Vectors I
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Represent 2D vectors algebraically using column vectors and component unit vectors (\(\mathbf{i}, \mathbf{j}\)), and solve vector addition and subtraction geometrically using the tip-to-tail method.
Interactive Tip-to-Tail Vector Visualizer
Adjust horizontal and vertical components of Vector A and Vector B to observe resultant displacement!
(a) Concrete Scenario: A boda-boda rider in Nairobi delivers a package by traveling \(6\text{ km}\) due east, then turning and riding \(8\text{ km}\) due north. Although the rider traveled a total road distance of \(14\text{ km}\), their straight-line displacement vector from the origin is the hypotenuse vector \(\mathbf{R}\), measuring exactly \(10\text{ km}\).
(b) Key Vector Concepts:
- Vector vs. Scalar: A scalar has magnitude only (e.g., speed = \(50\text{ km/h}\)); a vector has both magnitude and direction (e.g., velocity = \(50\text{ km/h}\) East).
- Tip-to-Tail Method: To add vectors geometrically, place the tail (start) of the second vector at the tip (arrowhead) of the first. The resultant vector \(\mathbf{R}\) points from the initial start to the final tip.
- Column Vector Notation: \(\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix} = x\mathbf{i} + y\mathbf{j}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along the horizontal and vertical axes.
(c) Vector Algebra Rules: \[\text{Addition: } \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \end{pmatrix}\] \[\text{Subtraction: } \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} - \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 - x_2 \\ y_1 - y_2 \end{pmatrix}\] \[\text{Magnitude: } |\mathbf{a}| = \sqrt{x^2 + y^2}\]
Key Formulas
Worked Examples
- Represent as column vectors: \(\mathbf{d}_1 = \begin{pmatrix} 60 \\ 0 \end{pmatrix}\), \(\mathbf{d}_2 = \begin{pmatrix} 0 \\ 80 \end{pmatrix}\).
- Sum component vectors: \(\mathbf{R} = \mathbf{d}_1 + \mathbf{d}_2 = \begin{pmatrix} 60 \\ 80 \end{pmatrix}\).
- Calculate magnitude: \(|\mathbf{R}| = \sqrt{60^2 + 80^2} = \sqrt{3,600 + 6,400} = \sqrt{10,000} = 100\text{ km}\).
- Answer: \(100\text{ km}\).
- Add components: \(\mathbf{R} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3+2 \\ 4+1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}\).
- Calculate magnitude: \(|\mathbf{R}| = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \approx 7.071\).
- Answer: \(7.07\).
- Subtract components: \(\vec{PQ} = (9 - 4)\mathbf{i} + (-5 - 7)\mathbf{j} = 5\mathbf{i} - 12\mathbf{j}\).
- Represent as column vector: \(\begin{pmatrix} 5 \\ -12 \end{pmatrix}\).
- Calculate magnitude: \(|\vec{PQ}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\).
- Answer: Vector = \(5\mathbf{i} - 12\mathbf{j}\) ; Magnitude = \(13\).
Common Mistakes
Real World
Practice