Vector Arithmetic

Are these the arrows of outrageous fortune of which we have heard?...

In measuring position by distance and direction we used a line with an arrowhead on it. This representation of position is called a "vector". Just as there is an arithmetic of numbers, there is an arithmetic of vectors. We are going to be interested in motion of particles and have already said that motion is change in position over time. To get a change in a number it is customary to subtract the initial value from the final. To get a change in position we will use the same technique, subtracting the starting position from the ending. So how would you subtract two positions? Well it is most convenient to think of the positions as vectors to do this.

We have identified a line segment with direction and magnitude as a vector. If V is a particular vector, then |V| symbolizes its magnitude. Quantities that can be specified by a magnitude only are called "scalar" quantities to distinguish them from vectors. Since scalar quantities are just numbers, they may be added in the ordinary way. An addition operation is also defined for vectors. Adding a mixture of vectors and scalars is not defined.

To add vectors there are two techniques available, geometric addition and algebraic addition. Both yield the same result. The choice of which technique to use in adding vectors depends on the application and is a matter of convenience. First we will discuss geometric vector addition. Since a vector is defined by its magnitude and direction, changing its location in our reference frame without changing its direction or magnitude leaves it the same vector. We are free to relocate a vector anywhere in our space where we find it convenient. To add vectors geometrically we just place the tail of one at the head of the other. The sum then is a vector from the tail of the first vector to the head of the last. Run the Geometric Vector Addition display to see how this works.

The algebraic addition of vectors involves simply adding up the like components of the vectors. Imagine a vector with its tail at the origin. The "scalar components" of that vector are just the coordinates of the head of the vector. Remember that the coordinates were the distances along each axis which defined the position of the head of the vector. To add two vectors, add all the x components together and all the y components. The algebraic addition of vectors works because the sums of the components are the components of the sum. Run the Algebraic Vector Addition display to see examples of this.

To break a vector down into its components is called vector resolution. What that means is that we will resolve a vector into its "vector components". I have already defined the scalar components of a vector as the coordinates of the head of the vector when its tail is at the origin. The vector components are the vectors lying along the axes that add up to the vector we are interested in. Run the Vector Resolution display to see an example in 3 dimensions. Just for variety we will use a 3 dimensional vector here. In general we will use the fewest number of dimensions needed to convey the desired information but I wanted to illustrate that vectors in 3 dimensions differ from 2 dimensional vectors only in that they involve an extra coordinate.

Since a vector along an axis has its direction fixed by definition, only its magnitude changes. That means we can uniquely define each component of a vector with just a number, knowing whether we are talking about the x, y or z component.

Remember that we could also specify the position of a particle with the scalars that we called coordinates. There is an obvious relationship between the coordinates of a position and the components of the vector pointing to that position. They are numerically equal.

Recognize that in vectors as in numbers, subtraction is just addition with the sign of the entity to be subtracted reversed. That of course brings up the question of what it means to have the sign of a vector reversed. To reverse the sign of a number we just multiply it by -1. To apply the same trick to a vector we must agree on what it means to multiply a vector by a scalar, in this case -1.

The product of a scalar times a vector is a vector whose components are the components of the original vector, each multiplied by the scalar. Except for some special cases which I will cover soon, we will use bold letters to symbolize vectors. Run the Scalar Vector Multiplication display to see how this multiplication works. Verify that when the scalar multiplier is -1, the product vector has the same magnitude as the original and points the other way.

Now that we know how to multiply vectors by scalars we can modify our idea of vector components slightly. Consider a vector of length 1, in the direction of the x-axis. This is called the x unit vector. There are also similar y and z unit vectors. Following tradition in these matters we will label the x unit vector ihat

(called i hat), the y unit vector jhat

(j hat) and the z unit vector khat

(k hat). Now if we have a vector V with its head at coordinates (a,b,c) it can be expressed as the sum of three vectors a* ihat

+b*

+c*

The vector a* ihat is the product of the unit vector ihat and the scalar component a. The vectors b* jhat and c* khat are also the products of unit vectors and their corresponding scalar components. The three vectors, a* ihat , b* jhat and c* khat are the vector components of V. The quantities a, b and c are the scalar components of V as well as the coordinates of head of the V vector. The unit vectors are the exception to the italicized rule. Run the Vector Components display to see the ideas of this paragraph illustrated.

We got into this discussion of multiplying a vector times a scalar by trying to find a way to subtract vectors. Another way to look at this subtraction business is to define the negative of a vector as that vector, which when added to the original yields a null result. That means the sum is a vector of zero length. Pretty clearly that is the case if the negative vector has the same length as the original and points the other way.

One last view of vector subtraction comes from our discussion of geometric vector addition. Remember that the sum of two vectors is a vector from the tail of the first to the head of the second. Now consider a vector P which is the difference between two vectors, P₂ and P₁. This means P=P₂-P₁, or by rearranging the terms, P₂=P₁+P. So P₂ must run from the tail of P₁ to the head of P. That makes P a vector from the head of P₁ to the head of P₂. To help visualize why this is so, Run the Geometric Vector Subtraction display. We will work in two dimensions since any three vectors which form a closed figure must be "co-planar" meaning all in one plane. (Think about it.)

