Chapter 07 — Vectors Part 4: Further Applications of Vectors

MATHEMATICS HIPHOP ACADEMY

CHAPTER 07 — VECTORS

PART 4: FURTHER APPLICATIONS OF VECTORS

Angles, Perpendicular Distances and Geometrical Proofs

Learning Objectives

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

• Find the angle between two straight lines using direction vectors.
• Determine whether two lines are perpendicular.
• Find the foot of a perpendicular from a point to a line.
• Calculate the perpendicular distance from a point to a line.
• Use vectors to solve geometrical problems involving points and lines.
• Use vector methods to prove collinearity and geometrical properties.

1. Revision: The Scalar Product

In Part 2, we studied the scalar product, also called the dot product. It is particularly useful when finding angles and working with perpendicular vectors.

For two vectors:

a = (a₁, a₂, a₃)

b = (b₁, b₂, b₃)

The scalar product is:

a · b = a₁b₁ + a₂b₂ + a₃b₃

It can also be expressed geometrically as:

a · b = |a||b| cos θ

Therefore:

cos θ = (a · b) / (|a||b|)

where θ is the angle between the two vectors.

2. The Angle Between Two Straight Lines

The angle between two straight lines can be found using their direction vectors.

Suppose two lines have direction vectors a and b. The angle between their directions is determined by:

cos θ = (a · b) / (|a||b|)

When a question asks for the smaller angle between two undirected lines, use the acute-angle convention:

cos θ = |a · b| / (|a||b|)

The absolute value ensures that the smaller angle is obtained, between 0° and 90°.

Worked Example 1

Find the acute angle between two lines whose direction vectors are:

a = (1, 2, 2) and b = (2, 1, −2).

Step 1: Find the scalar product.

a · b = (1 × 2) + (2 × 1) + (2 × −2)

= 2 + 2 − 4

= 0

Step 2: Find the magnitudes.

|a| = √(1² + 2² + 2²)

= √9 = 3

|b| = √(2² + 1² + (−2)²)

= √9 = 3

Step 3: Apply the angle formula.

cos θ = 0 / (3 × 3) = 0

θ = cos⁻¹(0)

θ = 90°

Answer: The lines are perpendicular.

3. Perpendicular Lines Using the Scalar Product

Two non-zero direction vectors are perpendicular when their scalar product is zero.

a · b = 0

For two lines with direction vectors (a₁, a₂, a₃) and (b₁, b₂, b₃), the condition becomes:

a₁b₁ + a₂b₂ + a₃b₃ = 0

Worked Example 2

Find the value of k if the vectors (2, k, 3) and (1, −2, 2) are perpendicular.

Solution

Since the vectors are perpendicular:

(2, k, 3) · (1, −2, 2) = 0

2(1) + k(−2) + 3(2) = 0

2 − 2k + 6 = 0

8 − 2k = 0

−2k = −8

k = 4

4. The Foot of a Perpendicular from a Point to a Line

Consider a fixed point P and a straight line passing through A with direction vector b.

We want to find the point H on the line such that PH is perpendicular to the line.

The vector equation of the line is:

r = a + tb

Since H lies on the line, its position vector is:

h = a + tb

Because PH is perpendicular to the line, the vector PH is perpendicular to b.

Therefore:

(p − h) · b = 0

Substituting h = a + tb:

(p − a − tb) · b = 0

(p − a) · b − t(b · b) = 0

Hence:

t = [(p − a) · b] / (b · b)

Once t is known, substitute it into h = a + tb to obtain the foot of the perpendicular.

Worked Example 3

Find the foot of the perpendicular from P(4, 3, 2) to the line:

r = (1, 1, 0) + t(1, 2, 2)

Step 1: Identify the vectors.

a = (1, 1, 0)

b = (1, 2, 2)

p = (4, 3, 2)

Step 2: Find p − a.

p − a = (4 − 1, 3 − 1, 2 − 0)

= (3, 2, 2)

Step 3: Calculate the scalar products.

(p − a) · b = (3 × 1) + (2 × 2) + (2 × 2)

= 3 + 4 + 4 = 11

b · b = 1² + 2² + 2²

= 9

Step 4: Find t.

t = 11/9

Step 5: Find the foot H.

h = a + tb

h = (1, 1, 0) + (11/9)(1, 2, 2)

h = (1 + 11/9, 1 + 22/9, 22/9)

H = (20/9, 31/9, 22/9)

Answer: The foot of the perpendicular is (20/9, 31/9, 22/9).

5. Perpendicular Distance from a Point to a Line

The perpendicular distance from a point to a straight line is the shortest distance from that point to any point on the line.

