Vectors are scalar commodities that act with direction. If we know a car was travelling at 65 mph when an accident occurred we have a scalar commodity (mph) and a value on that scale (65). If we also have knowledge of the direction the car was travelling then we have vector information.

We manipulate scalar data all the time using plus, minus, multiply and divide operators. But can we and what does it mean to do the same with scalar quantities, that may be of different commodities but whose product has meaning to us, which act in different directions.

Before we go any further, let us consider a practical example. The diagram shows the basic operation of an electrical motor. View the currents as going away from you and coming toward you. The role of a commutator is to arrange for such current flows. The force, and therefore the rotating torque, is related to a multiplication of the magnetic field strength and the current flow.

The field strength, current and force are all vectors. The resultant force vector is at right angles to the other two vectors and in this case, but not in all, they too are at right angles to one another. As such they deliver the greatest forces and hence the greatest rotary torque. I hope my inclusion of this example helps in establishing vector multiplication as useful.

Clearly one way of showing both size and directions is to draw the vectors on graph paper using a coordinate system. We use the length of an arrow to represents the size or scalar element of the vector and its alignment and arrow head to represent the direction of the vector.

If the vectors are about forces that you and I exert on an object, then when we push in the same direction the magnitudes of our pushes add and if we push in opposite directions they subtract. However our pushes on the same object point can be in all manner of directions. Each one can be in any compass direction and include an element of thrust upward or downward.

If I am pushing northward and you are pushing with an identical force toward the south west we can guess that the pushed object will move both to the north and west. Combining vectors geometrically gives us the combined force in both size and direction as long as we draw the vector lengths to some scale and with the right relative directions.

It may be more convenient to work out the result algebraically.The top triangle from the above is illustrated here and using the cosine rule from my blog on trigonometry we can use the equality shown to determine the size of the result.

Once we have R we can determine the part of it that acts to move the object west and the part that acts to move the object north. These are as shown in green. Like the M and Y forces that actually delivered R these two component vector forces would deliver the R result. We have if you like resolved the vector R into two components, an east/west one (x axis on a graph) and a north/south one (y axis on a graph). But we can look at these green component vectors in another way and consider them as scalar multiples of unit vectors. One is a scalar multiple of the unit west vector shown in red and the other a scalar multiple of the unit north vector in blue.

**Multiplication of Vectors**

In normal multiplication we multiply scalars together and get a scalar. We can do so because they are on the same scale. We have seen above how we can componetise vectors into north, south, east and west and we can likewise do so for up and down. All such component vectors have a scalar element and we can multiply those scalars together to get a scalar or **dot product, **as long as we confine ourselves to multiplying the scalar elements that lie on the same linear scale.

For example, consider two vectors A and B that have scalar components along the x, y and z axes as indicated by the suffixes that describe them and as converted to vectors via the unit vectors** i **,

**and**

*j***along those same axes. Be aware. in what follows, that we are doing scalar multiplication and that the scalars of the three unit vectors**

*k***,**

*i***and**

*j***are all equal to one.**

*k*When written out in full the dot product of the two vectors is as under

** A . B = ** (A

_{x}

**+ A**

*i*_{y}

**+ A**

*j*_{z}

**) . (B**

*k*_{x}

**+ B**

*i*_{y}

**+ B**

*j*_{z}

**) and to expand that product we have to multiply every term in the first bracket by that in the second bracket.**

*k*However, whilst we can multiply i scalars together, j scalars together and k scalars together because they are on the same scale all other scalar multiples are not possible because they are on scales orthogonal to one another (at right angles)

So the dot product is

**A**

*A . B =*_{x}B

_{x}+ A

_{y}B

_{y}+ A

_{z}B

_{z}

To multiply two vectors together and get a third vector we again set out to multiply all the bracketed components as above but changing the dot to a cross to signify that we are doing a **cross product**. We now multiply just the terms that are orthogonal. In our example above the motor force only arises when there are components of magnetic field an current acting a**cross **one another. so in the cross product we ignore **i** x **i**, **j** x **j** and **k** x **k**.

But there is a cyclic rule for multiplying the other unit vectors as shown. So for example **i x j = k** and **k x i = j** whereas in the opposite cycle **j x i = -k** and **i x k = -j**

** A x B = ** (A

_{x}

**+ A**

*i*_{y}

**+ A**

*j*_{z}

**) x (B**

*k*_{x}

**+ B**

*i*_{y}

**+ B**

*j*_{z}

**)**

*k*= A

_{x}B

_{y}

**ij**+ A

_{x}B

_{z}

**ik**+ A

_{y}B

_{x}

**ji +**A

_{y}B

_{z}

**jk**+ A

_{z}B

_{x}

**ki**+ A

_{z}B

_{y}

**k j**

= A

_{x}B

_{y}

**k**– A

_{x}B

_{z}

**j**– A

_{y}B

_{x}

**k +**A

_{y}B

_{z}

**i**+ A

_{z}B

_{x}

**j**– A

_{z}B

_{y}

**i**

=

**i**( A

_{y}B

_{z}– A

_{z}B

_{y}) –

**j**( A

_{x}B

_{z}– A

_{z}B

_{x}) +

**k**( A

_{x}B

_{y}– A

_{y}B

_{x})

The above result we express in the form of the matrix left. Right, a cross product vector is shown geometrically. It is at right angles to those multiplied and its magnitude is that of the yellow area and equal to AB sinθ.

In our motor example the angle between current and field is a right angle ( sinθ = 1) and the force that delivers the torque is at right angles (orthogonal) to both of them. The force is also related to the product of current and field. The yellow area is in this case a square whose sides are vectors with different scalar measures (field strength and current) and whose cross product is a vector that has another scalar measure (force).