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 123vvvvijk and 123wwwwijk . The cross product of v and w, denoted by v μ w, is a vector given by 123 2332 3113 1221123()()(vvv vwvw vwvw vwvwwww      ijkvw i )jk Example 1: Compute v μ w if 430vijk and 225 wijk . 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 and . (4)( 15) ( 3)( 20) (0)(2) 60 60 0 0  vvw2)( 15) (2)( 20) (5)(2) 30 40 10 0  (wvw 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. 1Cross Product (Geometrical Interpretation) Suppose q is the angle between the vectors v and w (so 0 § q § p). Then we have that Area of parallelogramsindetermined by and  vw vwvw To see why the first equality is true, we work through a bit of algebra. 222223 32 31 13 12 2122 23 22 22 22 2223 2323 32 31 1313 13 12 1212 212222 2 2 21231 2 3 11223322222222222()()()222()( )( )cos1cosvw vw vw vw vw vwvw vvww vw vw vvww vw vw vvww vwvvvwww vwvwvw            vwvw vwvw vwvwvw2sin Taking the square root of both sides, we have that sinvw vw , where we used the fact that since 0 § q § p, sin(q) ¥ 0. 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. qwvwsinq 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 sinvw . 2Since we have that sinvw vw , 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 v and w are parallel if and only if v μ w = 0 and v and w are perpendicular if and only if = 0 vw 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 triangle11sinformed by and 22vw vwvw 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. 3When 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 022PQvijk and 2PR  wijk 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. ab ba 2.   kk k  ab ab a b 3. abc abac 4.  ab cacbc 5. abc abc 6.  abc acbabc 4Property 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¥cqacb 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 s, vectors a and b μ c, then the height of the parallelepiped is given by h = ||a||cos(q). Thuthe volume of the parallelepiped is given by V = Ah = ||b