If H is the foot of the perpendicular from P to the line, then the distance is:

Distance = |p − h|

Alternatively, using the scalar product projection method, the distance can be calculated directly:

d² = |p − a|² − [(p − a) · b]² / |b|²

Here, a is the position vector of a point on the line, b is its direction vector, and p is the position vector of the external point.

Worked Example 4

Using the line and point from Worked Example 3, find the perpendicular distance from P(4, 3, 2) to the line.

Solution

From Worked Example 3, the foot of the perpendicular is:

H = (20/9, 31/9, 22/9)

The distance is:

PH = √[(4 − 20/9)² + (3 − 31/9)² + (2 − 22/9)²]

= √[(16/9)² + (−4/9)² + (−4/9)²]

= √[(256 + 16 + 16)/81]

= √(288/81)

= √(32/9)

Distance = 4√2/3 units

6. Using Vectors to Prove Collinearity

Three points are collinear if they lie on the same straight line.

Suppose A, B and C have position vectors a, b and c respectively.

Find:

AB = b − a

AC = c − a

The points are collinear if these two displacement vectors are scalar multiples of one another.

AC = kAB

where k is a real number.

Worked Example 5

Show that the points A(1, 2, 3), B(3, 6, 5) and C(4, 8, 6) are collinear.

Solution

AB = B − A

= (3 − 1, 6 − 2, 5 − 3)

= (2, 4, 2)

AC = C − A

= (4 − 1, 8 − 2, 6 − 3)

= (3, 6, 3)

Notice that:

AC = (3/2)(2, 4, 2)

Therefore:

AC = (3/2)AB

Hence, A, B and C are collinear.

7. Using Vectors to Prove Parallelism

Two line segments are parallel when their direction vectors are scalar multiples of one another.

For example, if:

AB = (2, 4, 6)

CD = (1, 2, 3)

Then:

AB = 2CD

Therefore, AB and CD are parallel.

Worked Example 6

A(1, 0, 2), B(4, 2, 5), C(0, 1, 1) and D(2, 2, 3) are four points.

Determine whether AB is parallel to CD.

Solution

AB = B − A

= (4 − 1, 2 − 0, 5 − 2)

= (3, 2, 3)

CD = D − C

= (2 − 0, 2 − 1, 3 − 1)

= (2, 1, 2)

Compare the components:

3/2, 2/1 and 3/2

The ratios are not all equal.

Therefore, AB is not parallel to CD.

8. Using Vectors to Prove a Parallelogram

A quadrilateral ABCD is a parallelogram if its opposite sides are parallel and equal in length, with corresponding directed vectors equal:

AB = DC

AD = BC

These vector equalities provide a useful method for proving that a quadrilateral is a parallelogram.

Worked Example 7

The position vectors of A, B, C and D are:

a = (1, 1, 0)

b = (4, 2, 1)

c = (5, 5, 3)

d = (2, 4, 2)

Show that ABCD is a parallelogram.

Solution

AB = b − a

= (4 − 1, 2 − 1, 1 − 0)

= (3, 1, 1)

DC = c − d

= (5 − 2, 5 − 4, 3 − 2)

= (3, 1, 1)

Therefore, AB = DC.

Now calculate AD:

AD = d − a

= (2 − 1, 4 − 1, 2 − 0)

= (1, 3, 2)

Calculate BC:

BC = c − b

= (5 − 4, 5 − 2, 3 − 1)

= (1, 3, 2)

Therefore, AD = BC.

Hence, ABCD is a parallelogram.

9. Important Examination Techniques

When finding the angle between lines:

• Extract the direction vectors from the line equations.
• Calculate their scalar product.
• Calculate both magnitudes.
• Use the cosine formula.
• Use the acute-angle convention if the question asks for the smaller angle between undirected lines.

When finding the foot of a perpendicular:

• Identify a point on the line and its direction vector.
• Write the unknown foot as a + tb.
• Use the perpendicular condition (p − h) · b = 0.
• Solve for t and substitute back into the line equation.

When proving a geometrical property:

• Find the relevant displacement vectors.
• Compare vectors for equality or scalar multiples.
• Use the scalar product to test perpendicularity.
• State the geometrical conclusion clearly.

10. Common Examination Mistakes

• Using the position vector of a point instead of the direction vector of the line when finding an angle.
• Forgetting to calculate the magnitudes in the scalar product formula.
• Confusing perpendicular vectors with parallel vectors.
• Assuming that a point on a line is automatically the foot of the perpendicular.
• Using an incorrect displacement vector by subtracting coordinates in the wrong order.
• Giving the foot of the perpendicular without calculating the requested distance.
• Failing to explain why a set of points is collinear or why a quadrilateral is a parallelogram.

