Chapter 07 — Vectors Part 3: Vector Equations of Lines, Parallel Lines and Intersections

MATHEMATICS HIPHOP ACADEMY

CHAPTER 07 — VECTORS

PART 3: VECTOR EQUATIONS OF LINES

Parallel Lines, Intersections and Skew Lines

Learning Objectives

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

• Understand the vector equation of a straight line.
• Identify the position vector and direction vector of a line.
• Find the equation of a line through a given point.
• Find the equation of a line passing through two points.
• Determine whether two lines are parallel.
• Determine whether two lines intersect and calculate their point of intersection.
• Recognise skew lines in three-dimensional space.

1. Introduction to Vector Equations of Lines

A straight line can be described using vectors. To define a line, we need two important pieces of information:

1. A fixed point through which the line passes.
2. A direction in which the line extends.

Suppose a line passes through a fixed point A whose position vector is a, and has direction vector b.

Any point P on the line has position vector r. The vector equation of the line is:

r = a + t b

where t is a real parameter.

Meaning of the symbols

• r — position vector of any point on the line.
• a — position vector of a fixed point on the line.
• b — direction vector of the line.
• t — a real number called the parameter.

As the value of t changes, the position vector r describes different points along the same straight line.

Important observation

When t = 0, r = a. Therefore, the line passes through the fixed point whose position vector is a.

When t = 1, r = a + b.

When t = 2, r = a + 2b.

Negative values of t describe points in the opposite direction along the same line.

2. Writing a Vector Equation in Component Form

In three dimensions, let:

a = (a₁, a₂, a₃) and b = (b₁, b₂, b₃)

Then:

r = (a₁, a₂, a₃) + t(b₁, b₂, b₃)

Expanding gives:

x = a₁ + tb₁

y = a₂ + tb₂

z = a₃ + tb₃

These are called the parametric equations of the line.

In two dimensions, the same method is used with only x and y coordinates.

3. Finding the Equation of a Line Through a Point

If a line passes through a point A and has a known direction vector, substitute the position vector of A and the direction vector into:

r = a + tb

Worked Example 1

Find the vector equation of the line passing through A(2, −1, 3) and having direction vector b = (4, 2, −1).

Solution

The position vector of A is:

a = (2, −1, 3)

The direction vector is:

b = (4, 2, −1)

Using r = a + tb:

r = (2, −1, 3) + t(4, 2, −1)

Answer: This is the required vector equation.

Its parametric equations are:

• x = 2 + 4t
• y = −1 + 2t
• z = 3 − t

4. Finding the Equation of a Line Through Two Points

Suppose a line passes through two points A and B, with position vectors a and b respectively.

The direction vector from A to B is:

AB = b − a

Therefore, the vector equation of the line through A and B is:

r = a + t(b − a)

Worked Example 2

Find the vector equation of the line passing through A(1, 2, 3) and B(4, 0, 5).

Step 1: Write the position vectors.

a = (1, 2, 3)

b = (4, 0, 5)

Step 2: Find the direction vector AB.

AB = b − a

AB = (4 − 1, 0 − 2, 5 − 3)

AB = (3, −2, 2)

Step 3: Form the equation.

r = a + tAB

r = (1, 2, 3) + t(3, −2, 2)

This is the required equation.

5. Finding a Point on a Line

When the parameter t is given, substitute it into the vector equation or the parametric equations to find the coordinates of the corresponding point.

Worked Example 3

A line has equation:

r = (2, 1, −3) + t(3, −2, 4)

Find the point on the line when t = 2.

Solution

Substitute t = 2:

r = (2, 1, −3) + 2(3, −2, 4)

r = (2, 1, −3) + (6, −4, 8)

r = (8, −3, 5)

Answer: The point is (8, −3, 5).

6. Determining Whether a Point Lies on a Line

A point lies on a line if its coordinates satisfy all the parametric equations for the same value of the parameter.

It is important to use one common parameter value for every coordinate.

Worked Example 4

Determine whether the point P(5, 1, 2) lies on the line:

r = (1, 3, 0) + t(2, −1, 1)

Solution

The parametric equations are:

• x = 1 + 2t
• y = 3 − t
• z = t

Using the x-coordinate, 5 = 1 + 2t.

Therefore, t = 2.

Check the y-coordinate:

3 − 2 = 1. This is correct.

Check the z-coordinate:

z = 2. This is also correct.

Therefore, P(5, 1, 2) lies on the line.

7. Parallel Lines

Two straight lines are parallel if their direction vectors are scalar multiples of one another.

