Vector Arithmetic and Representations

In mathematics, the starting position of a vector defines its specific type. A position vector describes the exact location of a point relative to a fixed origin ($O$), such as $\vec{OA} = \mathbf{a}$. Conversely, a displacement vector represents the movement between two specific points anywhere on a grid, such as $\vec{AB}$. In printed Edexcel exams, vectors are denoted by bold lowercase letters (like $\mathbf{a}$) or capital letters with an arrow (like $\vec{AB}$).

Multiplying Vectors by a Scalar

When you zoom in on a photograph on your phone, the image gets larger, but the proportions and orientation stay exactly the same. In vector mathematics, a scalar is a regular real number (a scale factor) that has magnitude but no direction. Multiplying a vector by a scalar scales its length while keeping it parallel to the original path.

The magnitude (length) of a vector is denoted by $|\mathbf{a}|$ and can be calculated using Pythagoras' Theorem:

$$|\begin{pmatrix} x \\ y \end{pmatrix}| = \sqrt{x^2 + y^2}$$

When multiplying a vector by a scalar $k$, the new magnitude becomes $|k|$ times the original. If $k$ is positive, the direction remains identical. If $k$ is negative, the direction is entirely reversed. If $0 < |k| < 1$, the vector shrinks.

Worked Example: Scalar Multiplication

Calculate the resultant column vectors when multiplying the vector $\mathbf{v} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ by the scalar $3$, and by the scalar $-0.5$.

Step 1: For the first scalar, multiply both the $x$ and $y$ components by $3$. $\mathbf{v}_1 = 3 \times \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \times 4 \\ 3 \times -2 \end{pmatrix}$

Step 2: State the final resultant vector. $\mathbf{v}_1 = \begin{pmatrix} 12 \\ -6 \end{pmatrix}$

Step 3: Repeat the process for the negative scalar, remembering that a negative multiplier will reverse the vector's direction. $\mathbf{v}_2 = -0.5 \times \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} -0.5 \times 4 \\ -0.5 \times -2 \end{pmatrix}$

Step 4: Calculate the final resultant vector, taking care with the double negative in the $y$ component. $\mathbf{v}_2 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$

Adding and Subtracting Vectors

Why does a plane flying in a crosswind end up travelling in a completely different direction to where its nose is pointing? When two or more vectors act together, they combine to form a single resultant vector. To add or subtract vectors algebraically, you simply perform the arithmetic on the corresponding $x$ and $y$ components separately.

Geometrically, addition is performed using the triangle law (often called the "nose-to-tail" method). You place the tail (start) of the second vector at the head (tip) of the first. The resultant vector is the shortcut drawn straight from the very start of the journey to the final destination. If a vector journey starts and ends at the exact same point, it forms a zero vector, which has 0 magnitude and no direction.

Subtracting a vector ($\mathbf{a} - \mathbf{b}$) is geometrically identical to adding its negative. You draw $\mathbf{a}$, and then draw $\mathbf{b}$ pointing in the opposite direction. A crucial rule for displacement vector journeys is "Finish minus Start", written algebraically as $\vec{AB} = \mathbf{b} - \mathbf{a}$.

Worked Example: Vector Addition

Given the vectors $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}$, calculate the resultant vector $\mathbf{a} + \mathbf{b}$.

Step 1: Write out the addition of the two column vectors. $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} + \begin{pmatrix} 5 \\ 4 \end{pmatrix}$

Step 2: Add the corresponding $x$ components (top numbers) and $y$ components (bottom numbers) together. $\begin{pmatrix} 3 + 5 \\ -2 + 4 \end{pmatrix}$

Step 3: State the final resultant column vector. $\begin{pmatrix} 8 \\ 2 \end{pmatrix}$

Worked Example: Vector Subtraction

Given the vectors $\mathbf{p} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$, calculate the resultant vector $\mathbf{p} - \mathbf{q}$.

Step 1: Write out the subtraction of the two column vectors. $\mathbf{p} - \mathbf{q} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ -5 \end{pmatrix}$

Step 2: Subtract the corresponding $x$ components (top numbers) and $y$ components (bottom numbers), taking care with double negatives. $\begin{pmatrix} 7 - 2 \\ 1 - (-5) \end{pmatrix}$

Step 3: State the final resultant column vector. $\begin{pmatrix} 5 \\ 6 \end{pmatrix}$