Math 261 Worksheet: The Cross-Product The Cross Product, Algebraically. Given two vectors u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) in R3 , we define the cross-product u × v by u2 v3 − u3 v2 u × v = u3 v1 − u1 v3 . (1) u1 v2 − u2 v1 We emphasize that the cross-product of two vectors is a vector, not a scalar, and we have only defined it for vectors in R3 . Note that the first component of u × v, u2 v3 − u3 v2 , involves the second and third components of u and v, but not their first components; analogous statements hold for the other two components of u × v. A mnemonic device for remembering the formula for the cross-product formally involves the determinant: i j k u2 u3 u1 u3 u1 u2 i− u × v = u1 u2 u3 = (2) v1 v3 j + v1 v2 k , v2 v3 v1 v2 v3 where i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1) are the standard unit vectors. Letting u, v, w ∈ R3 and α ∈ R be arbitrary, we mention a few key algebraic properties of the cross-product, which may be proven directly from the definition (1): 1. Orthogonality: (u × v) · u = (u × v) · v = 0. 2. Linearity: (a) u × (v + w) = (u × v) + (u × w) (b) (u + v) × w = (u × w) + (v × w) (c) (αu) × v = u × (αv) = α(u × v) 3. Lagrange’s Identity: ku × vk2 = kuk2 kvk2 − (u · v)2 4. Anti-symmetry: v × u = −(u × v) In fact, if we wanted an operation which takes two vectors u, v ∈ Rn and yields a third vector u × v in such a way that properties 1)–3) were satisfied, this is only possible when n = 3 or n = 7, and in both cases such a “cross-product” is almost uniquely defined by these three properties1 . 1 Massey, W. S. Cross products of vectors in higher-dimensional Euclidean spaces. Amer. Math. Monthly, Vol. 90, No. 10, PP. 697–701 (1983). 1 The Cross Product, Geometrically. We have already noted that the cross-product u × v is orthogonal/perpendicular to both u and v (Property 1), and we now use Lagrange’s Identity to determine its length. Let θ ∈ [0, π] be the angle between u and v. Recalling that u · v = kukkvk cos θ, we see that ku × vk2 = kuk2 kvk2 − (u · v)2 = kuk2 kvk2 − (kukkvk cos θ)2 = kuk2 kvk2 (1 − cos2 θ) = kuk2 kvk2 sin2 θ . Taking the square-root of both sides, and using the fact that sin θ ≥ 0 for θ ∈ [0, π], we determine that ku × vk = kukkvk sin θ . (3) Abusing language slightly, we can say that the length of u × v is equal to the area of the parallelogram determined by u and v, as pictured below: v Θ u The area of the parallelogram is “base length times height”; as pictured, the base length is kuk and the height is kvk sin θ. If u and v are not parallel, then they determine a plane through the origin, consisting of all the vectors in span{u, v}—the parallelogram determined by u and v lies in this plane. Since we know the length of u × v and we know that it is orthogonal to u and v (and hence to the plane they determine), this determines u × v up to orientation—on which side of the plane does it lie? The orientation is given by the Right-hand Rule: if your right hand is positioned so that your pointer finger is in the direction of u, your middle finger is in the direction of v, and your thumb is perpendicular to them both, then your thumb is pointing in the direction of u × v. In the parallelogram picture above, u × v would be sticking out the front of this page, and v × u would be sticking out the back of this page. Scalar Triple Product, Algebraically and Geometrically. Given three vectors u, v, w ∈ R3 , we define the scalar triple product [u, v, w] ∈ R by [u, v, w] = u · (v × w) . (4) Using the algebraic definitions of the cross-product and dot-product, we determine that the scalar triple product is identical with a determinant, u1 u2 u3 [u, v, w] = v1 v2 v3 . (5) w1 w2 w3 We mention two algebraic properties of the triple product: 2 1. Its value remains the same under cyclic permutations of its three arguments, [u, v, w] = [v, w, u] = [w, u, v] . 2. Its value changes by sign when any two of its three arguments change positions, [v, u, w] = [w, v, u] = [u, w, v] = −[u, v, w] . If {u, v, w} is a linearly independent set, then the three vectors determine a parallelepiped, as pictured below: w v u The volume of this parallelepiped is “base area times height”. If we take the base to be the parallelogram determined by u and v, then the base area is ku × vk. The height of the parallelepiped is kwk | cos φ|, where φ ∈ [0, π] is the angle between w and the normal vector u × v. Therefore, the volume of the parallelepiped is the absolute value of the scalar triple product, Volume = |[u, v, w]| . (6) Properties 1) and 2) for the scalar triple product, together with the use of the absolute values, implies that the order of u, v and w in (6) does not matter. Exercises 1. Fill in the following cross-product table for the standard unit vectors × i j k i j k For example, the entry in the row for j and column for k should be j × k, not k × j. 3 2. What is the value of u × v when u and v are parallel? You may use either (3) or the anti-symmetry property (and linearity) to determine the answer. 3. It is a fact that a plane in R3 may be uniquely described in terms of a point P = (x, y, z) in the plane and a vector n = (a, b, c) perpendicular to the plane (often called a normal vector ). In particular, X = (x, y, z) is in the plane if and only if (X − P ) · n = 0. So an equation for this plane is given by a(x − x) + b(y − y) + c(z − z) = 0 . (a) Give an equation for the plane containing the three points P = (0, −2, 1), Q = (3, 1, 1), R = (−5, 0, 2). (b) Find the area of the triangle determined by P , Q and R. Hint: It is half the area of a related parallelogram. 4. In both cases below, state whether or not the four given points, P, Q, R, S, all lie in the same plane. If they do not, give the volume of the tetrahedron having these four vertices. The scalar triple product can be used for this. Hint: The volume of the tetrahedron is one third of the volume of a related parallelepiped. (a) P = (0, −1, 1), Q = (1, 0, 0), R = (1/2, 1/2, 0), S = (−1/3, 1/4, 1/2). (b) P = (0, −2, 1), Q = (3, 1, 1), R = (−5, 0, 2), S = (−4, 1, 2). 4

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