Dot Product

Introduction to the Dot Product

Vectors are mathematical quantities that have both magnitude and direction. They are used to represent displacement, velocity, acceleration, force, and many other physical quantities.

When working with vectors, we often need to determine how two vectors are related. For example:

• Do two forces act in the same direction?

• Are two directions perpendicular?

• What angle separates two vectors?

• How much of one vector acts in the direction of another?

• How much work is performed by a force acting at an angle?

The dot product, also called the scalar product, is a mathematical operation that helps answer these questions.

Unlike vector addition, which produces another vector, the dot product produces a single number called a scalar.

The dot product connects the components of vectors with their magnitudes and the angle between them. This makes it useful in geometry, physics, engineering, navigation, and computer graphics.

Understanding the Dot Product

The dot product of two vectors A and B is written:

A · B

The dot symbol indicates the dot product operation.

The result is a scalar, not a vector.

For example:

A · B = 12

This means the dot product of vectors A and B is 12.

What Does the Dot Product Tell Us?

The dot product measures how strongly two vectors point in the same direction, taking their magnitudes into account.

• A positive dot product indicates that the angle between the nonzero vectors is acute.

• A zero dot product indicates that two nonzero vectors are perpendicular.

• A negative dot product indicates that the angle between the nonzero vectors is obtuse.

The magnitude of the result also depends on the lengths of the vectors.

A larger positive dot product does not necessarily mean a smaller angle unless the vector magnitudes are held constant.

Calculating the Dot Product Using Components

Vectors can be represented using horizontal and vertical components.

In two dimensions:

A = (Aₓ, Aᵧ)

B = (Bₓ, Bᵧ)

The dot product is calculated by multiplying corresponding components and adding the results.

A · B = AₓBₓ + AᵧBᵧ

Worked Example: Two-Dimensional Dot Product

Given:

A = (3, 4)

B = (2, 5)

Calculate A · B.

Step 1: Write the formula

A · B = AₓBₓ + AᵧBᵧ

Step 2: Substitute the components

A · B = (3)(2) + (4)(5)

Step 3: Multiply

A · B = 6 + 20

Step 4: Add

A · B = 26

The dot product is 26.

Because the result is positive, the angle between the vectors is acute.

Worked Example: Negative Components

Given:

A = (−2, 5)

B = (4, −3)

Calculate A · B.

A · B = (−2)(4) + (5)(−3)

A · B = −8 − 15

A · B = −23

The negative result indicates that the angle between the vectors is greater than 90°.

Worked Example: Zero Dot Product

Given:

A = (3, 4)

B = (−4, 3)

Calculate A · B.

A · B = (3)(−4) + (4)(3)

A · B = −12 + 12

A · B = 0

Since both vectors are nonzero, they are perpendicular.

Dot Product in Three Dimensions

The dot product can also be calculated for three-dimensional vectors.

Given:

A = (Aₓ, Aᵧ, Aᵤ)

B = (Bₓ, Bᵧ, Bᵤ)

where the third components represent the z-direction, the formula is:

A · B = AₓBₓ + AᵧBᵧ + AᵤBᵤ

More conventionally, writing the components as (Aₓ, Aᵧ, A_z) and (Bₓ, Bᵧ, B_z):

A · B = AₓBₓ + AᵧBᵧ + A_zB_z

Worked Example: Three-Dimensional Dot Product

Given:

A = (2, −1, 3)

B = (4, 5, −2)

Calculate A · B.

Step 1: Write the formula

A · B = AₓBₓ + AᵧBᵧ + A_zB_z

Step 2: Substitute

A · B = (2)(4) + (−1)(5) + (3)(−2)

Step 3: Multiply

A · B = 8 − 5 − 6

Step 4: Add

A · B = −3

The dot product is −3.

The vectors form an obtuse angle.

Important Observation

The same principle applies in two, three, or more dimensions:

Multiply corresponding components and add the products.

The Geometric Definition of the Dot Product

The dot product can also be calculated using the magnitudes of two vectors and the angle between them.

The geometric formula is:

A · B = |A||B| cos θ

where:

• A · B is the dot product.

• |A| is the magnitude of vector A.

• |B| is the magnitude of vector B.

