# Magnitude of a Vector

Lesson

## Magnitude

We use the word magnitude to mean the size, or length, of the vector.  The magnitude of a vector is indicated using the absolute value notation. So magnitude of of vector is indicated by .

How can we work out the length of a vector?  Well we use Pythagoras' Theorem.

To find the length of this vector

We first need the horizotonal and vertical components

Using pythagoras

 $\text{length}^2$length2 $=$= $4^2+3^2$42+32 $\text{length}^2$length2 $=$= $16+9$16+9 $\text{length}$length $=$= $\sqrt{25}$√25 $\text{length}$length $=$= $5$5

If we are just given the coordinates of the initial and terminal points we would work out the magnitude like this.

Find the magnitude of the vector with initial point
$$and terminal point$$

The horizontal component is the distance between the x-coordinates. Which is $$The vertical component is the distance between the y-coordinates. Which is$$

So $=\sqrt{3^2+9^2}=\sqrt{90}=3\sqrt{10}$=32+92=90=310

## Equality

Vectors are only equal if they have the same magnitude and direction.  They need not have the same initial and terminal points, just the same size and direction.

So these vectors are all equal

And these are not

## Unit Vector

A unit vector is a vector of of magnitude (length) 1.  There is specific notation we can use for a unit vector that uses i and j's and we will discuss this in later entries.  For now we just need to know that if the length is 1, it is a unit vector in that particular direction.

#### Worked Examples

##### Question 1

Consider the vector $\left(2\sqrt{3},2\right)$(23,2).

1. Find the magnitude of the vector.

2. Find the direction of the vector.

##### Question 2

Find the magnitude of $a$a.

 $-8$−8 $-15$−15