Vectors

In mathematics and physics vectors have both magnitude and direction. If we know a car was travelling at 65 mph when a collision occurred we regard that 65 mph as a scalar quantity. If we know the direction the car was travelling in when the collision occurred then we treat the combined information as a vector.

We manipulate scalar information all the time using plus, minus, multiply and divide operators. We can even add, subtract and multiply unlike scalar quantities if it suits our purpose. For example we could add a count of pears to a count of apples and give the answer as pieces of fruit. It would make little sense to multiply a quantity of apples by a quantity of pears but their are many unlike quantities we multiply together to get a result. For example we multiply voltage by current and then by time to get the energy we use in our homes. But can we and what does it mean to apply operators like multiply to vectors?

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 (into the screen) and coming toward you (out of the screen). The role of a commutator is to deliver such current flows.

The force, and therefore the rotating torque, has been found to be related to a multiplication of the magnetic field strength and the current flow. All are vectors as they have scale and direction. The 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. Being at right angles means they deliver the most effective force and thereby torque. I hope my inclusion of this example shows that vector multiplication has a practical purpose.

Clearly one way of showing both size and directions is to draw vectors on graph paper using a coordinate system. The length of an arrow represents the size or scalar part of the vector. The alignment of the arrow and its head represents 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. Our pushes on the same object point can be any direction in 3 dimensional space.

If I am pushing an object northward and you are pushing on the same object with an identical force but toward the south west we can guess that the pushed object will move, if it can do so, both to the north and west. More accurately by combining vectors geometrically as shown in the diagram we can get a combined force with both size and direction. Of course we must draw the vector lengths to some scale and with the right relative directions.

It may be more convenient and more accurate to work out the result algebraically. The top triangle from the above is illustrated here and we can use the equality of the cosine rule (see my blog on trigonometry) 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 same R force result. We have if you like replaced the M and Y forces with an east/west component and a north/south one that would deliver the same result. We call this process resolving. In this case we have resolved the M and Y forces into the convenient north/south and east/west directions but it should be realised that resolving can take place in any direction.

We can look at these resolved component vectors shown green 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.

Introducing the vector dot product

If in a 3d coordinate system we have a vector A with component lengths x1  yz1   and a vector B with component lengths x2  y2  z2   then a vector C = vector A – vector B will have component lengths of x1 –x2   y1 –y2  and z1 –z2. If A and B are perpendicular to one another then Pythagoras tells us||C||2 must equal ||A||2 +  ||B||2 or algebraically that (x1 –x2)+ (y1 –y2)+ (z1 –z2) must equal (x12 +y12 + z12 ) + (x22 +y22 + z22 ).
Simplified it means x1x2 + y1y2 + z1z2 = 0 when the associated vectors are perpendicular. We call this term the dot product of the two vectors A ⋅ B

For two vectors A and B with an angle Θ between them the Cosine rule tells us that ||C||2  = ||A||2  +  ||B||-2||A|| ||B||cosθ. Substituting all of the above paragraph values for ||C||2, ||A||2  and  ||B||2 into this formula we discover the more general formula in which the dot product x1x2  + y1y+ z1z2 = ||A|| ||B||cosθ (the scalar product of one vector length and the other vector length projected onto it).

Vectors and their cross product

Consider two vectors A and B, shown in red and yellow. Each has components along the x, y and z axes shown as Ax Ay Az and Bx By Bz. Those vector component can be regarded as scalars with their directions determined by i , j or k. So for example Ax i is vector A’s component in the i direction along the x axis.

Most will describe i , j and k as unit vectors but I think doing so adds to the confused thinking that sees them as numbers (see my blog on numbers and operators). I prefer to see them as related to and indicative of directions which are without number.

We saw above how a motor force only arose when there were components of magnetic field and current acting across one another. No force arose when the magnetic field and current aligned. So to get a resultant cross product vector A x B we must multiply each component of vector A by each component of vector B but ignore the products of aligned components (i.e we ignore any multiplications involving components with like i, j or k directions. Clearly if vector C is A x B then both C ⋅ A and C ⋅ B must be zero.

The very nature of a cross product is such that an i direction component multiplied by a j direction component will give us a k direction component. In fact i j = k, jk = i, ki = j, ji = -k, ik = -j and kj = -i as per the cyclic diagram shown. So we see that :-

A x B = (Ax i + Ay j + Az k) x (Bx i + By j + Bz k)
= Ax By ij + Ax Bz ik + Ay Bx ji + Ay Bz jk + Az Bx ki + Az By k j
= Ax By k – Ax Bz j – Ay Bx k + Ay Bzi + Az Bx j – Az By i
= i ( Ay Bz – Az By ) – j( Ax Bz – Az Bx) + k( Ax By – Ay Bx )

The above cross product can be shown as the matrix left. We will learn more about matrices in a later blog. On the right, we show a cross product vector geometrically. It is at right angles to the multiplied vectors and its magnitude is that of the yellow area AB sinθ.

In our motor example the angle between current (say A) and field (B) is a right angle so the yellow shape above would be a rectangle.The force delivering the torque ( A x B) is at right angles (orthogonal) to both current and field and the size of that force is related to the yellow area. Such a force will clearly reduce in value as the angle between current and field gets smaller.