# chap7 - Cross product 1 Chapter 7 Cross product We are...

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Cross product 1 Chapter 7 Cross product We are getting ready to study integration in several variables. Until now we have been doing only differential calculus. One outcome of this study will be our ability to compute volumes of interesting regions of R n . As preparation for this we shall learn in this chapter how to compute volumes of parallelepipeds in R 3 . In this material there is a close connection with the vector product, which we now discuss. A. Definition of the cross product We begin with a simple but interesting problem. Let x , y be given vectors in R 3 : x = ( x 1 , x 2 , x 3 ) and y = ( y 1 , y 2 , y 3 ). Assume that x and y are linearly independent; in other words, 0, x , y determine a unique plane. We then want to determine a nonzero vector z which is orthogonal to this plane. That is, we want to solve the equations x z = 0 and y z = 0. We are certain in advance that z will be uniquely determined up to a scalar factor. The equations in terms of the coordinates of z are x 1 z 1 + x 2 z 2 + x 3 z 3 = 0 , y 1 z 1 + y 2 z 2 + y 3 z 3 = 0 . It is no surprise that we have two equations but three “unknowns,” as we know z is not going to be unique. Since x and y are linearly independent, the matrix x 1 x 2 x 3 y 1 y 2 y 3 has row rank equal to 2, and thus also has column rank 2. Thus it has two linearly independent columns. To be definite, suppose that the first two columns are independent; in other words, x 1 y 2 - x 2 y 1 6 = 0 . (This is all a special case of the general discussion in Section 6B.) Then we can solve the following two equations for the “unknowns” z 1 and z 2 : x 1 z 1 + x 2 z 2 = - x 3 z 3 , y 1 z 1 + y 2 z 2 = - y 3 z 3 . The result is

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2 Chapter 7 z 1 = x 2 y 3 - x 3 y 2 x 1 y 2 - x 2 y 1 z 3 , z 2 = x 3 y 1 - x 1 y 3 x 1 y 2 - x 2 y 1 z 3 . Notice of course we have an undetermined scalar factor z 3 . Now we simply make the choice z 3 = x 1 y 2 - x 2 y 1 ( 6 = 0). Then the vector z can be written z 1 = x 2 y 3 - x 3 y 2 , z 2 = x 3 y 1 - x 1 y 3 , z 3 = x 1 y 2 - x 2 y 1 . This is precisely what we were trying to find, and we now simply make this a definition: DEFINITION. The cross product (or vector product ) of two vectors x , y in R 3 is the vector x × y = ( x 2 y 3 - x 3 y 2 , x 3 y 1 - x 1 y 3 , x 1 y 2 - x 2 y 1 ) . DISCUSSION. 1. Our development was based on the assumption that x and y are linearly independent. But the definition still holds in the case of linear dependence, and produces x × y = 0. Thus we can say immediately that x and y are linearly dependent ⇐⇒ x × y = 0 . 2. We also made the working assumption that x 1 y 2 - x 2 y 1 6 = 0. Either of the other two choices of independent columns produces the same sort of result. This is clearly seen in the nice symmetry of the definition. 3. The definition is actually quite easily memorized. Just realize that the first component of z = x × y is z 1 = x 2 y 3 - x 3 y 2 and then a cyclic permutation 1 2 3 1 of the indices automatically produces the other two components.
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