Having added, subtracted, multiplied by a scalar and resolved vectors, we may as well finish the story with the multiplication of one vector by another. We will look at two ways to multiply vectors, both of which have physical significance. First we will consider the scalar product of two vectors. It is called the scalar product because the result of the multiplication is a scalar quantity. This is distinctly different from the scalar multiplication that we already covered. Sorry about the use of the word scalar in both cases. In fact that may be why an alternate name for this kind of multiplication has come into use. It is also called the "dot product" after the dot symbol that distinguishes this kind of multiplication. The dot product of two vectors V₁ and V₂ is written V₁ · V₂ where the "·" between the vectors is the scalar product operator.

To get the dot product of two vectors, multiply the two lengths together and then multiply by the cosine of the angle between them. This may be thought of as the product of the length of one vector times the component of the other vector in the direction of the first. Run the Dot Product display to see some examples.

An alternative way to calculate the dot product is to write each vector in terms of its components and multiply them as you multiply any polynomials in algebra. For example suppose we had the vectors V₁=4* ihat

+3*

and V₂=4* ihat

+0*

. The dot product would be:
(4 * ihat

+ 3 *

) (4 *

+ 0 *

) = 4 * 4 * ihat

+ 4 * 0 *

+ 3 * 4 *

+ 3 * 0 *

.
Remember that ihat

and

are vectors perpendicular to each other, of length 1.0. So the dot product of any different unit vectors is zero and the dot product of a unit vector with itself is 1. Replacing the unit vector dot products with their values gives us:
V₁ · V₂ = 4 * 4 * 1 + 4 * 0 * 0 + 3 * 4 * 0 + 3 * 0 * 1 = 16.

Go back to the Dot Product display to verify that these two ways of calculating a dot product are equivalent.

Any terms in our polynomial product which involve dissimilar unit vectors will be zero because the unit vectors are perpendicular and the dot product includes as a factor the cosine of the angle between the vectors. The dot product then boils down to just the sum of the products of the scalar components. In our example above that would be 4 * 4 + 3 * 0. This gives us an easy way to get the dot product of vectors in three dimensions where the cosine of the angle between the vectors is not that easy to get a grip on. If V₁ = a * ihat

+ b *

+ c *

and V₂ = d * ihat

+ e *

+ f *

. The dot product V₁ · V₂ = a * d + b * e + c * f. Now if we want to know the angle between any two vectors in 3D space, we can divide the dot product by the product of the lengths and find the angle that has that quotient as its cosine. We will use |V| to symbolize the length of the vector V. So (V₁ · V₂)/|V₁| * |V₂| is the cosine of the angle between the vectors. Run the Dot Product 3D display to see how this works.

Arrrgh! I promised you two ways to multiply vectors together. I know this seems excessive, two different ways to multiply objects together, but we are going to need both ways when we come to the ideas of work and torque.

The second process yields a vector so it is called the vector product. It is also known as the "cross product" due to the operation being symbolized by an X, as in V = V₁ X V₂ where V is the cross product of the vectors V₁ and V₂. V₁ X V₂ is defined as a vector whose length is the product of the lengths of V₁ and V₂ times the sine of the angle between them, and whose direction is perpendicular to both V₁ and V₂. The sense of the cross product is determined by a right hand rule. The right hand rule in this case works like this. Place the fingers of your right hand in the direction of V₁ and curl them toward V₂. The cross product will point to the side where your thumb is located. Run the Cross Product display for an illustration. Notice that V₁ X V₂ = -(V₂ X V₁).

As we did for the dot product, we can calculate the cross product without knowing explicitly the angle between the vectors. First look at the cross products of the unit vectors like this: ihat

= 0 and

devilish hard To help you remember what equals what in the above relationships, notice that the order of the unit vectors remains the same in each equation. khat follows jhat which follows ihat if you loop around to the beginning from the end of each equation.

Then suppose V₁ = a * ihat + b * jhat + c * khat and V₂ = d * ihat + e * jhat + f * khat .

Next multiply V₁ X V₂ by multiplying each term in V₁ by each term in V₂.

V₁ X V₂ =
a * d * ihat

+ a * e *

+ a * f *

+
b * d * jhat

+ b * e *

+ b * f *

+
c * d * khat

+ c * e *

+ c * f *

Next replace all the cross products with their equivalent value.

V₁ X V₂ =
a * d * 0 + a * e * khat

+ a * f * (- jhat

) +
b * d * (- khat

) + b * e * 0 + b * f * ihat

+
c * d * jhat

+ c * e * (- ihat

) + c * f * 0

Now collect the coefficients of the unit vectors to find V₁ X V₂ = (b * f - c * e) * ihat

+ (c * d - a * f) * jhat

+ (a * e - b * d) * khat

For those of you who remember determinants from your high school algebra days, you might recognize the formula we just developed for the cross product as the expansion of a determinant that looks like this:

V₁ X V₂ = cross product by determinant

This is as far as we will go for now with vector arithmetic. The next section of the course puts much of this vector arithmetic into action.

Are there any questions?

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