Introduction: Measuring the Gap
Welcome to the world of vectors! In this chapter, we are going to learn how to calculate the exact distance between two points in a 2D plane. Whether you are designing a level in a video game or calculating the shortest flight path between two cities, knowing the distance between two coordinates is a fundamental skill. If you have ever used a ruler to measure the space between two dots on a page, you’ve already done this in real life. Now, we are just going to learn the Mathematics behind it using Vectors. Don't worry if vectors feel a bit strange at first; by the end of these notes, you'll see that it’s mostly just a clever application of Pythagoras' Theorem!1. Understanding Position Vectors (Syllabus 1.10e)
Before we can find the distance between two points, we need to know where they are. In vector notation, we use Position Vectors to describe the location of a point relative to the origin \( (0,0) \). What is a Position Vector? Imagine you are standing at the origin. A position vector is like a set of instructions telling you how to get to a specific point. For a point \(A\) with coordinates \( (a, b) \), the position vector is usually written as: \(\mathbf{a} = a\mathbf{i} + b\mathbf{j}\) or as a column vector \(\begin{pmatrix} a \\ b \end{pmatrix}\). Key Terms to Know: • Displacement Vector: The vector that takes you from one point to another. • Component Vector: The individual horizontal (\(\mathbf{i}\)) and vertical (\(\mathbf{j}\)) parts of a vector. • Unit Vector: A vector with a magnitude (length) of exactly 1. Example: If point \(A\) is at \((2, 3)\), its position vector is \(2\mathbf{i} + 3\mathbf{j}\).2. Finding the "Bridge" Between Two Points
To find the distance between point \(A\) and point \(B\), we first need to find the Displacement Vector \( \vec{AB} \). Think of this as the "bridge" connecting the two points. If point \(A\) has position vector \(\mathbf{a}\) and point \(B\) has position vector \(\mathbf{b}\), the vector from \(A\) to \(B\) is found by: \(\vec{AB} = \mathbf{b} - \mathbf{a}\) Step-by-Step Logic: 1. Start with the "destination" vector (\(\mathbf{b}\)). 2. Subtract the "starting" vector (\(\mathbf{a}\)). 3. The result is the vector that tells you how to travel directly from \(A\) to \(B\). Quick Review: To get the vector between two points, it’s always "Finish minus Start".3. The Distance Formula (Syllabus 1.10f)
Now that we have the vector \(\vec{AB}\), how do we find the actual length (distance)? We calculate the Magnitude of the vector. If point \(A = a\mathbf{i} + b\mathbf{j}\) and point \(B = c\mathbf{i} + d\mathbf{j}\), the displacement vector is: \(\vec{AB} = (c - a)\mathbf{i} + (d - b)\mathbf{j}\) The distance is the Modulus (length) of this vector, calculated using: Distance = \(\sqrt{(c - a)^2 + (d - b)^2}\) Did you know? This formula is just Pythagoras' Theorem (\(a^2 + b^2 = c^2\)) in disguise! The term \((c-a)\) is the horizontal distance, and \((d-b)\) is the vertical distance.4. Step-by-Step Example
Let's find the distance between point \(P (1, 2)\) and point \(Q (5, 5)\). Step 1: Write down the position vectors. \(\mathbf{p} = 1\mathbf{i} + 2\mathbf{j}\) \(\mathbf{q} = 5\mathbf{i} + 5\mathbf{j}\) Step 2: Find the displacement vector \(\vec{PQ}\). Subtract \(\mathbf{p}\) from \(\mathbf{q}\): \((5 - 1)\mathbf{i} + (5 - 2)\mathbf{j} = 4\mathbf{i} + 3\mathbf{j}\) Step 3: Calculate the magnitude (the distance). Square the components, add them together, and take the square root: Distance = \(\sqrt{4^2 + 3^2}\) Distance = \(\sqrt{16 + 9}\) Distance = \(\sqrt{25} = 5\) units. Key Takeaway: The vector \(4\mathbf{i} + 3\mathbf{j}\) tells us to go 4 right and 3 up. The distance formula tells us the direct "as the crow flies" path is 5 units long.5. Memory Aids and Tips
The "4 S" Rule for Distance: To remember the process for the magnitude, use the 4 S's: 1. Subtract the coordinates. 2. Square the results. 3. Sum (add) them up. 4. Square root the answer. Analogy: The GPS Mistake Imagine your GPS tells you to drive 3 miles East and 4 miles North. That’s your vector. If you could fly a helicopter directly to the destination across the fields, the distance you fly would be the magnitude (5 miles).6. Common Mistakes to Avoid
• The Negative Trap: Students often worry when they get a negative number after subtracting (e.g., \(2 - 5 = -3\)). Don't panic! When you square a negative number, it always becomes positive (\((-3)^2 = 9\)). Distance is never negative. • Mixing up \( \mathbf{i} \) and \( \mathbf{j} \): Always keep your horizontal components (\(\mathbf{i}\)) together and your vertical components (\(\mathbf{j}\)) together. Don't mix them! • Forgetting the Root: It’s very common to do the squaring and adding but forget to take the final square root. Always check if your answer looks sensible compared to the points.7. Summary Checklist
Quick Review Box
• A position vector starts from the origin \((0,0)\).• The displacement vector between two points \(A\) and \(B\) is \(\mathbf{b} - \mathbf{a}\).
• The distance is the magnitude of that displacement vector.
• Use the formula: \(\sqrt{x^2 + y^2}\) for the components of the displacement vector.
• Square every component before adding them. Top Tip: If you find this tricky, draw a small sketch of the two points on a graph. It will help you visualize the right-angled triangle you are solving!