• θ is the angle between the vectors.

• cos θ is the cosine of the angle.

The angle θ is normally taken between 0° and 180°.

How the Angle Affects the Dot Product

The cosine function determines whether the dot product is positive, zero, or negative.

Vectors Pointing in the Same Direction

When θ = 0°:

cos 0° = 1

Therefore:

A · B = |A||B|

The dot product is as large as possible for the given magnitudes.

Perpendicular Vectors

When θ = 90°:

cos 90° = 0

Therefore:

A · B = 0

The dot product is zero.

Vectors Pointing in Opposite Directions

When θ = 180°:

cos 180° = −1

Therefore:

A · B = −|A||B|

The dot product is as negative as possible for the given magnitudes.

Calculating the Dot Product from Magnitudes and Angles

Worked Example: Acute Angle

Two vectors have magnitudes:

|A| = 6

|B| = 8

The angle between them is 60°.

Calculate the dot product.

Step 1: Write the formula

A · B = |A||B| cos θ

Step 2: Substitute

A · B = (6)(8) cos 60°

Step 3: Evaluate the cosine

cos 60° = 0.5

Step 4: Calculate

A · B = 48(0.5)

A · B = 24

Worked Example: Obtuse Angle

Two vectors have magnitudes:

|A| = 5

|B| = 10

The angle between them is 120°.

A · B = (5)(10) cos 120°

Since:

cos 120° = −0.5

A · B = 50(−0.5)

A · B = −25

The negative result is consistent with the obtuse angle.

Calculating Vector Magnitudes

To determine the angle between vectors, we must often calculate their magnitudes first.

For a two-dimensional vector:

A = (Aₓ, Aᵧ)

its magnitude is:

|A| = √(Aₓ² + Aᵧ²)

For a three-dimensional vector:

A = (Aₓ, Aᵧ, A_z)

its magnitude is:

|A| = √(Aₓ² + Aᵧ² + A_z²)

Worked Example

Find the magnitude of:

A = (6, 8)

|A| = √(6² + 8²)

|A| = √(36 + 64)

|A| = √100

|A| = 10

Three-Dimensional Example

Find the magnitude of:

B = (2, −3, 6)

|B| = √(2² + (−3)² + 6²)

|B| = √(4 + 9 + 36)

|B| = √49

|B| = 7

Finding the Angle Between Two Vectors

The geometric dot product formula can be rearranged to determine the angle between vectors.

Starting with:

A · B = |A||B| cos θ

Divide by the product of the magnitudes:

cos θ = (A · B)/(|A||B|)

Take the inverse cosine:

θ = cos⁻¹[(A · B)/(|A||B|)]

This formula applies when both vectors are nonzero.

Worked Example: Finding an Acute Angle

Given:

A = (1, 2)

B = (3, 1)

Find the angle between the vectors.

Step 1: Calculate the dot product

A · B = (1)(3) + (2)(1)

A · B = 3 + 2

A · B = 5

Step 2: Find the magnitude of A

|A| = √(1² + 2²)

|A| = √5

Step 3: Find the magnitude of B

|B| = √(3² + 1²)

|B| = √10

Step 4: Substitute

cos θ = 5/(√5 × √10)

cos θ = 5/√50

cos θ ≈ 0.7071

Step 5: Take the inverse cosine

θ = cos⁻¹(0.7071)

θ = 45°

Interpretation

The vectors form an angle of 45°.

Since the dot product is positive, an acute angle is expected.

Worked Example: Finding an Obtuse Angle

Given:

A = (2, 1)

B = (−3, 2)

Find the angle between the vectors.

Step 1: Calculate the dot product

A · B = (2)(−3) + (1)(2)

A · B = −6 + 2

A · B = −4

Step 2: Calculate the magnitudes

|A| = √(2² + 1²) = √5

|B| = √((−3)² + 2²) = √13

Step 3: Substitute

cos θ = −4/(√5 × √13)

cos θ = −4/√65

cos θ ≈ −0.4961

Step 4: Calculate the angle

θ = cos⁻¹(−0.4961)

θ ≈ 119.7°

Reasonableness Check

The dot product is negative.

Therefore, the angle should be greater than 90°.

The calculated angle of approximately 119.7° is reasonable.

