Vectors and magnitude
Vectors in two dimensions (Extended)
- A vector has size and direction, written as a column $\begin{pmatrix} x \\ y \end{pmatrix}$, as $\overrightarrow{AB}$, or in bold $\mathbf{a}$.
- Add / subtract: work on top and bottom separately. Scalar multiply: multiply both numbers.
- Worked example: $\mathbf{a}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}$, $\mathbf{b}=\begin{pmatrix} 2 \\ -4 \end{pmatrix}$ → $\mathbf{a}+\mathbf{b}=\begin{pmatrix} 5 \\ -3 \end{pmatrix}$, $3\mathbf{a}=\begin{pmatrix} 9 \\ 3 \end{pmatrix}$.
Practice
a = (3, 1) and b = (2, −4). The top number of a + b is?
3 + 2 = 5.
Practice
a = (3, 1) and b = (2, −4). The bottom number of a + b is?
1 + (−4) = −3.
Magnitude (Extended)
- The magnitude (length) uses Pythagoras:
$$\left| \begin{pmatrix} x \\ y \end{pmatrix} \right| = \sqrt{x^2 + y^2}$$
- Worked example: $\left|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right| = \sqrt{9 + 16} = 5$.
Practice
Find the magnitude (length) of the vector (3, 4).
√(3² + 4²) = √25 = 5.
You've got it
Key idea
- add/subtract vectors component by component; scalar multiply scales both
- $\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}$
- magnitude $= \sqrt{x^2 + y^2}$, so $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ has length $5$