Cross product 1 Chapter 7 Cross product We are getting ready to study integration in several variables. Until now we have been doing only di®erential calculus. One outcome of this study will be our ability to compute volumes of interesting regions of Rn. As preparation for this we shall learn in this chapter how to compute volumes of parallelepipeds in R3. In this material there is a close connection with the vector product, which we now discuss. A. De¯nition of the cross product We begin with a simple but interesting problem. Let x, y be given vectors in R3: x = (x1; x2; x3) and y = (y1; y2; y3). 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 ² z = 0 and y ² 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 x1z1 + x2z2 + x3z3 = 0; y1z1 + y2z2 + y3z3 = 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 µ ¶ x1 x2 x3 y1 y2 y3 has row rank equal to 2, and thus also has column rank 2. Thus it has two linearly independent columns. To be de¯nite, suppose that the ¯rst two columns are independent; in other words, x1y2 ¡ x2y1 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" z1 and z2: x1z1 + x2z2 = ¡x3z3; y1z1 + y2z2 = ¡y3z3: The result is 2 Chapter 7 x2y3 ¡ x3y2 z1 = z3; x1y2 ¡ x2y1 x3y1 ¡ x1y3 z2 = z3: x1y2 ¡ x2y1 Notice of course we have an undetermined scalar factor z3. Now we simply make the choice z3 = x1y2 ¡ x2y1(6= 0). Then the vector z can be written 8 < z1 = x2y3 ¡ x3y2; z = x y ¡ x y ; : 2 3 1 1 3 z3 = x1y2 ¡ x2y1: This is precisely what we were trying to ¯nd, and we now simply make this a de¯nition: DEFINITION. The cross product (or vector product) of two vectors x, y in R3 is the vector x £ y = (x2y3 ¡ x3y2; x3y1 ¡ x1y3; x1y2 ¡ x2y1): DISCUSSION. 1. Our development was based on the assumption that x and y are linearly independent. But the de¯nition 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 x1y2 ¡ x2y1 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 de¯nition. 3. The de¯nition is actually quite easily memorized. Just realize that the ¯rst component of z = x £ y is z1 = x2y3 ¡ x3y2 and then a cyclic permutation 1 ! 2 ! 3 ! 1 of the indices automatically produces the other two components. 4. A convenient mnemonic for the de¯nition is given by the formal determinant expression, 0 1 ^{ |^ k^ @ A x £ y = det x1 x2 x3 : y1 y2 y3 Cross product 3 In this expression the entries in the ¯rst row are the standard unit coordinate vectors, and the \determinant" is to be calculated by expansion by the minors along the ¯rst row. 5. We are left with one true ambiguity in the de¯nition, and that is which sign to take. In our development we chose z3 = x1y2 ¡ x2y1, but we could have of course chosen z3 = x2y1 ¡ x1y2. In this case, the entire mathematical community agrees with the choice we have made. 6. Nice special cases: ^{ £ |^ = k;^ |^£ k^ = ^{; k^ £ ^{ =|: ^ 7. By the way, two vectors in R3 have a dot product (a scalar) and a cross product (a vector). The words \dot" and \cross" are somehow weaker than \scalar" and \vector," but they have stuck. ALGEBRAIC PROPERTIES. The cross product is linear in each factor, so we have for example for vectors x, y, u, v, (ax + by) £ (cu + dv) = acx £ u + adx £ v + bcy £ u + bdy £ v: It is anticommutative: y £ x = ¡x £ y: It is not associative: for instance, ^{ £ (^{ £ |^) = ^{ £ k^ = ¡|^; (^{ £ ^{) £ |^ = 0 £ ^j = 0: PROBLEM 7{1. Let x 2 R3 be thought of as ¯xed. Then x £ y is a linear function from R3 to R3 and thus can be represented in a unique way as a matrix times the column vector y. Show that in fact 0 1 0 ¡x3 x2 x £ y = @ x3 0 ¡x1A y: ¡x2 x1 0 4 Chapter 7 PROBLEM 7{2. Assuming x 6= 0 in the preceding