Three Dimensional Vectors and Curves in Space

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Our objective on this page is to show that for any curve in with non-vanishing velocity, and acceleration that is not a multiple of velocity, there is a "moving 3-frame" called a Frenet Frame, that is analogous to the moving 2-frames we studied in the section titled Plane Curvature. In order to do that, we will introduce first the idea of a determinant, and then the construction of the "cross product" of a pair of vectors. We will use this fact on the next page to establish Kepler's second law: that planets moving under Newtonian gravity move in a plane that contains the Sun, and that they sweep out equal areas in equal times.

When we come to study the algebra of vectors in three-dimensional space we discover few surprises. We learn that we can associate three coordinates to points in space, once we establish a system of three unit vectors that do not all lie in a single plane. Thus we represent vectors in as triples of numbers and these vectors represent parallel translations of all of space, just as vectors did in the plane. We define addition and scalar multiplication of vectors in the same way as we did earlier, and with reference to a system of axes, we calculate these component-wise.

Inner Product in

New features emerge, however, when we examine the inner product and Euclidean structure of . We define an inner product in exactly as we did before. It is a rule that associates numbers to pairs of vectors in space and satisfies:

We will see below that the Gram-Schmidt orthonormalization procedure extends to three dimensions and guarantees the existence of a triple of unit vectors that satisfy the properties:

Orthonormal Basis

Such a system is called an orthonormal basis. We will for convenience write these vectors . Once we have such a basis, we can represent any vector in coordinates in terms of it

and the inner product has the simple form:

We can also represent any other orthonormal basis with a matrix in the following way. The represents

. Orientation in space

When we ask which orientation-preserving transformations preserve this Euclidean structure, we discover that they are rotations, and may be represented by matrices and the columns satisfy conditions (5) and (6) above, with one other condition that expresses the fact that they do not reverse orientation. That condition is an important topic in this part.

The task ahead is to define the determinant of a linear transformation from to itself. When we defined determinants of linear transformations of the plane, we associated those transformations with matrices by establishing a set of unit vectors and associating the transformation with the matrix. Now a general property of determinants will be this. Whenever one sets up a new system of unit vectors, and so associates a new matrix with the original linear transformation the determinant of the new matrix representative of the linear transformation will be the same number as the one computed from the original matrix representative. Since we usually calculate determinants from the matrix representatives of the linear transformation, this is a good fact to know. It will not be difficult to prove.

When the system of unit vectors is an orthonormal system with respect to the inner product structure in the plane, we see that the absolute value of the determinant of a matrix is the area of the parallelogram spanned by its columns. In this case, the determinant of the transformation is a geometric property that we interpreted as a ratio of signed areas. We want to make a similar interpretation in three dimensions.

We will associate a linear transformation from with a matrix that represents the transformation in terms of an orthonormal set of unit vectors . In that case, we can speak either of the determinant of the transformation, or of the determinant of the matrix. Geometrically, the absolute value of this determinant will be the volume of the parallelopiped spanned by the three columns. It is a geometric invariant associated with the triple of vectors.

. Alternating 3-linear map

In order to define the determinant, we return to the definition of an alternating multilinear map.

Definition 1: Let be a function that associates a number with each ordered triple of vectors in . Say the function is alternating if interchanging any two of changes the sign of the value of . The function is also multilinear if

and

Similar relations hold for the other 2 indices since is alternating. Such a function will be called alternating and 3-linear.

End of Definition

Question 1: Suppose a set of unit vectors is fixed in advance. Show that any alternating 3-linear function is completely determined by the number . Therefore show that if for some alternating 3-linear function , then if is any other alternating 3-linear function, with , then

End of Question

Now there is an alternating 3-linear function such that . Simply extend this definition in the only way possible ( ) recalling that if two variables are the same, as in the last example, must vanish. In this way we see that the alternating 3-linear functions are in one to one correspondence with the real number line.

. Determinant of a linear transformation of space

Suppose now that is a linear transformation. That is, if X, Y are elements of and then

Definition 2: If is an arbitrary alternating 3-linear mapping then define to be the alternating 3-linear mapping

The mapping from that carries is linear. So for some

This is the number which is the determinant of linear transformation T.

End of Definition

An immediate consequence of this definition is that if S and T are linear transformations then

It was a little more difficult to see this for linear transformations of the plane because we took a more concrete approach there.

