Chapter 07 — Vectors (Part 2): Scalar Product, Angles and Perpendicular Vectors

CHAPTER 07: VECTORS

PART 2: THE SCALAR PRODUCT, ANGLES AND PERPENDICULAR VECTORS

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

• Define and calculate the scalar product of two vectors.
• Use the component formula for the scalar product in two and three dimensions.
• Use the scalar product to calculate the angle between two vectors.
• Determine whether two vectors are perpendicular.
• Find unknown constants using conditions involving the scalar product.
• Apply vector methods to geometrical problems involving triangles and lines.
• Interpret positive, negative and zero scalar products.

1. INTRODUCTION TO THE SCALAR PRODUCT

In Part 1, we studied how to add and subtract vectors, multiply them by scalars, calculate their magnitudes and find unit vectors. We now introduce another important operation called the scalar product. The scalar product is also known as the dot product. Unlike vector addition, which produces another vector, the scalar product of two vectors produces an ordinary number called a scalar. It is written as:

a⃗ · b⃗

The dot between the two vectors represents the scalar product. Important: The scalar product is not the same as ordinary multiplication of corresponding components to produce another vector. It involves multiplying corresponding components and adding the products.

2. THE COMPONENT FORMULA

Consider two vectors in two dimensions:

a⃗ = ( a₁, a₂ )
b⃗ = ( b₁, b₂ )

Their scalar product is:

a⃗ · b⃗ = a₁b₁ + a₂b₂

For vectors in three dimensions:

a⃗ = ( a₁, a₂, a₃ )
b⃗ = ( b₁, b₂, b₃ )

The scalar product is:

a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃

In unit-vector notation, if:

a⃗ = a₁i⃗ + a₂j⃗ + a₃k⃗
b⃗ = b₁i⃗ + b₂j⃗ + b₃k⃗

then:

a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃

Worked Example 1

Given:

a⃗ = ( 3, 4 )
b⃗ = ( 2, 5 )

Calculate a⃗ · b⃗.

Solution Multiply corresponding components and add:

a⃗ · b⃗ = 3(2) + 4(5)

= 6 + 20

= 26

Therefore: a⃗ · b⃗ = 26.

Worked Example 2

Calculate the scalar product of:

p⃗ = ( 2, -3, 4 )
q⃗ = ( 5, 1, -2 )

Solution

p⃗ · q⃗ = 2(5) + (-3)(1) + 4(-2)

= 10 - 3 - 8

= -1

Therefore: p⃗ · q⃗ = -1.

3. THE GEOMETRICAL DEFINITION OF THE SCALAR PRODUCT

The scalar product is also related to the angle between two non-zero vectors. If two vectors a⃗ and b⃗ have magnitudes |a⃗| and |b⃗|, and the angle between their directions is θ, then:

a⃗ · b⃗ = |a⃗||b⃗| cos θ

This is the geometrical definition of the scalar product. Where:

• |a⃗| is the magnitude of the first vector.
• |b⃗| is the magnitude of the second vector.
• θ is the angle between the two vectors, measured from 0° to 180°.

The component formula and geometrical formula describe the same scalar product. Therefore:

a₁b₁ + a₂b₂ = |a⃗||b⃗| cos θ

For three-dimensional vectors:

a₁b₁ + a₂b₂ + a₃b₃ = |a⃗||b⃗| cos θ

4. FINDING THE ANGLE BETWEEN TWO VECTORS

Starting with:

a⃗ · b⃗ = |a⃗||b⃗| cos θ

Divide both sides by |a⃗||b⃗|:

cos θ = (a⃗ · b⃗) / (|a⃗||b⃗|)

Hence:

θ = cos⁻¹ [ (a⃗ · b⃗) / (|a⃗||b⃗|) ]

This formula is used to calculate the angle between two non-zero vectors.

Worked Example 3

Find the angle between:

a⃗ = ( 3, 4 )
b⃗ = ( 4, 3 )

Give your answer to one decimal place.

Solution Step 1: Calculate the scalar product.

a⃗ · b⃗ = 3(4) + 4(3)

= 12 + 12 = 24

Step 2: Find the magnitudes.

|a⃗| = √(3² + 4²) = 5

|b⃗| = √(4² + 3²) = 5

Step 3: Apply the angle formula.

cos θ = 24/(5 × 5)

cos θ = 24/25 = 0.96

Step 4: Find the angle.

θ = cos⁻¹(0.96)

θ ≈ 16.3°

Therefore, the angle between the vectors is: 16.3°.

