Vectors in 2D and 3D

Introduction

Assalamu alaikum, bachon! I am Ustad Bilal Ahmed, and today we will tackle one of the most crucial topics in your A Level journey: Vectors. Think of vectors not just as arrows on a page, but as a powerful language to describe movement, forces, and positions in the world around us. From mapping the flight path of a PIA aircraft to designing the supports for a bridge over the Indus River, vectors provide the mathematical framework.

In your Pure Mathematics 1 (9709) paper, vector questions are a guaranteed feature. They test your understanding of both abstract geometric concepts and your ability to perform precise algebraic manipulations. Mastering this topic means you can solve complex problems involving lines and angles in space with confidence. We will break down every concept from the ground up, from basic notation to finding the shortest distance between a point and a line, ensuring you are fully prepared for your Cambridge examinations.

Core Theory

Let's begin with the fundamentals. A vector has both magnitude (size) and direction, unlike a scalar which only has magnitude.

Vector Notation and Basic Operations

We can represent a vector in two primary ways:

1. Column Vector: A vector from the origin O to a point A(x, y, z) is written as a position vector `OA` or a.

a = `( (x), (y), (z) )`

1. Unit Vector Notation: Using the base unit vectors i, j, and k (which are vectors of magnitude 1 along the x, y, and z axes respectively).

a = xi + yj + zk

The magnitude (or modulus) of a vector a is its length, denoted by |a|.

|a| = √(x² + y² + z²)

A unit vector in the direction of a, denoted â, is found by dividing the vector by its magnitude. It has a magnitude of 1.

â = a / |a|

Position and Displacement Vectors

A position vector starts from the origin (O). For a point P, the position vector is p = `OP`.

A displacement vector is the vector between two points, say A and B. It represents the journey *from* A *to* B.

`AB` = b - a = `OB` - `OA`

(Remember: Position of end point minus position of start point. This is a very common place to make a mistake!)

Scalar (Dot) Product

The scalar product, or dot product, is a way of multiplying two vectors to get a scalar result. It has two definitions which we use interchangeably:

1. Geometric Definition: a · b = |a||b|cosθ, where θ is the angle between the two vectors when placed tail-to-tail.
2. Algebraic Definition: If a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k, then:

a · b = a₁b₁ + a₂b₂ + a₃b₃

This gives us a powerful tool to find the angle between two vectors:

cosθ = (a · b) / (|a||b|)

A crucial property: If two non-zero vectors a and b are perpendicular (orthogonal), the angle between them is 90°, and cos(90°) = 0. Therefore:

a · b = 0 (Condition for perpendicularity)

Vector Equation of a Line

To define a line in 3D space, we need two things: a point on the line and the direction of the line.

The vector equation of a line is given by:

r = a + tb

Where:

• r is the position vector of *any* general point on the line.
• a is the position vector of a *known* point on the line.
• b is a direction vector parallel to the line.
• t is a scalar parameter. For each value of t, you get a different point on the line.

Problems Involving Lines

1. Intersection of Two Lines: Let two lines be r₁ = a₁ + sb₁ and r₂ = a₂ + tb₂. If they intersect, there is a point where r₁ = r₂.

a₁ + sb₁ = a₂ + tb₂

Equate the i, j, and k components to get three simultaneous equations in s and t. Solve two equations to find s and t, then substitute into the third equation to check for consistency. If it's consistent, they intersect. If not, they are skew. (If their direction vectors b₁ and b₂ are parallel, the lines are parallel and will not intersect unless they are the same line).

1. Angle Between Two Lines: The angle between two intersecting lines is the angle between their direction vectors, b₁ and b₂. Use the dot product formula on the direction vectors.

1. Shortest Distance from a Point to a Line: Let P be a point with position vector p, and the line be r = a + tb. Let N be the point on the line closest to P. The vector `PN` will be perpendicular to the direction vector b.

• The position vector of N, `ON`, can be written as a + tb for some value of t.
• The vector `PN` = `ON` - `OP` = (a + tb) - p.
• Since `PN` is perpendicular to b, their dot product is zero: `PN` · b = 0.
• ((a - p) + tb) · b = 0. Solve this equation to find the specific value of t for point N.
• Substitute t back to find the position vector of N, and then find the distance |`PN`|.

Key Definitions

• Vector: A quantity with both magnitude and direction.
• Scalar: A quantity with only magnitude.
• Position Vector (`OA`): A vector from the origin O to a point A.
• Displacement Vector (`AB`): The vector from point A to point B, calculated as `OB - OA`.
• Magnitude (|a|): The length of a vector, calculated as √(x² + y² + z²).
• Unit Vector (â): A vector with a magnitude of 1 in the same direction as the original vector, found by **a** / |**a**|.
• Scalar Product (a · b): Defined as |**a**||**b**|cosθ or a₁b₁ + a₂b₂ + a₃b₃. It results in a scalar.
• Perpendicular Vectors: Two vectors **a** and **b** are perpendicular if and only if **a** · **b** = 0.
• Vector Equation of a Line (r = a + tb): Defines a line using a point on the line (**a**) and its direction vector (**b**).
• Skew Lines: Two lines in 3D space that do not intersect and are not parallel.

Worked Examples (Pakistani Context)

Example 1: Pure Algebraic Problem

The points A and B have position vectors a = 2i - j + 3k and b = 3i + 2j - k. The line L has the equation r = ( i + 4j + k ) + t( i + j + 2k ).