Suppose their direction vectors are b₁ and b₂. The lines are parallel if:

b₁ = kb₂

where k is a non-zero scalar.

Parallel lines may be:

• Distinct parallel lines: They have different positions and never meet.
• Coincident lines: They lie exactly on top of each other and represent the same line.

Having parallel direction vectors alone does not prove that two lines are distinct. We must also check whether they are the same line.

Worked Example 5

Determine whether the lines below are parallel:

Line 1: r = (1, 2, 3) + s(2, −4, 6)

Line 2: r = (3, 1, 0) + t(1, −2, 3)

Solution

The direction vectors are:

b₁ = (2, −4, 6)

b₂ = (1, −2, 3)

Notice that:

(2, −4, 6) = 2(1, −2, 3)

Therefore, the lines are parallel.

To check whether they are coincident, test whether the fixed point (3, 1, 0) of the second line lies on the first line.

For the first line:

• 1 + 2s = 3 gives s = 1.
• 2 − 4s = 1 gives s = 1/4.

The values of s do not agree. Therefore, the point is not on the first line.

The lines are distinct and parallel.

8. Intersecting Lines

Two lines intersect if they have at least one common point.

Suppose the equations are:

Line 1: r = a + sb

Line 2: r = c + td

At their point of intersection, their position vectors are equal:

a + sb = c + td

We then compare the x, y and z components to form simultaneous equations.

Remember to use different parameters for the two lines. One line uses s and the other uses t.

Worked Example 6

Find the point of intersection of the lines:

Line 1: r = (1, 0, 2) + s(1, 2, −1)

Line 2: r = (3, 4, 0) + t(−1, 0, 1)

Step 1: Write the parametric equations.

For Line 1:

• x = 1 + s
• y = 2s
• z = 2 − s

For Line 2:

• x = 3 − t
• y = 4
• z = t

Step 2: Equate corresponding coordinates.

From the x-coordinates:

1 + s = 3 − t ... (1)

From the y-coordinates:

2s = 4 ... (2)

From the z-coordinates:

2 − s = t ... (3)

Step 3: Solve.

From equation (2):

s = 2.

Substitute into equation (3):

t = 2 − 2 = 0.

Check equation (1):

1 + 2 = 3 − 0.

3 = 3, which is true.

Step 4: Find the intersection point.

Substitute s = 2 into Line 1:

r = (1, 0, 2) + 2(1, 2, −1)

r = (1, 0, 2) + (2, 4, −2)

r = (3, 4, 0)

Answer: The lines intersect at (3, 4, 0).

9. Skew Lines

In three-dimensional space, two lines can fail to meet without being parallel. Such lines are called skew lines.

Skew lines:

• Are not parallel.
• Do not intersect.
• Do not lie in the same plane.

This is different from two-dimensional geometry, where two non-parallel straight lines always intersect.

Worked Example 7

Determine the relationship between the lines:

Line 1: r = (0, 0, 0) + s(1, 1, 0)

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

Solution

For Line 1:

• x = s
• y = s
• z = 0

For Line 2:

• x = 1
• y = t
• z = 1 + t

Step 1: Check whether they are parallel.

The direction vectors are (1, 1, 0) and (0, 1, 1).

They are not scalar multiples. Therefore, the lines are not parallel.

Step 2: Check whether they intersect.

Equating x-coordinates gives:

s = 1.

Equating y-coordinates gives:

s = t, so t = 1.

Now check the z-coordinates:

0 = 1 + t.

This requires t = −1.

But we already obtained t = 1. The conditions are inconsistent.

Therefore, the lines do not intersect.

The lines are skew.

10. A Systematic Method for Classifying Two Lines

When given two vector equations of lines in three dimensions, use the following method.

Step 1 — Compare direction vectors.

If one is a scalar multiple of the other, the lines are parallel or coincident.

Step 2 — If the direction vectors are not scalar multiples, test for intersection.

Equate the three coordinate expressions and solve using two different parameters.

Step 3 — Check all three equations.

If the same parameter values satisfy all three equations, the lines intersect.

If they do not, the lines are skew.

Step 4 — If the direction vectors are parallel, test whether a point from one line lies on the other.

If it does, the lines are coincident. Otherwise, they are distinct parallel lines.