Identifying Perpendicular Vectors

Two vectors are perpendicular when they meet at an angle of 90°.

For two nonzero vectors:

A · B = 0 if and only if A and B are perpendicular.

This is one of the most useful applications of the dot product.

Worked Example: Testing Perpendicularity

Given:

A = (4, 3)

B = (−3, 4)

Determine whether the vectors are perpendicular.

Step 1: Calculate the dot product

A · B = (4)(−3) + (3)(4)

A · B = −12 + 12

A · B = 0

Step 2: Interpret

Both vectors are nonzero, and their dot product is zero.

Therefore:

A and B are perpendicular.

Worked Example: Vectors That Are Not Perpendicular

Given:

A = (5, 2)

B = (3, −4)

A · B = (5)(3) + (2)(−4)

A · B = 15 − 8

A · B = 7

Since the dot product is not zero, the vectors are not perpendicular.

Finding an Unknown Component

The perpendicularity condition can be used to solve algebraic problems.

Worked Example

Given:

A = (3, 4)

B = (x, 6)

Find x if the vectors are perpendicular.

Step 1: Apply the perpendicularity condition

A · B = 0

Step 2: Substitute

3x + 4(6) = 0

Step 3: Simplify

3x + 24 = 0

Step 4: Solve

3x = −24

x = −8

Verification

B = (−8, 6)

A · B = (3)(−8) + (4)(6)

A · B = −24 + 24

A · B = 0

The vectors are perpendicular.

Orthogonal Vectors

Vectors with a dot product of zero are called orthogonal vectors.

For nonzero vectors, orthogonal means perpendicular.

The term orthogonal is particularly common in linear algebra and higher-dimensional vector spaces.

Example

A = (1, 0, 0)

B = (0, 1, 0)

A · B = (1)(0) + (0)(1) + (0)(0)

A · B = 0

The vectors are orthogonal.

The Zero Vector

The zero vector has all components equal to zero.

For example:

O = (0, 0)

Its dot product with every vector is zero.

However, the zero vector has no defined direction, so an angle between the zero vector and another vector is undefined.

This is why angle calculations require nonzero vectors.

Understanding Vector Projection

A vector projection describes the part of one vector that acts along the direction of another vector.

Imagine shining a light perpendicular to a line. The shadow of a vector onto that line represents its projection.

Projection is useful when a vector must be separated into components acting along particular directions.

Examples include:

• Finding the component of a force along a slope.
• Determining motion in a particular direction.
• Calculating work done by a force.
• Analysing velocities.
• Resolving vectors in engineering.

Scalar Projection

The scalar projection of A onto B gives the signed length of the component of A in the direction of B.

The formula is:

Scalar projection of A onto B = (A · B)/|B|

It can also be written as:

|A| cos θ

The scalar projection may be positive, negative, or zero.

Worked Example

Given:

A = (6, 8)

B = (1, 0)

Find the scalar projection of A onto B.

Step 1: Calculate the dot product

A · B = (6)(1) + (8)(0)

A · B = 6

Step 2: Calculate the magnitude of B

|B| = √(1² + 0²)

|B| = 1

Step 3: Apply the formula

Scalar projection = 6/1

Scalar projection = 6

Interpretation

The component of A in the positive horizontal direction is 6 units.

Vector Projection

The vector projection gives both the magnitude and direction of the component.

The formula for the projection of A onto nonzero B is:

proj_B(A) = [(A · B)/|B|²]B

Since:

|B|² = B · B

the formula may also be written:

proj_B(A) = [(A · B)/(B · B)]B

Worked Example: Vector Projection

Given:

A = (4, 3)

B = (2, 0)

Find the projection of A onto B.

Step 1: Calculate the dot product

A · B = (4)(2) + (3)(0)

A · B = 8

Step 2: Calculate |B|²

|B|² = 2² + 0²

|B|² = 4

Step 3: Apply the formula

proj_B(A) = (8/4)(2, 0)

Step 4: Simplify

proj_B(A) = 2(2, 0)

proj_B(A) = (4, 0)

Interpretation

The projection of A onto B is the horizontal component of A.