Worked Example 4

Find the angle between the vectors:

u⃗ = ( 1, 2, 2 )
v⃗ = ( 2, -1, 2 )

Solution First calculate the scalar product:

u⃗ · v⃗ = 1(2) + 2(-1) + 2(2)

= 2 - 2 + 4 = 4

Calculate the magnitudes:

|u⃗| = √(1² + 2² + 2²) = 3

|v⃗| = √(2² + (-1)² + 2²) = 3

Apply the angle formula:

cos θ = 4/(3 × 3) = 4/9

θ = cos⁻¹(4/9)

θ ≈ 63.6°

Therefore, the angle between the vectors is approximately 63.6°.

5. PERPENDICULAR VECTORS

Two non-zero vectors are perpendicular if the angle between them is 90°. From the geometrical definition:

a⃗ · b⃗ = |a⃗||b⃗| cos θ

When θ = 90°:

cos 90° = 0

Therefore:

a⃗ · b⃗ = 0

Hence, the important condition is:

a⃗ · b⃗ = 0 ⇔ a⃗ ⟂ b⃗

This equivalence applies to non-zero vectors.

Worked Example 5

Show that the vectors:

a⃗ = ( 2, 3 )
b⃗ = ( 3, -2 )

are perpendicular.

Solution Calculate their scalar product:

a⃗ · b⃗ = 2(3) + 3(-2)

= 6 - 6 = 0

Since the scalar product is zero, the vectors are perpendicular.

a⃗ ⟂ b⃗

Worked Example 6

Determine whether the vectors:

p⃗ = ( 1, 2, 3 )
q⃗ = ( 2, -1, 0 )

are perpendicular.

Solution

p⃗ · q⃗ = 1(2) + 2(-1) + 3(0)

= 2 - 2 + 0 = 0

Therefore, the vectors are perpendicular.

6. FINDING UNKNOWN CONSTANTS USING THE SCALAR PRODUCT

In examination questions, you may be given two vectors containing an unknown constant and told that the vectors are perpendicular. The condition for perpendicularity is:

a⃗ · b⃗ = 0

You then form an equation and solve for the unknown.

Worked Example 7

The vectors:

a⃗ = ( k, 3 )
b⃗ = ( 2, -4 )

are perpendicular. Find k.

Solution Since the vectors are perpendicular:

a⃗ · b⃗ = 0

Therefore:

k(2) + 3(-4) = 0

2k - 12 = 0

2k = 12

k = 6

Therefore, k = 6.

Worked Example 8

Given:

p⃗ = ( 2, k, 1 )
q⃗ = ( 3, 1, -4 )

If p⃗ and q⃗ are perpendicular, find k.

Solution

p⃗ · q⃗ = 0

2(3) + k(1) + 1(-4) = 0

6 + k - 4 = 0

k + 2 = 0

k = -2

Therefore, k = -2.

7. THE SIGN OF THE SCALAR PRODUCT

The sign of the scalar product gives useful information about the angle between two non-zero vectors. Recall:

a⃗ · b⃗ = |a⃗||b⃗| cos θ

Because magnitudes are positive, the sign depends on cos θ.

Scalar product	Angle	Interpretation
Positive	0° ≤ θ < 90°	Acute angle, or zero angle
Zero	θ = 90°	Perpendicular vectors
Negative	90° < θ ≤ 180°	Obtuse angle, or opposite directions

Special cases:

• If the vectors point in exactly the same direction, θ = 0° and the scalar product is |a⃗||b⃗|.
• If they are perpendicular, θ = 90° and the scalar product is zero.
• If they point in exactly opposite directions, θ = 180° and the scalar product is -|a⃗||b⃗|.

Worked Example 9

Without calculating the angle, state whether the angle between the vectors is acute, right or obtuse if their scalar product is -15.

Solution The scalar product is negative. Therefore, the angle is obtuse. Answer: Obtuse angle.

8. IMPORTANT ALGEBRAIC PROPERTIES

The scalar product obeys several useful laws.

Property	Formula
Commutative law	a⃗ · b⃗ = b⃗ · a⃗
Distributive law	a⃗ · (b⃗ + c⃗) = a⃗ · b⃗ + a⃗ · c⃗
Scalar multiplication	(ka⃗) · b⃗ = k(a⃗ · b⃗)
Self product	a⃗ · a⃗ = |a⃗|²
Perpendicularity	a⃗ · b⃗ = 0

Worked Example 10

Given:

a⃗ = ( 2, 1 )
b⃗ = ( 3, 4 )
c⃗ = ( 1, -2 )

Calculate a⃗ · (b⃗ + c⃗).

Solution 1: Add the vectors first.

b⃗ + c⃗ = (3+1, 4-2) = (4,2)

Now calculate:

a⃗ · (b⃗ + c⃗) = 2(4) + 1(2)

= 8 + 2 = 10

Solution 2: Use the distributive law.

a⃗ · b⃗ + a⃗ · c⃗

= [2(3)+1(4)] + [2(1)+1(-2)]

= 10 + 0 = 10

Both methods give the same answer.