11. Formula Summary

Scalar product:

a · b = a₁b₁ + a₂b₂ + a₃b₃

Angle between vectors:

cos θ = (a · b) / (|a||b|)

Perpendicular vectors:

a · b = 0

Foot of perpendicular:

h = a + tb

t = [(p − a) · b] / |b|²

Distance from point to line:

d = |p − h|

Collinearity:

AC = kAB

Parallelogram:

AB = DC and AD = BC

12. Practice Exercises

Attempt all the questions before opening the worked solutions.

Exercise A — Angles and Perpendicularity

1. Find the angle between the vectors (1, 0, 1) and (1, 1, 0).

2. Find the acute angle between the direction vectors (2, 1, 2) and (1, 2, −2).

3. Find k if (k, 2, 1) is perpendicular to (2, −1, 4).

4. Determine whether the lines with direction vectors (3, 2, −1) and (1, −2, −1) are perpendicular.

Exercise B — Foot of a Perpendicular

1. Find the foot of the perpendicular from P(3, 4, 1) to the line:

r = (1, 1, 1) + t(1, 1, 0).

2. Find the perpendicular distance from P(3, 4, 1) to the line in Question 1.

3. Find the foot of the perpendicular from P(2, 3, 4) to the line:

r = (0, 0, 0) + t(1, 2, 2).

Exercise C — Geometrical Proofs

1. Show that A(1, 1, 1), B(2, 3, 4) and C(3, 5, 7) are collinear.

2. Determine whether the line segments joining A(0, 0, 0) to B(2, 4, 6) and C(1, 2, 3) to D(2, 4, 6) are parallel.

3. The position vectors of A, B, C and D are:

a = (0, 0, 0), b = (2, 1, 0), c = (3, 3, 1), d = (1, 2, 1).

Show that ABCD is a parallelogram.

13. Worked Answers to Practice Exercises

Open each section after attempting the questions.

Exercise A — Show Answers

1. Let a = (1, 0, 1) and b = (1, 1, 0).

a · b = 1(1) + 0(1) + 1(0) = 1.

|a| = √2 and |b| = √2.

cos θ = 1/2.

θ = 60°.

2. The scalar product is:

2(1) + 1(2) + 2(−2) = 0.

Therefore, the vectors are perpendicular.

θ = 90°.

3. Apply the perpendicular condition:

2k + 2(−1) + 1(4) = 0.

2k − 2 + 4 = 0.

2k + 2 = 0.

k = −1.

4. Calculate the scalar product:

3(1) + 2(−2) + (−1)(−1) = 3 − 4 + 1 = 0.

Therefore, the lines are perpendicular.

Exercise B — Show Answers

1. Here a = (1, 1, 1), b = (1, 1, 0) and p = (3, 4, 1).

p − a = (2, 3, 0).

(p − a) · b = 2 + 3 = 5.

b · b = 1 + 1 = 2.

Therefore, t = 5/2.

h = (1, 1, 1) + (5/2)(1, 1, 0).

H = (7/2, 7/2, 1).

2. From the previous answer:

PH = (3 − 7/2, 4 − 7/2, 1 − 1).

= (−1/2, 1/2, 0).

Distance = √(1/4 + 1/4).

Distance = √2/2 units.

3. Here a = (0, 0, 0), b = (1, 2, 2) and p = (2, 3, 4).

p · b = 2 + 6 + 8 = 16.

b · b = 1 + 4 + 4 = 9.

t = 16/9.

h = (16/9)(1, 2, 2).

H = (16/9, 32/9, 32/9).

Exercise C — Show Answers

1.

AB = (2 − 1, 3 − 1, 4 − 1) = (1, 2, 3).

AC = (3 − 1, 5 − 1, 7 − 1) = (2, 4, 6).

AC = 2AB.

Therefore, A, B and C are collinear.

2.

AB = (2, 4, 6).

CD = (2 − 1, 4 − 2, 6 − 3) = (1, 2, 3).

AB = 2CD.

Therefore, AB is parallel to CD.

3.

AB = b − a = (2, 1, 0).

DC = c − d = (3 − 1, 3 − 2, 1 − 1) = (2, 1, 0).

Therefore, AB = DC.

AD = d − a = (1, 2, 1).

BC = c − b = (3 − 2, 3 − 1, 1 − 0) = (1, 2, 1).

Therefore, AD = BC.

Hence, ABCD is a parallelogram.

End of Part 4

You have completed the main applications developed in this lesson.

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