More on vectors and transformations

1. Scalar products and vector products.

We can take the scalar product of a force and a displacement for calculating work done.  Since moving perpendicular to a force involves no work, the only contributions to work done will be from each component of force multiplied by its corresponding component of distance.

If we represent the force as vector (F_x, F_y, F_z) and the distance as vector (x, y, z) we have


This is the 'dot product' of the vectors (F_x, F_y, F_z) and (x, y, z).

It is not hard to show that the 'dot product' is the product of the vector's magnitudes, multiplied by the cosine of the angle between them.  An alternative description is that it is the magnitude of the first vector, multiplied by the component of the second vector resolved in the direction of the first.

But how about calculating the moment of a force?

Suppose that the force (F_x, F_y, F_z) operates through the point (x, y, z) and we wish to know the moment about the origin.

We could try dropping normals and other complicated things, but it is easier to consider the result component by component.

A force (0, F_y, 0) applied at (x, 0, 0) is easy to imagine.

In the x-y plane, the force is "up the page" and acts to the right of the origin.  It therefore screws anticlockwise - a corkscrew would rise upwards.  That is the positive z-direction, and we have a convention that we consider that to be a positive component of the couple.

So we have a contribution  x F_y to the z component of the couple.

On the other hand, a force (F_x, 0, 0) applied at (0, y, 0) would have a clockwise effect - a negative couple about the z axis.

All other products are either zero - because the force component is through the origin - or in some other axis direction.

So if we reintroduce our unit vectors i, j and k, we have a component

and for the combined result of a force acting in three dimensions we will get


which we could write as the determinant of the matrix

i     j     k x     y     z Fx   Fy   Fz

We have an expression for the scalar product in terms of the angle between the vectors. The vector product is only a little more awkward. 

It is the product of the vector magnitudes, multiplied by the sine of the angle between them.  It is a vector, orthogonal to both of the input vectors.

If you think of the two vectors as defining two sides of a parallelogram in space, the vector product is equal to the area of the parallelogram and is a vector perpendicular to its surface.


So if we use subscript notation we have

b x c =  det

i      j     k b1   b2   b3 c1   c2   c3

Now let us introduce another vector a.  With all three vectors acting through the origin, we can regard them as edges of a parallelepiped - a sort of three-dimensional parallelogram.

We know the area of the b c face - it is the magnitude of b x c. 

To find the volume, we must multiply this by the the resolved component of a perpendicular to the face. 

But b x c is already a vector in that direction, so we just have to take the scalar product of a with b x c.

It is not hard to see that we can get this result by replacing i, j and k in the determinant above with the components of a, to get

volume  =  det

a1   a2   a3 b1   b2   b3 c1   c2   c3

(We would get the same numerical result if we transposed the matrix so that the vectors appeared as columns instead of rows)

Now we can start to make all sorts of deductions about the vectors if the determinant is zero.

One of the vectors might have magnitude zero.

Two of the vectors might be equal - or at least in the same direction

More generally, the three vectors might lie in a single plane.


Let us call the matrix made of the three column vectors  A. 

2.  Rotations

Let us suppose that a, b and c are three very healthy unit vectors, orthogonal to each other.  They could become an alternative set of coordinate axes.

That means that a.a = 1, b.b = 1 and c.c = 1

Also, since the vectors are orthogonal, the scalar product of any two different vectors is zero, e.g. a.b = 0.


Let us consider the product of A with its transpose:

A' A  =

a1   a2   a3 b1   b2   b3 c1   c2   c3

a1    b1   c1 a2    b2   c2 a3    b3   c3

Remember the 'scalar products' way to look at matrix multiplication.  We see that

A' A  =

a.a   a.b   a.c b.a   b.b   b.c c.a   c.b   c.c

But from what we know of these scalar products

A' A  =

1     0     0 0     1    0 0     0    1

So


or


It is extremely easy to invert!

So how did we get this wonderful matrix?  It has columns composed of three othogonal unit vectors - just what we need for an alternative frame of reference.  So a vector which has coordinates y in this frame of reference will have coordinates

in the original frame.  On the other hand a point with coordinates x in the original frame will in this new frame be

which is the same as

for this unitary matrix

This matrix A has a good chance of representing a rotation of the coordinate frame from the first set of orthogonal vectors i, j and k to this new set, through the same origin.  But we still have to take care.  Consider the matrix

A  =

-1     0     0 0     1    0 0     0    1

Its rows are orthogonal and of unit size, but this A will make a reflection of the x-axis.  A left-hand glove would become a right-hand glove, not a rotated glove.  So to avoid this, we must insist that if A is to be a rotation, its determinant must be 1.

3.  Translations

When looking at three-dimensional coordinates, we can rotate the frame of axes or we can move it with a displacement.  Althernatively we can leave the frame alone and move the point about - the mathematics will be the same.

When a point (x, y, z)' is moved a vector distance (1, 2, 3)' it arrives at
(x + 1, y + 2, z + 3)' - it's not really difficult!  To find the new vector we simply add the displacement to it.

The problem is that we now have to use two different processes to deal with the two types of movement, rotation and translation.  Can we find some way of glueing them together into a single operation?  If we can, we can start to deal with combinations of transformations, such as 'screwing' where the object is rotated at the same time as being moved along the rotation axis.

We have to appease the mathematicians!  Rotation is a transformation given by a simple multiplication of a vector by a matrix, but the ability to add a constant to the result requires an affine transformation.

However there is a way around the problem.

Suppose that instead of writing our vector as (x, y, z)' we write it as (x, y, z, 1)'.

What is the 1 for?  It gives something for a matrix to grab onto to add a translation d to the vector!  But now the vector has four components and the matrix is 4 x 4.

We can 'partition' a matrix to see its various parts in action, so if we write T x for the product of our point with a transformation matrix, now 4 x 4, we can break it down as follows.

	A	d		x	=	A x + d
	0  0  0	1		1		1

So at the expense of changing our matrices to 4 x 4, where the bottom row is always (0, 0, 0, 1), we can apply any combination of rotations and translations, just by multiplying the T matrices together.

This transformation is called the Denavit Hartenberg (or D-H) matrix.