Vector Magnitudes & Lengths

What is the Magnitude of a Vector?

The magnitude (or modulus) of a vector is its length or size.

It is always a positive value.

Notation: The magnitude of vector $\mathbf{a}$ is written as $|\mathbf{a}|$.

The magnitude of vector $\vec{AB}$ is written as $|\vec{AB}|$.

Finding the Magnitude from Components

If a vector is written as a column vector: $\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}$ Then the magnitude is found using Pythagoras' Theorem:

$$|\mathbf{a}| = \sqrt{x^2 + y^2}$$

Example: $$\mathbf{a} = \begin{pmatrix} 6 \\ -2 \end{pmatrix} \Rightarrow |\mathbf{a}| = \sqrt{6^2 + (-2)^2} \\ = \sqrt{36 + 4} = \sqrt{40} \approx 6.32$$

Example

Points: $A(8, 2), B(11,-4)$

(a) Find the column vector $\vec{AB}$

$$\vec{AB} = \begin{pmatrix} 8 - 11 \\ 2 - (-4) \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$$

(b) Find the magnitude $|\vec{AB}|$

$$|\vec{AB}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.71$$

(c) Explain why $|\vec{BA}| = |\vec{AB}|$

Because magnitude only measures size, not direction

So even though $\vec{BA}$ points the opposite way, it is the same length

Scaling Magnitudes

(d) Vector $\vec{CD}$ is 4 times the magnitude of $\vec{AB}$. Find a possible column vector for $\vec{CD}$

We know $\vec{AB} = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$, so:

$$\vec{CD} = 4 \times \begin{pmatrix} -3 \\ 6 \end{pmatrix} = \begin{pmatrix} -12 \\ 24 \end{pmatrix}$$