This is an introduction to

mathematical vectors, and how to use them. You will benefit from some

graph paper (or

squared paper), a

pencil and a

ruler; I can't render

vector triangles very well in ASCII and thus will be asking you to do it yourself.

A vector is a pair of values - in mathematical language this is as opposed to a scalar, which is a single value - and is commonly represented like this -

|- -|
| x |
| |
| y |
|- -|

There are obvious problems with that representation in text, so from now I will use (x,y) notation to denote a vector.

A vector is said to have magnitude and direction. On your squared paper, draw axis from 0 to 10 on the x axis, and 0 to 10 on the y axis. Then place a point at (1,1) on the graph.

We will now use vectors for the first practical purpose: to apply a transformation to this point. Presume you are asked to apply the transformation vector (2,3) to this point - what do you think you should do? The answer is that you should place a new point at (3,4) - you've added the transformation vector to the original vector which represented the position.

We refer to this vector as displacement because it has both magnitude and direction - when a scalar value is used to represent distance there is usually no direction given.

We keep saying that the vector has magnitude and direction, but where exactly do these values come from? They're certainly not just the x and y values. We find the magnitude by applying a piece of maths you've probably heard of - the Pythagorean Theorem. This states simply that in a right-handed triangle, the square on the hypotenuse is equal to the sum of the squares of the other two sides. A right-angled triangle looks like this, with the sides labeled -

|\
| \
b | \ h
| \
|----
a

Now, think about your vector - what does it represent? Well, if you draw it ( first draw a line from (1,1) to (3,1), then a line from (1,1) to (1,4), and finally a line from (1,1) to (3,4) ) you'll see you have a right-angled

triangle! What's more, the

hypotenuse is the side of the triangle you need to find to find the

magnitude of the

vector. If we put the

Pythagorean Theorem in algebraic terms, it's -

h^{2} = a^{2} + b^{2}

Consult the diagram above and measure your a and b - a is 2, and b is 3. So -

h^{2} = a^{2} + b^{2}
h^{2} = 2^{2} + 3^{2}
h^{2} = 4 + 9
h^{2} = 13
h = √13
h = 3.6

The magnitude of your vector is 3.6 (to one

decimal place, do the calculation yourself for further accuracy).

The direction of the vector is the angle from the north line to the hypotenuse. We can calculate this with some simple trigonometry. Remember SOHCAHTOA? We need to find the angle that's inbetween the **adjacent** and the **hypotenuse**, so we want the CAH bit of this mnemonic. The relationship defined here is - (the angle is θ ):

cos θ = A
---
H
Therefore -
θ = cos^{-1(}^{2}/_{3.6}) = 56.25°

We're not quite done yet - the direction was the angle from the north line remember - we just have the

internal angle of the triangle. Simply taking the angle above away from 90 will give us our answer: and it's 33.75°.

And that's how you find the magnitude and direction of a displacement vector.

Remember the two types of vectors discussed in this write-up: the position vector used to designate a particular point, and the displacement vector used to apply to it and move it (or transform in correct mathematical language).