (5 kg), Time (3 hours), Temperature (37°C), Distance (10 m), Speed (60 km/h)

Vector Quantities

A vector is a quantity that has both magnitude AND direction.

Examples: Displacement (5 m East), Velocity (60 km/h North), Force (10 N downward), Acceleration (9.8 m/s² downward)

Notation and Representation

Vectors can be represented in several ways:

1. Bold letter: \( \mathbf{a} \), \( \mathbf{F} \), \( \mathbf{v} \)

2. Arrow above letter: \( \vec{a} \), \( \vec{F} \), \( \vec{v} \)

3. Underlined letter: \( \underline{a} \)

In this lesson, we will mostly use the arrow notation: \( \vec{a} \), \( \vec{b} \), etc.

Graphical Representation

A vector is drawn as a directed line segment (an arrow). The length of the arrow represents the magnitude, and the arrowhead shows the direction.

The vector from point A to point B is written as \( \vec{AB} \). Here, A is the initial point (tail) and B is the terminal point (head).

Magnitude of a Vector

The magnitude (length) of vector \( \vec{a} \) is written as \( |\vec{a}| \) or simply \( a \). It is always a positive number (or zero).

Types of Vectors

Before we do calculations, let us learn the different types of vectors you must know for your exam:

Vector Addition

Now let us learn how to add vectors. This is one of the most important operations in this unit. There are two main methods: the Triangle Law and the Parallelogram Law.

Triangle Law of Vector Addition

Statement: If two vectors \( \vec{a} \) and \( \vec{b} \) are represented as two sides of a triangle taken in order, then their sum (resultant) \( \vec{a} + \vec{b} \) is represented by the third side of the triangle taken from the tail of the first to the head of the second.

In simple words: Place the tail of \( \vec{b} \) at the head of \( \vec{a} \). Then the vector from the tail of \( \vec{a} \) to the head of \( \vec{b} \) is the sum.

Worked Example 1: A person walks 4 km East and then 3 km North. Find the resultant displacement.

Solution:

Let \( \vec{a} = 4\hat{i} \) (4 km East) and \( \vec{b} = 3\hat{j} \) (3 km North).

Magnitude of the resultant:

Direction (angle with East):

So the person is 5 km away from the starting point, at an angle of approximately 36.87° North of East.

Did you notice? This is just like the Pythagorean theorem! The triangle formed is a right triangle with sides 3 and 4, so the hypotenuse is 5. That is the beauty of vectors.

Parallelogram Law of Vector Addition

Statement: If two vectors \( \vec{a} \) and \( \vec{b} \) are represented as two adjacent sides of a parallelogram, then their resultant \( \vec{a} + \vec{b} \) is represented by the diagonal of the parallelogram passing through the common point.

Magnitude of the Resultant:

where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \).

Direction of the Resultant:

where \( \alpha \) is the angle the resultant makes with vector \( \vec{a} \).

Worked Example 2: Two forces of 6 N and 8 N act on a body at an angle of 60° to each other. Find the magnitude and direction of the resultant force.

Solution:

Given: \( |\vec{F}_1| = 6 \) N, \( |\vec{F}_2| = 8 \) N, \( \theta = 60° \)

Direction:

Practice Question 1: Two vectors of magnitudes 5 and 12 are added. What is the maximum possible magnitude of their resultant? What is the minimum?

Practice Question 2: If \( |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}| \), what can you say about the angle between \( \vec{a} \) and \( \vec{b} \)?

Vector Subtraction

Vector subtraction is actually a special case of addition. Subtracting a vector is the same as adding its negative.

Graphical method: To find \( \vec{a} – \vec{b} \), place the tail of \( \vec{b} \) at the tail of \( \vec{a} \). Then draw a vector from the head of \( \vec{b} \) to the head of \( \vec{a} \). This gives \( \vec{a} – \vec{b} \).

Magnitude:

Worked Example 3: If \( \vec{a} \) has magnitude 10 and \( \vec{b} \) has magnitude 6 and the angle between them is 60°, find \( |\vec{a} – \vec{b}| \).

Solution:

Multiplication of a Vector by a Scalar

When we multiply a vector \( \vec{a} \) by a scalar (real number) \( k \), we get a new vector \( k\vec{a} \) whose:

• Magnitude: \( |k\vec{a}| = |k| \cdot |\vec{a}| \)

• Direction: Same as \( \vec{a} \) if \( k > 0 \); opposite to \( \vec{a} \) if \( k < 0 \); zero vector if \( k = 0 \).

Properties of Scalar Multiplication:

Worked Example 4: If \( \vec{a} = 3\hat{i} – 2\hat{j} + \hat{k} \), find \( 4\vec{a} \) and \( -2\vec{a} \).

Solution:

Practice Question 3: Find a vector in the direction of \( \vec{a} = 2\hat{i} – \hat{j} + 2\hat{k} \) that has magnitude 7.