11. Common Examination Mistakes

• Writing the direction vector incorrectly when subtracting two position vectors. Remember AB = b − a.
• Using the same parameter for both lines when testing an intersection.
• Checking only two coordinate equations and forgetting the third.
• Assuming that non-parallel lines must intersect. In three dimensions, they may be skew.
• Failing to distinguish between coincident and distinct parallel lines.
• Giving only the intersection parameters without stating the coordinates of the intersection point.
• Writing a vector expression without the equation sign r =.

12. Formula Summary

Vector equation of a line:

r = a + tb

Direction vector through two points:

AB = b − a

Line through two points:

r = a + t(b − a)

Parallel direction vectors:

b₁ = kb₂

Condition for intersection:

a + sb = c + td

For an intersection in three dimensions, the same values of s and t must satisfy all three coordinate equations.

13. Practice Exercises

Attempt the following questions before opening the solutions. Show all your working clearly.

Exercise A — Vector Equations of Lines

1. Find the vector equation of the line passing through A(3, −2, 1) with direction vector (2, 1, −3).

2. Find the vector equation of the line passing through P(1, 2, 0) and Q(4, −1, 6).

3. A line is given by r = (2, −1, 4) + t(3, 2, −1). Find the point corresponding to t = −2.

4. Determine whether the point (7, 1, 2) lies on the line r = (1, 5, 4) + t(2, −1, −1).

Exercise B — Parallel Lines

1. Determine whether the lines r = (1, 2, 3) + s(2, 4, −6) and r = (4, 0, 1) + t(1, 2, −3) are parallel.

2. Determine whether the lines r = (0, 1, 2) + s(1, 2, 3) and r = (2, 5, 8) + t(2, 4, 6) are coincident or distinct parallel lines.

Exercise C — Intersections

1. Find the point of intersection of:

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

r = (2, 3, 4) + t(−1, −1, −2)

2. Determine whether the lines intersect:

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

r = (3, 2, 2) + t(1, −1, 0)

Exercise D — Skew Lines

1. Determine whether the following lines are parallel, intersecting or skew:

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

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

2. Explain why two non-parallel lines in three-dimensional space do not necessarily intersect.

14. Worked Answers to Practice Exercises

Use these solutions to check your method and final answers after attempting the exercises.

Exercise A — Show Answers

1. The position vector is (3, −2, 1) and the direction vector is (2, 1, −3).

r = (3, −2, 1) + t(2, 1, −3).

2. Direction vector PQ = Q − P = (4 − 1, −1 − 2, 6 − 0) = (3, −3, 6).

r = (1, 2, 0) + t(3, −3, 6).

3. Substitute t = −2:

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

4. From x = 1 + 2t = 7, t = 3. Then y = 5 − 3 = 2, which is not 1. Therefore, the point does not lie on the line.

Exercise B — Show Answers

1. The direction vectors are (2, 4, −6) and (1, 2, −3).

(2, 4, −6) = 2(1, 2, −3).

Therefore, the lines are parallel.

2. The direction vectors are scalar multiples. Test whether (2, 5, 8) lies on the first line.

From x = s = 2, we obtain s = 2. The y-coordinate becomes 1 + 2(2) = 5 and the z-coordinate becomes 2 + 3(2) = 8.

All coordinates agree. Therefore, the lines are coincident.

Exercise C — Show Answers

1. Equating coordinates gives:

s = 2 − t, 1 + s = 3 − t, and 2 = 4 − 2t.

The third equation gives t = 1. The first gives s = 1. The second is then satisfied.

Substitute s = 1 into the first line:

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

Intersection: (1, 2, 2).

2. The parametric equations are:

First line: x = 1 + 2s, y = s, z = 1 + s.

Second line: x = 3 + t, y = 2 − t, z = 2.

From z-coordinates, 1 + s = 2, so s = 1.

From y-coordinates, 1 = 2 − t, so t = 1.

Check x: 1 + 2(1) = 3 + 1, so 3 = 4, which is false.

Therefore, the lines do not intersect. Their direction vectors (2, 1, 1) and (1, −1, 0) are not scalar multiples, so the lines are skew.

Exercise D — Show Answers

1. The direction vectors (1, 2, 1) and (0, 1, 1) are not scalar multiples.

The parametric equations are:

First line: x = s, y = 2s, z = s.

Second line: x = 1, y = t, z = 2 + t.

Equating x gives s = 1. Equating y gives t = 2. But equating z gives 1 = 2 + t, so t = −1.

The conditions are inconsistent. The lines do not intersect and are not parallel.

Answer: The lines are skew.

2. In three dimensions, lines can lie in different planes. They may have different directions but still never meet. Such lines are called skew lines.

End of Part 3

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