Vector Multiplication

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In mathematics the word vector denotes an entity that possesses both numerical magnitude and direction, subject to certain rules concerning how we express them in different coordinate frames. For calculation we usually represent a vector as the set of its components, its projections onto the axes of the coordinate grid in which we have put it. Think of the projection on one axis as a shadow that the vector casts upon the axis under light moving perpendicular to the axis. So when we multiply vectors together, we use their components in appropriate algebraic formulae to carry out the actual calculation.

To begin, consider the inner (or dot) product of two vectors with each other. We define the dot product as a scalar (a single number) produced by multiplying the magnitudes of the two vectors by each other and then by the cosine of the angle between the vectors; so for vectors A=(Ax, Ay, Az) and B=(Bx, By, Bz) we have

(Eq'n 1)

in which θ represents the angle between the vectors. In effect we project one vector onto the other and multiply the magnitude of the projection by the magnitude of that other vector.

To see how that works out with the vectors' components imagine that A points in some arbitrary direction. If B points entirely parallel to the x-axis, then it consists of a single component, Bx, and the product of that with A's projection onto it equals AxBx (the appropriate cosines are left implicit in the relationships between the components and the vectors themselves). If B points entirely parallel to the y-axis, we get the product as AyBy and we get AzBz if B points entirely parallel to the z-axis. If B manifests itself with all three components, then those partial products simply add and we get

(Eq'n 2)

To understand how we use that equation consider the work that I do when I mow the lawn with an old-fashioned reel-type lawnmower. To make the lawnmower move forward horizontally I must push the handle partly down and partly forward. Thus the vector describing the lawnmower's movement points entirely in the horizontal direction and the vector describing the force that I exert has both a horizontal component and a vertical component. The amount of work that I do equals the dot product of those vectors, which means that the vertical component of the force that I exert does no work because the lawnmower doesn't move in that direction. If we represent the distance that the lawnmower moves by the vector S=(Sx, 0, 0) and the force that I exert by the vector F=(Fx, Fy, 0), then the work that I do equals W=FA S=FxSx.

Next consider the outer (or cross) product of two vectors with each other. We define the cross product as a pseudovector produced by multiplying the magnitudes of the two vectors by each other and by the sine of the angle between the two vectors; so for vectors A and B we have

(Eq'n 3)

in which ep represents a unit vector that points in a direction perpendicular to both A and B. We can conceive that unit vector as representing the plane in which A and B lie and we define its direction through the right hand rule: if you curl the fingers of your right hand through the smaller angular displacement from A to B, then your extended thumb will point (more or less) in the direction of the unit vector of the cross product. That fact also necessitates that

(Eq'n 4)

Equation 3 tells us that in calculating the cross product we in effect project one of the vectors onto a plane to which the other vector stands perpendicular and then multiply that other vector's magnitude by the magnitude of the projection. Imagine that A points entirely in the positive x-direction and that B points entirely in the positive y-direction. In that case their cross product will point entirely in the positive z-direction and have the magnitude AB=AxBy. Next imagine rotating the coordinate frame clockwise about the z-axis through an angle θ<90 degrees. In the rotated frame we have A=[Ax=Acosθ, Ay=Asinθ] and B=[Bx=-Bsinθ, By=Bcosθ], in which the x-component of B has taken a negative value because it represents the projection of B onto the frame's negative x-axis. Rotating the frame can't change the value of our product, so in this case we must have

(Eq'n 5)

We can repeat that analysis with A extending entirely in the positive y-direction and B extending entirely in the positive z-direction and with rotating the frame about the x-axis. And then we can do the same with A extending entirely in the positive z-direction and B extending entirely in the positive x-direction and with rotating the frame about the y-axis. Finally, we can assume that A and B extend in random directions and assemble our results as the components of a general description of a cross product;

(Eq'n 6)

If that seems like a lot of trouble to endure, consider one of the uses to which we put that formula. If I stand on Earth's equator, my feet rest on the head of a vector that extends from Earth's center to the soles of my shoes, one of Earth's radius vectors, and my feet also rest on the base of a vector that points due east and represents the motion I have in that direction due to Earth's rotation. For a complete description of those vectors we must include a mathematical acknowledgment of the fact that they move, that seen from the south they rotate like the hands of a clock and make one full rotation once every twenty-four hours. It can come as an inconvenience, having to know what time it is in order to carry out a calculation. However, if I multiply the velocity vector by my mass and then form the cross product with the radius vector,

(Eq'n 7)

I get a pseudovector that points from Earth= s center through the North Pole. That pseudovector, my angular momentum relative to Earth's center, does not move at all. That fact makes some calculations in physics easier than others.

Next we combine Equations 2 and 6 to calculate the dot product of a vector C with the cross product of A on B; we get

(Eq'n 8)

Further multiplications of three or more vectors will get more complicated and direct manipulation of the vectors' components will give us some impressive algebraic messes. In addition to creating improved opportunities for error, those messes also obscure what the multiplication actually does. Fortunately, we now have enough information that we can abandon the brute force approach and use techniques that involve a little more finesse.

Consider the triple cross product, Ax(BxC). We know that our product vector must point in a direction perpendicular to BxC, so it must lie in the same plane occupied by B and C. That fact means that we can represent our product as a linear sum of certain multiples of B and C; that is,

(Eq'n 9)

in which f and g represent functions that we must now determine. We also know that our product vector must point in a direction perpendicular to A, so we also have

(Eq'n 10)

That equation necessitates that f=A•C and g=-A•B. Equation 10 necessitates that I put a minus sign on one of the functions, either f or g, so why did I choose g?

Actually, I didn't have a choice: the right-hand rule necessitates that g take the minus sign. Imagine that B points entirely in the positive x-direction and that C points entirely in the positive y-direction. Their cross product, BxC, points entirely in the positive z-direction. If A points entirely in the positive y-direction, its cross product with BxC points in the positive x-direction and is thus proportional to B. If A points entirely in the positive x-direction, its cross product with BxC point in the negative y-direction and is thus proportional to negative C. The essence of those facts doesn't change when we allow the vectors to be oriented at random, so we must have as true to mathematics

(Eq'n 11)

If we permute the order of the vectors, we also get

(Eq'n 12)

and

(Eq'n 13)

Adding together Equations 11, 12, and 13 then gives us

(Eq'n 14)

Next we consider the dot product of two cross products, (AxB)A (CxD). We recognize that multiplication as a variation on the product in Equation 8, so we make the necessary replacement and get

(Eq'n 15)

in which I used Equation 13 to make the substitution in the second line for the triple cross product.

Finally we have the cross product of two cross products, (AxB)x(CxD). We recognize that equation as variation on Equation 11, so we can write immediately

(AxB)x(CxD)=((AxB)A D)C-((AxB)A C)D.

(Eq'n 16)

If we want to multiply even more vectors together, then we can use the equations above to reduce them into sums of simple products and thereby make our calculations easier and less prone to error.

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