Components of a Vector

This is a very important concept. Any vector in 2D or 3D can be broken into components along the coordinate axes.

2D Vectors (in the XY-plane)

A vector \( \vec{a} \) in the XY-plane can be written as:

where \( \hat{i} \) is a unit vector along the x-axis and \( \hat{j} \) is a unit vector along the y-axis.

Here, \( a_x \) is the x-component and \( a_y \) is the y-component.

If the vector makes an angle \( \theta \) with the positive x-axis and has magnitude \( |\vec{a}| = a \):

3D Vectors (in space)

A vector in 3D is written as:

where \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along x, y, and z axes respectively.

Worked Example 5: Find the magnitude and direction (angle with x-axis) of the vector \( \vec{a} = 4\hat{i} + 3\hat{j} \).

Solution:

Worked Example 6: A vector \( \vec{a} \) has magnitude 10 and makes an angle of 30° with the x-axis. Find its components.

Solution:

Position Vector of a Point

The position vector of a point \( P(x, y, z) \) with respect to the origin \( O(0, 0, 0) \) is:

If point A has position vector \( \vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} \) and point B has position vector \( \vec{b} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k} \), then:

Worked Example 7: If A(2, -1, 3) and B(5, 2, -1), find \( \vec{AB} \) and its magnitude.

Solution:

Direction Cosines and Direction Ratios

This is another very important topic for your exam. Pay close attention!

Direction Cosines

If a vector \( \vec{a} \) makes angles \( \alpha, \beta, \gamma \) with the positive x, y, and z axes respectively, then:

These are called the direction cosines of the vector, denoted by \( l, m, n \):

Important relation:

Why is this important? Because if you know two direction cosines, you can always find the third one! Let me show you…

Worked Example 8: A vector has direction cosines \( l = \frac{1}{2} \) and \( m = \frac{1}{2} \). Find \( n \).

Solution:

Note: There are TWO possible values for \( n \) — positive and negative. Both are valid!

Direction Ratios

Direction ratios (also called direction numbers) are three numbers \( a, b, c \) that are proportional to the direction cosines \( l, m, n \).

Worked Example 9: Find the direction cosines of the vector \( \vec{a} = 2\hat{i} – 3\hat{j} + 6\hat{k} \).

Solution:

Verification: \( l^2 + m^2 + n^2 = \frac{4}{49} + \frac{9}{49} + \frac{36}{49} = \frac{49}{49} = 1 \) ✓

Practice Question 5: If a line makes angles 90° and 60° with the x-axis and y-axis respectively, find the angle it makes with the z-axis.

Scalar (Dot) Product of Two Vectors

Now we come to one of the two types of vector multiplication. The dot product (also called scalar product or inner product) of two vectors gives a scalar (a number) as the result.

Definition

The dot product of \( \vec{a} \) and \( \vec{b} \) is:

where \( \theta \) is the angle between the two vectors (0° ≤ θ ≤ 180°).

Component Form (for 3D vectors)

If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then:

Worked Example 10: Find \( \vec{a} \cdot \vec{b} \) if \( \vec{a} = 2\hat{i} + 3\hat{j} – \hat{k} \) and \( \vec{b} = \hat{i} – \hat{j} + 2\hat{k} \).

Solution:

Finding the Angle Between Two Vectors

From the definition, we can find the angle:

Worked Example 11: Find the angle between \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} – \hat{j} \).

Solution:

The vectors are perpendicular! (This makes sense because the dot product is zero.)

Properties of the Dot Product

Projection of a Vector

The projection of vector \( \vec{a} \) on vector \( \vec{b} \) is:

The vector projection of \( \vec{a} \) on \( \vec{b} \) is:

Worked Example 12: Find the projection of \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \) on \( \vec{b} = \hat{i} – \hat{j} + \hat{k} \).

Solution:

The projection is zero because the vectors are perpendicular!

Practice Question 6: If \( \vec{a} = 3\hat{i} – \hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} – \hat{k} \), find the angle between them and state whether they are acute or obtuse.

Vector (Cross) Product of Two Vectors

The second type of vector multiplication gives a vector as the result. This is why it is called the vector product or cross product.

Definition

The cross product of \( \vec{a} \) and \( \vec{b} \) is:

where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), and \( \hat{n} \) is a unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) (determined by the right-hand rule).

Right-Hand Rule: Point your index finger along \( \vec{a} \), middle finger along \( \vec{b} \), then your thumb points in the direction of \( \vec{a} \times \vec{b} \).

Component Form (Determinant Method)

If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \):

Worked Example 13: Find \( \vec{a} \times \vec{b} \) if \( \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \) and \( \vec{b} = \hat{i} + 3\hat{j} – \hat{k} \).

