Notice that v v w (4)( 15) ( 3)( 20) (0)(2) ( 2)( 15) (2)( 20) (5)(2)

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1 The Cross Product When discussing the dot product, we showed how two vectors can be combined to get a number. Here, we shall see another way of combining vectors, this time resulting in a vector. This operation is called the cross product. The cross product of two vectors v and w is a third vector which is perpendicular to both v and w. Unlike the dot product (which is defined for any dimension of the vector), the cross product is only defined for three-dimensional vectors. The cross product is more easily remembered in terms of a linear algebra result, as the determinant of a 3 μ 3 matrix. Cross Product Let v v1iv2j v3k and w w1iw2j w3k. The cross product of v and w, denoted by v μ w, is a vector given by i j k vw v v v ( v w v w ) i ( v w vw ) j( vw v w ) k w w w Example 1: Compute v μ w if v 4i3j0k and w 2i2j 5k. Verify that v μ w is perpendicular to both v and w. Solution: v μ w = [( 3)(5) (0)(2)]i + [(0)( 2) (4)(5)]j + [(4)(2) ( 3)( 2)]k = 15i 20j + 2k. Notice that vvw (4)( 15) ( 3)( 20) (0)(2) (2)( 15) (2)( 20) (5)(2) w v w and And so, we see that both v and w are perpendicular to v μ w. Similar to the case of the dot product, there is a nice geometrical interpretation of the cross product, this time in terms of areas of parallelograms. To that end, we make the following claim. 1

2 Cross Product (Geometrical Interpretation) Suppose q is the angle between the vectors v and w (so 0 q p). Then we have that Area of parallelogram vw v w sin determined by v and w To see why the first equality is true, we work through a bit of algebra ( vw 2 3 vw 3 2) ( vw 3 1 vw 1 3) ( vw 1 2 vw 2 1) vw vw 2vvwwvw vw 2vvwwvw vw 2vvwwvw ( v v v )( w w w ) ( vw v w v w ) v w vw 2 2 v w v w v v w w 1cos sin cos 2 Taking the square root of both sides, we have that used the fact that since 0 q p, sin(q) 0. vw v w sin, where we To see why the second equality is true, we recall that the area of a parallelogram is given by A = bh, where b is the length of the base and h is the height. w wsinq q v Figure 1: A Parallelogram In Figure 1, notice that the base, b, is given by v and the height, h, is given by w sin(q). Thus, the area of the parallelogram is given by v w sin. 2

3 Since we have that vw v w sin, we have another way to determine if two vectors are parallel. (Recall, the first way was to determine if the entries were all scaled by the same amount.) That is, two vectors are parallel if and only if q = 0 or q = p. In both instances, we have that sin(q) = 0, so we have that v μ w = 0. Said another way, we have that v μ w = 0. Determining if Two Vectors are Parallel or Perpendicular Suppose q is the angle between the (non-zero) vectors v and w. Then we have that and v and w are parallel if and only if v μ w = 0 v and w are perpendicular if and only if vw = 0 Since the area of the triangle formed by the vectors v and w is precisely half of the area of the parallelogram, we have the following formula. Area of the Triangle Formed by v and w Suppose q is the angle between the vectors v and w (so 0 q p). Then we have that The area of triangle 1 1 v w sin vw formed by v and w 2 2 Example 2: Find the area of the triangle formed by the vectors v = i + 0j + 0k and w = 0i + j + 0k. Solution: This forms a triangle with base 1 and height 1 in the first quadrant, so its area is 1/2. We can verify that with the formulas, though. v μ w = [(0)(0) (0)(1)]i + [(0)(0) (1)(0)]j + [(1)(1) (0)(0)]k = k. From that, we see that v μ w = k = 1, and thus the area of the triangle formed by the two vectors is indeed equal to 1/2. 3

4 When we introduced the dot product, we showed the connection between that and the equation of the plane. In particular, the normal vector, n, was perpendicular to the plane. Thus, if we can find two vectors that are contained in the plane, then we can use the cross product to find the normal vector. To see how this can work, consider the following example. Example 3: Find the equation of the plane that contains the points P = (1, 2, 3), Q = (1, 0, 1) and R = (-1, 3, 4). Solution: Notice that unlike the examples we considered when we first introduced planes, we cannot (easily) determine the slope in the x- and y-directions. Instead, we shall use vectors. The vectors v PQ 0i2j2k and w PR 2ijk are in the plane. Thus, their cross product, v μ w, will be perpendicular to the plane. This is normal vector, n, to the plane. That is, n = v μ w = [( 2)(1) ( 2)(1)]i + [( 2)( 2) (0)(1)]j + [(0)(1) ( 2)( 2)]k = 4j 4k. Now that we have the normal vector to the plane, all that we need is a point on the plane. We have three of them, so we choose point P = (1, 2, 3) and we have that the equation of the plane is given by 0(x 1) + 4(y 2) 4(z 3) = 0, i.e. 4y 4z = -4. We have the following properties that hold for cross products and dot products. Properties of the Cross Products and Dot Products For any vectors a, b, and c and any scalar k, we have that 1. abba ka bk ab a kb abcabac 4. ab cacbc 5. abcabc 6. a b c a c b a b c 4

5 Property 5 above is referred to as the scalar triple product of the vectors a, b, and c and it has a nice geometric interpretation: it is the volume of the parallelepiped constructed by the vectors a, b, and c. b c q a c b Figure 2: The Parallelepiped formed by a, b and c The area of the parallelogram is given by A = b μ c. If q is the angle between the vectors a and b μ c, then the height of the parallelepiped is given by h = a cos(q). Thus, the volume of the parallelepiped is given by V = Ah = b μ c a cos(q), which is a bc. precisely the formula for the dot product, Now, it is possible that the angle q is greater than p/2. In that case, the value of cos(q) would be negative and the result would not make sense. We can correct this problem by replacing cos(q) with cos(q). Thus, we have the following formula for the volume of a parallelepiped. Volume of a Parallelepiped The volume of the parallelepiped formed by the vectors a, b, and c is given by the absolute value of their scalar triple product: V a bc Note: If the volume is equal to 0, then we have that a, b, and c all lie in the same plane. Example 4: Show that the vectors a = i + 2j + 3k, b = 4i + 5j + 6k, and c = 7i + 8j + 9k all lie in the same plane. 5

6 Solution: To show that the three vectors lie in the same plane, we shall take their triple scalar product and show that it is equal to zero. b μ c = [(5)(9) (6)(8)]i + [(6)(7) (4)(9)]j + [(4)(8) (5)(7)]k = 3i + 6j 3k. c (1)( 3) 2(6) 3( 3) a b So, we conclude that the three vectors lie in the same plane.. 6