9. USING THE SCALAR PRODUCT IN GEOMETRY

The scalar product can be used to calculate angles in triangles and other geometrical figures. When finding an angle at a vertex, choose the two vectors that both start at that vertex and point towards the other two vertices. For example, to find angle ABC, use:

BA⃗ and BC⃗

Do not use AB⃗ and BC⃗ for angle ABC because AB⃗ points towards B rather than away from B.

Worked Example 11

Triangle ABC has coordinates: A(1, 1), B(4, 1) and C(2, 4). Find angle ABC.

Solution The angle is at B. We therefore use BA⃗ and BC⃗. First:

BA⃗ = (1-4, 1-1) = (-3,0)

Next:

BC⃗ = (2-4, 4-1) = (-2,3)

Calculate the scalar product:

BA⃗ · BC⃗ = (-3)(-2) + 0(3)

= 6

Find the magnitudes:

|BA⃗| = √[(-3)²+0²] = 3

|BC⃗| = √[(-2)²+3²] = √13

Apply the angle formula:

cos B = 6/(3√13)

cos B = 2/√13

B = cos⁻¹(2/√13)

B ≈ 56.3°

Therefore, angle ABC is approximately 56.3°.

10. ANGLES BETWEEN TWO LINES

The angle between two intersecting lines can be found using their direction vectors. Suppose the direction vectors are a⃗ and b⃗. First calculate:

cos θ = (a⃗ · b⃗)/(|a⃗||b⃗|)

When the question asks for the acute angle between two lines, use the smaller angle between the lines. If the calculated angle between the chosen direction vectors is obtuse, subtract it from 180°.

Worked Example 12

Two lines have direction vectors:

a⃗ = (1, 2)
b⃗ = (-2, 1)

Determine the angle between the lines.

Solution Calculate the scalar product:

a⃗ · b⃗ = 1(-2) + 2(1)

= -2 + 2 = 0

Since the scalar product is zero, the direction vectors are perpendicular. Therefore, the lines meet at:

90°

11. COMMON EXAMINATION ERRORS

Be careful about the following:

1. Do not multiply only the first components. Multiply all corresponding components and add the results.
2. Do not confuse the scalar product with vector addition.
3. Calculate both magnitudes correctly before finding an angle.
4. Use the correct vectors starting from the vertex when finding an angle in a triangle.
5. Check whether the question asks for the angle between vectors or the acute angle between lines.
6. Use the correct calculator mode: degrees or radians, as required by the question.
7. Do not divide by the magnitude of a zero vector. The angle formula requires two non-zero vectors.

12. EXERCISES

Attempt the questions before opening the answers.

Exercise A: Calculating scalar products

1. Given a⃗ = (2, 3) and b⃗ = (4, 5), calculate a⃗ · b⃗.
2. Find p⃗ · q⃗ if p⃗ = (6, -2) and q⃗ = (3, 4).
3. Calculate the scalar product of (1, 2, 3) and (4, -1, 2).
4. Given u⃗ = (-2, 5, 1) and v⃗ = (3, 2, -4), find u⃗ · v⃗.
5. If a⃗ = 3i⃗ + 2j⃗ and b⃗ = 4i⃗ - 5j⃗, calculate a⃗ · b⃗.

Exercise B: Angles between vectors

1. Find the angle between (1, 0) and (0, 1).
2. Find the angle between (3, 4) and (4, 3), correct to one decimal place.
3. Find the angle between (1, 2, 2) and (2, -1, 2), correct to one decimal place.
4. Find the angle between (2, 1) and (1, 3), correct to one decimal place.
5. Find the angle between (2, 0, 0) and (1, 1, 0), correct to one decimal place.

Exercise C: Perpendicular vectors

1. Show that (2, 3) and (3, -2) are perpendicular.
2. Determine whether (1, 2, 3) and (2, -1, 0) are perpendicular.
3. Determine whether (4, 2) and (1, 2) are perpendicular.
4. Find k if (k, 4) and (2, -3) are perpendicular.
5. Find k if (1, k, 2) and (4, 2, -3) are perpendicular.

Exercise D: Unknown components and scalar products

1. Given a⃗ = (k, 2) and b⃗ = (3, 4), find k if a⃗ · b⃗ = 17.
2. If a⃗ = (2, k) and b⃗ = (3, 5), find k if a⃗ · b⃗ = 16.
3. Given p⃗ = (k, 1, 2) and q⃗ = (2, 3, 1), find k if p⃗ and q⃗ are perpendicular.
4. If a⃗ = (2, k) and b⃗ = (k, 3) are perpendicular, find the possible values of k.