The original vector is:

A = (4, 3)

Its horizontal projection is:

(4, 0)

The remaining perpendicular component is:

(0, 3)

Worked Example: Projection onto a Diagonal Vector

Given:

A = (3, 4)

B = (1, 1)

Find the vector projection of A onto B.

Step 1: Calculate A · B

A · B = (3)(1) + (4)(1)

A · B = 7

Step 2: Calculate |B|²

|B|² = 1² + 1²

|B|² = 2

Step 3: Apply the formula

proj_B(A) = (7/2)(1, 1)

Step 4: Multiply

proj_B(A) = (3.5, 3.5)

Verification

The remaining component is:

A − proj_B(A)

= (3, 4) − (3.5, 3.5)

= (−0.5, 0.5)

Check whether this component is perpendicular to B:

(−0.5, 0.5) · (1, 1)

= −0.5 + 0.5

= 0

Therefore, the remaining component is perpendicular to B, confirming that the projection is correct.

Decomposing a Vector into Parallel and Perpendicular Components

Any vector A can be separated into:

• A component parallel to a nonzero vector B.

• A component perpendicular to B.

The parallel component is:

A_parallel = proj_B(A)

The perpendicular component is:

A_perpendicular = A − proj_B(A)

Therefore:

A = A_parallel + A_perpendicular

Worked Example

Given:

A = (5, 4)

B = (3, 0)

Find the parallel and perpendicular components of A relative to B.

Step 1: Calculate the dot product

A · B = (5)(3) + (4)(0)

A · B = 15

Step 2: Calculate |B|²

|B|² = 3² = 9

Step 3: Find the parallel component

A_parallel = (15/9)(3, 0)

A_parallel = (5, 0)

Step 4: Find the perpendicular component

A_perpendicular = (5, 4) − (5, 0)

A_perpendicular = (0, 4)

Final Answer

Parallel component = (5, 0)

Perpendicular component = (0, 4)

Verification

(5, 0) + (0, 4) = (5, 4)

The original vector has been reconstructed.

Properties of the Dot Product

The dot product has several important algebraic properties.

Commutative Property

A · B = B · A

The order of the vectors does not change the result.

Example:

A = (2, 3)

B = (4, 5)

A · B = 2(4) + 3(5) = 23

B · A = 4(2) + 5(3) = 23

Both results are equal.

Distributive Property

A · (B + C) = A · B + A · C

This allows dot products to be expanded across vector addition.

Scalar Multiplication

(kA) · B = k(A · B)

where k is a scalar.

Dot Product of a Vector with Itself

A · A = |A|²

For example:

A = (3, 4)

A · A = 3² + 4²

A · A = 25

Therefore:

|A| = √25 = 5

Non-Negativity

For every vector A:

A · A ≥ 0

The result equals zero only when A is the zero vector.

Important Observation

Although the dot product is commutative, vector projection is generally not.

In general:

proj_B(A) ≠ proj_A(B)

The direction of projection matters.

The Dot Product and the Law of Cosines

The dot product is closely related to the Law of Cosines.

For vectors A and B:

|A − B|² = (A − B) · (A − B)

Expanding:

|A − B|² = A · A − 2(A · B) + B · B

Since:

A · A = |A|²

B · B = |B|²

and:

A · B = |A||B| cos θ

we obtain:

|A − B|² = |A|² + |B|² − 2|A||B| cos θ

This is the vector form of the Law of Cosines.

Why This Matters

The Law of Cosines generalises the Pythagorean Theorem.

When θ = 90°:

cos 90° = 0

Therefore:

|A − B|² = |A|² + |B|²

This is the Pythagorean relationship.

The dot product therefore connects vector geometry, trigonometry, and triangle geometry.

Real-World Application: Work Done by a Force

In physics, work is the energy transferred when a force causes displacement.

When the force and displacement are vectors, work is calculated using the dot product.

W = F · s

or:

W = |F||s| cos θ

where:

• W is work done, measured in joules (J).

• F is the force vector, measured in newtons (N).

• s is the displacement vector, measured in metres (m).

• θ is the angle between force and displacement.

Worked Example: Pulling an Object

A person pulls a box with a force of 50 N at an angle of 30° above the horizontal.

The box moves 8 m horizontally.

Calculate the work done by the pulling force.