## (5) Finding the magnitude or length
of a vector

This drawing shows you how we set about doing this.

Notice the right-hand coordinate system but with the z-axis vertical
this time.)
The formula for the length OP or |**p**| is a kind of 3 dimensional
form of Pythagoras' theorem.

Here's how the working would go if we had
a = 4 and b = 3 and c = 12.

Using Pythagoras' theorem in the yellow right-angled triangle we get

OQ = the square root of (16 + 9) = 5

and then, using Pythagoras' theorem in the vertical green right-angled
triangle, we get

OP = the square root of (25 + 144) = 13.
Alternatively, in just one step of working, we have

OP = the square root of (16 + 9 + 144) = 13.

**A reminder on unit vectors**
Unit vectors are vectors which have unit length. They are
important in many applications. In three dimensions, the vectors **i**,
**j** and **k** which run along the x, y and z axes respectively, all
have unit length.

Sometimes, for example when working with planes, it is necessary to find
a unit vector in the same direction as a given vector.

Suppose you need the unit vector in the same direction
as **u** = 2**i** - **j** + 2**k** or
**u** = (2, -1, 2) written as a row vector.

The length or magnitude of **u** is given by the square root
of (4 + 1 + 4) = 3.

Also, the required vector must be parallel to **u**.

So we can get the vector we want by just scaling down **u** by a
factor of 3.

It is 1/3(2**i** - **j** + 2**k**) = 1/3(2, -1, 2) =
(2/3, -1/3, 2/3).

Unit vectors are often written with a little hat on top, so here we
would have found **û**.

Here's a practical application which uses components and magnitude.

If a body is acted on by three forces **P**, **Q** and **S**
whose lines of action
all pass through the same point, find the resultant force **R**, and its
magnitude |**R**| if

**P** = 3**i** - 5**j** + 2**k**

**Q** = - 2**i** + 4**j** + 4**k**

**S** = 3**i** - 3**j** - 4**k**.
**P** + **Q** + **S** = **R** = (3 - 2 + 3)**i** +
(- 5 + 4 - 3)**j**
+ ( 2 + 4 - 4)**k** = 4**i** - 4**j** + 2**k**

and |**R**| = the square root of (16 + 16 + 4) = 6.

We might also need to know what angles this resultant makes with the
x, y and z axes. Finding these angles will come as a spin-off from
discovering how we can multiply two vectors together.

But how can we actually *do* this since the two vectors can have
different directions?

**Next**

or **back to the previous section**
or **return to the vector homepage.**