. Determinant of a 3x3 matrix

Now let us calculate the determinant. Let the set of unit vectors be fixed in advance and represent a linear transformation with matrix so that the j column also represents the vector

Suppose that . Then the discussion above tells us that the determinant of A is simply

Expanding by alternating multilinearity we see that this is

All the other terms are zero.

Let us show that the absolute value of the determinant of A is the volume of the parallelopiped spanned by the columns . We assume that they do not all lie in the same plane, hence are non-zero, and we extend the Gram-Schmidt procedure.

. Gram-Schmidt orthonormalization in 3 dimensions

Suppose then that where and . We assume that

Now consider the vector

It is easy to see that .

We see that the parallelopiped spanned by has the same volume as the rectangular parallelopiped (box) spanned by mutually perpendicular where

Now suppose that we have an alternating 3-linear map such that

Then . But

Now

Determinant as signed volume of a parallelopiped

And it follows from our characterization of the determinant of a linear transformation that

and

therefore

Suppose we fix an orthonormal basis of vectors . We represent vectors in as triples of numbers where as before . Recall that the inner product has the simple form:

We can also represent any other orthonormal basis with a matrix in the following way. The represents

.

The orientation-preserving linear transformations that preserve this Euclidean structure are rotations, and may be represented by matrices and if we call these columns , they satisfy the conditions:

In the plane, we were able to represent the rotations as complex numbers of length 1. In we see that the rotations have a more elaborate description.

Cross product of vectors

Now we will define the "cross product" of a pair of vectors. This will be a new vector that is perpendicular to the original two. Suppose that

is a pair of vectors. Then it is not difficult to see that the function by the rule

=

is a linear function from .

There is therefore a unique vector such that for all

Question 2: If the last statement is not obvious, prove it, using the fact that satisfies the conditions: If X, Y are elements of and then

1.     

End of Question

It is easy to see in this case that

This vector is defined to be the cross product:

The defining equation is therefore:

It follows that

From these, all others can be calculated.

A simple calculation shows that for

We use orthogonal projection.

Suppose that where and . Assume that , i.e. that is not a multiple of . Then

Since are mutually perpendicular,

Now if then but we can extend to a basis for . So . It follows that

and so

If either , the result is still true.

Question 3: Show that

End of Question

Question 4: Prove, using properties of determinants the following facts:

1. 
2. 
3. (So the cross product is perpendicular to )
4. The cross product is not associative.

Find an example that shows that

End of Question

Frenet Frame

We are now ready to define the moving frame associated with a space curve that is called the Frenet Frame. In Chapter 3, Curves in Art and Nature, Section 1, Plane Curvature, Part 3, Arc Length and the Fundamental Theorem of Calculus, we defined the arc length parameterization of a plane curve with non-vanishing velocity. We could define the arc length parameterization of a curve in by extension, but will use a simpler method to define the frame.

If we have a space curve where are smooth (infinitely differentiable) functions of their parameter t, then a different parameterization of , say

would yield the same points along the curve.

As in the plane, each smooth curve defines another curve and

We call the velocity curve. As in the plane curve case, we will assume that

That is, the velocity never vanishes. Represent the velocity curve

and .

While we could reproduce the arc-length parameterization as we did in the plane curve case, let us instead define the acceleration curve and use orthogonal projection to write

So is the "normal component" of the acceleration with respect to . Inspection shows that

and so

Now it is easy to see that on the other hand,

Therefore if (that is, and generate a plane) a unit vector perpendicular to the curve is

We may therefore construct an orthonormal frame at if and do not lie in the same line. That frame is

Also, it is not difficult to see that

and

so the third unit vector in the frame is perpendicular to the plane spanned by and . It is also obvious that is perpendicular to the plane spanned by and for each t. That fact will be important in the next part on Kepler's Second Law because we will show that for a central force like Newtonian gravity, is constant, therefore the planetary orbit lies in a plane.

Exploration: Frenet Frames of Space Curves

In this exploration, you may experiment with Frenet Frames. First define your curve in 2 steps. Specify the domain for the t parameter by entering the endpoints in the fields:

Next, define the component functions for the curve in the fields:

When you press you will see the space curve:

Now to see the Frenet Frames move along the curve, press . The speed field will determine how quickly they move. Choose values between 1 and 100.

The red, blue and white arrows are the unit vectors respectively.

You may select a point by using the slider or, if you prefer to set the slider value by typing the parameter, just type the number in the field and press Enter:

In the next section we will prove Kepler's Second Law.