Vectors and Geometric Proofs for the TMUA

Understanding Vector Basics

Vectors are used to represent quantities that have both a size (magnitude) and a specific direction. In the TMUA, you must distinguish between scalars, which only have magnitude (such as distance or speed), and vectors (such as displacement or velocity).

Geometrically, a vector is represented by an arrow. The length of the arrow represents the magnitude, and the arrow's orientation represents the direction. Notationally, a vector can be written as a bold lower-case letter, such as $\mathbf{a}$, or by using the start and end points with an arrow above them, such as $\vec{AB}$, which represents the vector starting at point $A$ and ending at point $B$.

Vector Representations

There are two primary ways to represent vectors in the TMUA syllabus:

1. Diagrammatic Representation: This involves drawing vectors as directed line segments in a plane. To find the sum of two vectors $\mathbf{a}$ and $\mathbf{b}$ diagrammatically, we use the triangle law. We place the start of vector $\mathbf{b}$ at the end of vector $\mathbf{a}$. The resultant vector $\mathbf{a} + \mathbf{b}$ is the single vector that completes the triangle, going from the start of $\mathbf{a}$ to the end of $\mathbf{b}$.

2. Column Representation: A vector can be expressed in terms of its horizontal and vertical components. In two dimensions, this is written as $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$, where $x$ is the horizontal displacement and $y$ is the vertical displacement.

Vector Arithmetic and Scalar Multiplication

Arithmetic with vectors follows specific rules:

Addition and Subtraction: To add or subtract vectors in column form, you simply add or subtract their corresponding components. If $\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$, then $\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix}$. Diagrammatically, $\mathbf{a} - \mathbf{b}$ is interpreted as $\mathbf{a} + (-\mathbf{b})$, where $-\mathbf{b}$ is a vector with the same magnitude as $\mathbf{b}$ but pointing in the exactly opposite direction.

Scalar Multiplication: Multiplying a vector by a scalar $k$ changes its magnitude. If $k > 0$, the direction remains the same. If $k < 0$, the direction is reversed. In column form, $k\mathbf{a} = \begin{pmatrix} ka_1 \\ ka_2 \end{pmatrix}$.

Parallel Vectors: Two vectors are parallel if and only if one is a scalar multiple of the other. That is, $\mathbf{a}$ is parallel to $\mathbf{b}$ if $\mathbf{a} = k\mathbf{b}$ for some non zero scalar $k$.

Geometric Arguments and Proofs

Vectors are particularly useful for proving geometric properties within shapes.

Position Vectors: The position vector of a point $P$ is the displacement of $P$ from a fixed origin $O$, denoted $\vec{OP}$ or $\mathbf{p}$. The displacement vector between any two points $A$ and $B$ can be found using their position vectors: $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$.

Example Proof: Midpoints and Ratios

Consider a triangle $OAB$ where $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. Let $M$ be the midpoint of $OA$ and $N$ be the midpoint of $OB$. We can find the vector $\vec{MN}$ to see how it relates to the base $\vec{AB}$.

1. First, express $\vec{MN}$ as a path: $\vec{MN} = \vec{MO} + \vec{ON}$.
2. Since $M$ is the midpoint of $OA$, $\vec{MO} = -\frac{1}{2}\mathbf{a}$.
3. Since $N$ is the midpoint of $OB$, $\vec{ON} = \frac{1}{2}\mathbf{b}$.
4. Thus, $\vec{MN} = -\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$.
5. We know $\vec{AB} = \mathbf{b} - \mathbf{a}$.
6. Therefore, $\vec{MN} = \frac{1}{2}\vec{AB}$.

This proof shows that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Collinearity: To prove that three points $A$, $B$, and $C$ lie on a straight line (are collinear), you must show that $\vec{AB}$ is parallel to $\vec{BC}$. Since they share a common point $B$, if the vectors are parallel, the points must be collinear.