(i) Find the vector equation of the line passing through A and B.

(ii) Find the acute angle between the line L and the line AB.

Solution:

(i) First, find the direction vector of the line through A and B.

Direction vector `AB` = b - a = (3i + 2j - k) - (2i - j + 3k)

`AB` = (3-2)i + (2-(-1))j + (-1-3)k = i + 3j - 4k

The line passes through point A (or B). Using A's position vector as the point on the line:

r = a + s(`AB`)

r = (2i - j + 3k) + s(i + 3j - 4k) [This is the equation of the line AB]

(ii) The angle between the lines is the angle between their direction vectors.

Direction vector of L: d₁ = i + j + 2k

Direction vector of AB: d₂ = i + 3j - 4k

Use the dot product formula: cosθ = (d₁ · d₂) / (|d₁||d₂|)

d₁ · d₂ = (1)(1) + (1)(3) + (2)(-4) = 1 + 3 - 8 = -4

|d₁| = √(1² + 1² + 2²) = √6

|d₂| = √(1² + 3² + (-4)²) = √(1 + 9 + 16) = √26

cosθ = -4 / (√6 * √26) = -4 / √156

θ = arccos(-4 / √156) ≈ 108.7°

The question asks for the *acute* angle. The angle between lines is always taken as the acute one.

Acute angle = 180° - 108.7° = 71.3°

Example 2: Lahore Metro Orange Line

An engineering team is analyzing a section of the Lahore Metro Orange Line. Relative to an origin O at Kalma Chowk, the track passing through the Anarkali station (A) can be modelled as a straight line L₁. Anarkali station is at A(2, 5, 1) and the track moves in the direction d₁ = 3i + 4j. Another track, L₂, for a maintenance line passes through point B(9, 3, 1) with direction d₂ = -i + j. All coordinates are in km.

(i) Write down the vector equations for both tracks.

(ii) Determine if the paths of the tracks intersect. If they do, find the coordinates of the intersection point.

Solution:

(i) The vector equations are of the form r = a + tb.

For track L₁ (Anarkali):

r = (2i + 5j + k) + s(3i + 4j)

For track L₂ (Maintenance):

r = (9i + 3j + k) + t(-i + j)

Note: The z-coordinate is constant at 1, meaning the tracks are at a constant elevation of 10m (if 1 unit = 10m) in this model.

(ii) To find if they intersect, we set the equations equal to each other:

(2i + 5j + k) + s(3i + 4j) = (9i + 3j + k) + t(-i + j)

Now, we equate the components:

i: 2 + 3s = 9 - t (1)

j: 5 + 4s = 3 + t (2)

k: 1 = 1 (This is consistent, as expected since they are co-planar)

We have a system of two linear equations. Let's solve them.

From (2), t = 2 + 4s.

Substitute this into (1):

2 + 3s = 9 - (2 + 4s)

2 + 3s = 7 - 4s

7s = 5

s = 5/7

Now find t:

t = 2 + 4(5/7) = 2 + 20/7 = 14/7 + 20/7 = 34/7

Since we found unique values for s and t, the paths do intersect. To find the coordinates of the intersection point, substitute s back into the equation for L₁ (or t into L₂).

Using L₁:

r = (2i + 5j + k) + (5/7)(3i + 4j)

r = (2 + 15/7)i + (5 + 20/7)j + k

r = (14/7 + 15/7)i + (35/7 + 20/7)j + k

r = (29/7)i + (55/7)j + k

The intersection point has coordinates (29/7, 55/7, 1). This is approximately (4.14, 7.86, 1). The engineers can use this point to design the junction safely.

Exam Technique

For your Paper 1 exam, vector questions are often worth a significant number of marks (8-12 marks). Here is how you should approach them.

1. Read and Underline: Read the entire question carefully. Underline key terms like "position vector", "direction vector", "perpendicular", "intersection", "acute angle". Misreading one word can change the entire question.

1. Distinguish Points and Directions: Be crystal clear about what is a position vector (a, representing a point) and what is a direction vector (b, representing the line's orientation). Don't mix them up in the line equation r = a + tb.

1. Show Your Working: For "Show that..." questions, you must provide every single step of your logic. You are being marked on the process. For "Find..." or "Calculate..." questions, clear working is essential for method marks (M marks). If you make a small calculation error but your method is correct, you will still get most of the credit.

1. Common Mistakes to Avoid:

• Calculating `AB` as a - b. It is always b - a.
• Using position vectors to find the angle between lines instead of direction vectors.
• Algebraic errors when solving simultaneous equations for intersection. Double-check your work.
• Forgetting to find the acute angle if the dot product gives you an obtuse one. Angle between lines is conventionally acute.
• In shortest distance problems, making sure the vector you dot with the direction vector (to get zero) is the one connecting the general point on the line to the external point (`PN` in our theory example).

1. Mark Scheme Insights:

• An M1 mark is often given for correctly setting up the dot product formula.
• An M1 is given for equating the components of two line equations.
• An A1 (Accuracy) mark is given for the correct final answer.
• B1 (Independent) marks are often given for stating a correct vector or equation.
• By showing the correct formula and substitution, you secure method marks even if your final answer is incorrect. Never leave a question blank; attempt to set up the problem.