Solution:

Properties of the Cross Product

Magnitude of Cross Product and Area

This gives us important geometric results:

• Area of parallelogram with adjacent sides \( \vec{a} \) and \( \vec{b} \): \( |\vec{a} \times \vec{b}| \)

• Area of triangle with adjacent sides \( \vec{a} \) and \( \vec{b} \): \( \frac{1}{2}|\vec{a} \times \vec{b}| \)

Worked Example 14: Find the area of the triangle with vertices A(1, 2, 3), B(2, 0, 1), C(3, 1, 2).

Solution:

Practice Question 7: If \( \vec{a} = 2\hat{i} – \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} – 2\hat{k} \), find \( \vec{a} \times \vec{b} \) and verify that \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \).

Unit Vector

A unit vector in the direction of a given vector \( \vec{a} \) is obtained by dividing \( \vec{a} \) by its magnitude:

This always has magnitude 1.

Worked Example 15: Find the unit vector in the direction of \( \vec{a} = 3\hat{i} – 4\hat{j} \).

Solution:

Verification: \( |\hat{a}| = \sqrt{\frac{9}{25}+\frac{16}{25}} = \sqrt{\frac{25}{25}} = 1 \) ✓

Vector Operations Summary with Component Form

Let \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \).

Collinearity and Coplanarity Tests

Test for Collinearity (Parallel Vectors)

Two vectors \( \vec{a} \) and \( \vec{b} \) are collinear (parallel) if and only if:

Worked Example 16: Determine whether \( \vec{a} = 2\hat{i} + 4\hat{j} – 6\hat{k} \) and \( \vec{b} = -\hat{i} – 2\hat{j} + 3\hat{k} \) are collinear.

Solution:

Notice that \( \vec{a} = -2\vec{b} \):

Since \( \vec{a} = \lambda\vec{b} \) with \( \lambda = -2 \), the vectors are collinear.

Test for Coplanarity

Three vectors \( \vec{a}, \vec{b}, \vec{c} \) are coplanar if and only if:

This is called the scalar triple product.

Worked Example 17: Show that the vectors \( \hat{i} + 2\hat{j} + 3\hat{k} \), \( 2\hat{i} + \hat{j} – \hat{k} \), and \( 4\hat{i} + 5\hat{j} + \hat{k} \) are coplanar.

Solution:

Wait — 12 ≠ 0, so these vectors are actually NOT coplanar! This shows you must always compute carefully.

Practice Question 8: Are the vectors \( \hat{i} – \hat{j} + \hat{k} \), \( 2\hat{i} + \hat{j} – \hat{k} \), and \( 3\hat{i} – 3\hat{j} + 3\hat{k} \) coplanar?

Scalar Triple Product

The scalar triple product of three vectors \( \vec{a}, \vec{b}, \vec{c} \) is defined as:

Properties

• The scalar triple product gives the volume of the parallelepiped formed by the three vectors.

• \( [\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}] \) (cyclic permutation doesn’t change the value)

• \( [\vec{a}\;\vec{b}\;\vec{c}] = -[\vec{b}\;\vec{a}\;\vec{c}] \) (swapping two vectors changes the sign)

• If any two vectors are equal, the scalar triple product is zero.

• \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \) means the vectors are coplanar.

Worked Example 18: Find the volume of the parallelepiped formed by vectors \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} – \hat{j} \), and \( \vec{c} = \hat{i} + 2\hat{j} + 3\hat{k} \).

Solution:

Vector Triple Product

The vector triple product is defined as:

This is called the “BAC-CAB” rule — easy to remember!

Worked Example 19: Find \( \hat{i} \times (\hat{j} \times \hat{k}) \) using the BAC-CAB rule.

Solution:

Applications of Vectors — Exam-Style Worked Examples

Worked Example 20: If \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) and \( |\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7 \), find the angle between \( \vec{a} \) and \( \vec{b} \).

Solution:

Since \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), we have \( \vec{c} = -(\vec{a}+\vec{b}) \).

Worked Example 21: Prove that the diagonals of a rhombus are perpendicular using vectors.

Solution:

Let the rhombus be ABCD with \( \vec{AB} = \vec{a} \) and \( \vec{AD} = \vec{b} \), where \( |\vec{a}| = |\vec{b}| \).

The diagonals are: \( \vec{AC} = \vec{a} + \vec{b} \) and \( \vec{BD} = \vec{b} – \vec{a} \)

Since the dot product is zero, the diagonals are perpendicular. ✓

Practice Question 9: If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}| = 3, |\vec{b}| = 4 \), and \( \vec{a} \times \vec{b} = 6\hat{k} \), find the angle between \( \vec{a} \) and \( \vec{b} \).

Section Review — Key Exam Notes

Practice Question 10: If \( |\vec{a}| = 2, |\vec{b}| = 3 \) and the angle between them is 60°, find: (a) \( \vec{a}\cdot\vec{b} \) (b) \( |\vec{a}\times\vec{b}| \) (c) the angle between \( \vec{a} \) and \( \vec{a}+\vec{b} \).