We immediately see two differences between the dot product operation that we added to our vector space and the cross product operation that we now include: The first is that the result of a cross product is a vector and not a scalar. The second is that the vector operation does not apply to vectors in a plane; it is an operation on vectors in 3-space.
The dot product gave us a computation for "checking" on orthogonality. If we are given two vectors in space, say (2,3,4) and (-1,3,6) we can produce a vector say (x,y,z) that is perpendicular to both of them; but we can not do it without getting out of the plane containing the two given vectors. We will need to have 2x + 3y + 4z = 0 at the same time that we have -x + 3y + 6z = 0 and this can be done in many ways.
For example, x = 6 y = -16 and z = 9 will work as will x = 12 y = -32 and z = 18 . I will be "defining" an operation for you that will "explain" how I got the first set of numbers. I think that you can "see" how I got the second set.
To summarize, we have a "new" vector, namely, ( 6, -16, 9 ) that is orthogonal to the given vectors (2,3,4) and (-1,3,6) . I hope that it is clear to you when I say, " These three vectors would not all be in the same plane, given that they are mutually ( that means, taken in pairs ) perpendicular ." This idea will explain for us why the cross product does not apply to vectors in a plane.
Then are we saying that the cross product of two vectors is to be a vector perpendicular to both of them? Since we are saying this, we are left with the possibility of many answers. But, suppose we further require the "sense" of the cross product vector satisfy a special property and the magnitude or norm also satisfy a special property, then might we reduce the cross product result to a single, a unique, vector? The answers to this question will be "Yes".
The formal work on "Cross Product" is well done in our text book and as you begin to "look" at it notice the use of i,j,and k in the definition ( they give us the vector answer rather than scalar ) and notice how important the "diagonal pattern"or "digital pattern" , as your author calls it, is to the operation of Cross Product; also, we see that the "diagonal pattern" might well have the name "Cross pattern" ---thus the name "Cross Product".
See 11.4 Definition of Cross Product
See determinant form with cofactor expansion( the naming of this is of no consequence to us at this time)
Your authors have included Theorems for properties of Cross Product( Thm 11.7) and for geometric properties of Cross Product( Thm. 11.8 ) I am including the proof for only one of the items in these two theorems and it may not be completely understandable for you without collaboration. I include the proof to emphasize the property and the other proofs are omitted because they would clutter ( their proofs are not difficult ). The property that I am emphasizing( #2Thm.11.8) is the one that represents a decision on length for the Cross Product vector. This property can then a used to prove property #4 of Thm.11.8 which addresses the issue of length very explicitly, namely, that the norm of the Cross Product of two vectorsis the positive real number(scalar because of the definition of norm of a vector) that is the measure of the area of a parallelegram formed by using the two vectors as adjacent sides.
Now we should practice some cross product computations and introduce the "triple scalar product" which is nothing more than a computation involving our two new operations on vector space, namely, dot and cross product.
Finally, "triple scalar product" has an intesting propert that makes it useful for us in determining if three vectors are coplanar or not. We'll be especially interested in this when studing Linear Algebra later in the course.
See 11.4 Triple Scalar Product
If the volume of the box is not 0, then the vectors for the adjacent sides were not coplanar.
Look at the following solved problems
in the Exercises at the end of
11.4 : 7,13,15,23,25,35,37 Then
do the following unsolved
problems as homework: 8,24,26,36,38
.