Exercise E: Geometry and examination practice

1. Given A(1, 1), B(4, 1) and C(2, 4), calculate angle ABC.
2. The vectors a⃗ = (3, 4) and b⃗ = (4, -3) are given. Show that they are perpendicular.
3. Find the angle between the vectors AB⃗ and AC⃗ where A(0,0), B(3,0) and C(1,2).
4. Two vectors have magnitudes 5 and 8, and their scalar product is 20. Find the angle between them.
5. The scalar product of two non-zero vectors is -12. What can you conclude about the angle between them?

13. EXPANDABLE ANSWERS AND SOLUTIONS

Exercise A — Worked answers

1. 2(4) + 3(5) = 8 + 15 = 23. 2. 6(3) + (-2)(4) = 18 - 8 = 10. 3. 1(4) + 2(-1) + 3(2) = 4 - 2 + 6 = 8. 4. (-2)(3) + 5(2) + 1(-4) = -6 + 10 - 4 = 0. 5. 3(4) + 2(-5) = 12 - 10 = 2.

Exercise B — Worked answers

1. The scalar product is zero. Therefore, the vectors are perpendicular. Answer: 90°. 2. Scalar product = 3(4) + 4(3) = 24. Both magnitudes are 5. cos θ = 24/25 = 0.96. θ = cos⁻¹(0.96) ≈ 16.3°. 3. Scalar product = 1(2) + 2(-1) + 2(2) = 4. Both magnitudes are 3. cos θ = 4/9. θ ≈ 63.6°. 4. Scalar product = 2(1) + 1(3) = 5. Magnitudes: |a⃗| = √5 and |b⃗| = √10. cos θ = 5/(√5 × √10) = 1/√2. θ = 45.0°. 5. Scalar product = 2(1) + 0(1) + 0(0) = 2. Magnitudes are 2 and √2. cos θ = 2/(2√2) = 1/√2. θ = 45.0°.

Exercise C — Worked answers

1. 2(3) + 3(-2) = 6 - 6 = 0. Therefore, the vectors are perpendicular. 2. 1(2) + 2(-1) + 3(0) = 0. Therefore, the vectors are perpendicular. 3. 4(1) + 2(2) = 8. The scalar product is not zero. Therefore, the vectors are not perpendicular. 4. For perpendicular vectors: 3k + 4(-3) = 0. 3k - 12 = 0. k = 4. 5. 4(1) + 2k + 2(-3) = 0. 4 + 2k - 6 = 0. 2k - 2 = 0. k = 1.

Exercise D — Worked answers

1. 3k + 2(4) = 17. 3k + 8 = 17. 3k = 9. k = 3. 2. 2(3) + 5k = 16. 6 + 5k = 16. 5k = 10. k = 2. 3. 2k + 1(3) + 2(1) = 0. 2k + 5 = 0. k = -5/2. 4. 2(k) + k(3) = 0. 2k + 3k = 0. 5k = 0. k = 0.

Exercise E — Full solutions

Question 1 For angle ABC, use BA⃗ and BC⃗. BA⃗ = (1-4, 1-1) = (-3,0). BC⃗ = (2-4, 4-1) = (-2,3). Scalar product: (-3)(-2) + 0(3) = 6. Magnitudes: |BA⃗| = 3. |BC⃗| = √13. Therefore: cos B = 6/(3√13) = 2/√13. B = cos⁻¹(2/√13) ≈ 56.3°. Question 2 a⃗ · b⃗ = 3(4) + 4(-3). = 12 - 12 = 0. Therefore, the vectors are perpendicular. Question 3 AB⃗ = (3,0). AC⃗ = (1,2). Scalar product: AB⃗ · AC⃗ = 3(1) + 0(2) = 3. Magnitudes: |AB⃗| = 3. |AC⃗| = √5. Therefore: cos θ = 3/(3√5) = 1/√5. θ = cos⁻¹(1/√5) ≈ 63.4°. Question 4 Given: a⃗ · b⃗ = 20, |a⃗| = 5 and |b⃗| = 8. cos θ = 20/(5 × 8). cos θ = 1/2. θ = 60°. Question 5 The scalar product is negative. Therefore, the angle between the two non-zero vectors is obtuse. Hence: 90° < θ ≤ 180°.

14. FORMULA SUMMARY

Component formula in 2D:
a⃗ · b⃗ = a₁b₁ + a₂b₂

Component formula in 3D:
a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃

Geometrical formula:
a⃗ · b⃗ = |a⃗||b⃗| cos θ

Angle formula:
cos θ = (a⃗ · b⃗)/(|a⃗||b⃗|)

Perpendicular vectors:
a⃗ · b⃗ = 0

Self product:
a⃗ · a⃗ = |a⃗|²

END OF CHAPTER 07 — PART 2
You should now be able to calculate scalar products, find angles between vectors and use the scalar product to establish perpendicularity. Practise these skills carefully because they form an important foundation for more advanced vector geometry.

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