problem, ¯nd the characteristic polynomial of the 3 £ 3 matrix given there. What are its eigenvalues? B. The norm of the cross product The approach I want to take here goes back to the Schwarz inequality on p. 1{15, for which we are now going to give an entirely di®erent proof. Suppose then that x, y 2 Rn. We are going to prove jx ² yj · kxk kyk by calculating the di®erence of the squares of the two sides, as follows: à ! Xn Xn Xn 2 2 2 2 2 2 kxk kyk ¡ (x ² y) = xi yi ¡ xiyi i=1 i=1 i=1 Xn Xn Xn Xn 2 2 = xi yj ¡ xiyi xjyj i=1 j=1 i=1 j=1 Xn Xn 2 2 = xi yj ¡ xiyixjyj i;j=1 i;j=1 X Xn X 2 2 2 2 2 2 = xi yj + xi yi + xi yj i<j i=1 i>j X Xn X 2 2 ¡ xiyixjyj ¡ xi yi ¡ xiyixjyj i<j i=1 i>j X X X 2 2 2 2 = xi yj + xj yi ¡ 2 xiyixjyj i<j j>i i<j X ¡ 2 2 2 2¢ = xi yj ¡ 2xiyixjyj + xj yi i<j X 2 = (xiyj ¡ xjyi) : i<j This then proves that kxk2 kyk2 ¡ (x ² y)2 ¸ 0. If we specialize to R3, we have 2 2 2 2 2 2 kxk kyk ¡ (x ² y) = (x1y2 ¡ x2y1) + (x2y3 ¡ x3y2) + (x1y3 ¡ x3y1) : 2 2 2 But the right side is precisely z3 + z1 + z2 in the above notation. Thus we have proved the Cross product 5 wonderful Lagrange's identity kxk2 kyk2 = (x ² y)2 + kx £ yk2: This identity relates norms, dot products, and cross products. In terms of the angle θ between x and y, we have from p. 1{17 the formula x ² y = kxk kyk cos θ. Thus, kx £ yk = kxk kyk sin θ: This result completes the geometric description of the cross product, up to sign. The vector x £ y is orthogonal to the plane determined by 0, x and y, and its norm is given by the formula we have just derived. An even more geometric way of saying this is that the norm of x £ y is the area of the parallelogram with vertices 0, x, y, x + y: x y y x+y θ x 6 Chapter 7 PROBLEM 7{3. The Lagrange identity can easily be generalized. Using the following outline, prove that for x, y, u, v 2 R3, (x £ y) ² (u £ v) = (x ² u)(y ² v) ¡ (x ² v)(y ² u): Outline: the right side equals à !à ! à !à ! X X X X xiui yjvj ¡ xivi yjuj i j i j X = xiyj(uivj ¡ ujvi) (why?) i;j X = xjyi(ujvi ¡ uivj) (why?) i;j 1 X = (x y ¡ x y )(u v ¡ u v ) (why?) 2 i j j i i j j i i;j X = (xiyj ¡ xjyi)(uivj ¡ ujvi) (why?) i<j = (x £ y) ² (u £ v): C. Triple products The formal representation of x £ y as a determinant enables us to calculate inner products readily: for any u 2 R3 0 1 ^{ |^ k^ ³ ´ @ A ^ (x £ y) ² u = det x1 x2 x3 ² u1^{ + u2|^+ u3k y y y 0 1 2 3 1 u1 u2 u3 = det @x1 x2 x3A : y1 y2 y3 (Notice how clearly this formula displays the fact that (x £ y) ² x = (x £ y) ² y = 0.) This number (x £ y) ² u is called the scalar triple product of the vectors x, y, and u. The formula we have just given makes it clear that Cross product 7 (x £ y) ² u = (u £ x) ² y = (y £ u) ² x = ¡(y £ x) ² u = ¡(x £ u) ² y = ¡(u £ y) ² x: Thus we have the interesting phenomenon that writing x, y, u in order gives (x £ y) ² u = x ² (y £ u): So the triple product doesn't care which you call £ and which you call ². For this reason, it's frequently written [x; y; u] = (x £ y) ² u: PROBLEM 7{4. The vector triple product is (x £ y) £ u. It can be related to dot products by the identity (x £ y) £ u = (x ² u)y ¡ (y ² u)x: Prove this by using Problem 7{3 to calculate the dot product of each side of the proposed formula with an arbitrary v 2 R3. PROBLEM 7{5. Prove quickly that the other vector triple product satis¯es x £ (y £ u) = (x ² u)y ¡ (x ² y)u: The identities in Problems 7{4 and 7{5 can be remembered by expressing the right side of each as vector triple product = (outer²far) near ¡ (outer²near) far: The scalar triple product is exactly the device that enables us to compute volumes of parallelepipeds in R3.