Section.3. The Cross Product Recall that a vector can be uniquely determined by its length and direction. De nition. The cross product of two vectors u and v, denote by u v, is a vector with length j u vj = j uj j vj sin θ (θ is the angle between u and v, 0 θ π) and the direction that is perpendicular to both u and v following the righthand rule: Suppose that u and v lie on the screen as you see. Then u v is perpendicular to the screen and is pointing towards you. W = x The cross product is also referred to as Outer Product or ector Product. Example 3.1. (a) i j =, (b) j = i, (c) i = j. 1
In physics, the measurement for turning e ect, torque, is de ned as the cross product of length vector R and force vector F. R F = R F sin θ = (length of the wrench) perpendicular component of F T = R x F F F sin θ F When using a wrench to tighten a screw, the strength is measured by Torque T= R x F R θ Example 3.. A bolt is tightened by applying 40 Newton force to a 0.05m wrench at 75degree angle. Find the magnitude of torque. Solution: Let F be the force, r be length vector. Then F = 40, j rj = 0.05, θ = 75 0. F r = F j rj sin θ = 40 0.05 sin 75 o = 9.66 (J = Nm) Geometrically, let u and v be two generating edges of a parallelogram as shown in the following gure.
h = sinθ θ θ The area of a parallelogram is The area of a triangle is Area = length of the base height = j uj h = j uj j vj sin θ = j u vj. Area = area of parallelogram = j u vj. Property of Cross product: (1) u v? u, v This is by de nition. () u v = 0 i u // v (parallel) This is due to the fact that u // v i sin θ = 0. (3) u v = v u This is because the right-hand rule. 3
W = x x (4) (λ u) v = λ( u v) = u (λ v) (5) u ( v + w ) = u v+ u w Example 3.3. Let u =< 1, 3, 4 >, v =<, 7, 5 >. Find u v. Solution: We shall use the facts that (From Example 3.1) i j =, j = i, i = j, and above properties. u v = i + 3 j + 4 i + 7 j 5 = i i + 7 j 5 + 3 j i + 7 j 5 + 4 i + 7 j 5 = i i + i 7 j i 5 + 3 j i + 3 j 7 j 3 j 5 + 4 i + 4 7 j 4 5 = 7 + 5 j 6 15 i + 8 j 8 i = 43 i + 13 j +. 4
In general, let u =< x 1, y 1, z 1 >, v =< x, y, z >.Then u v = x 1 i + y 1 j + z 1 x i + y j + z = (y 1 z y z 1 ) i + (z 1 x z x 1 ) j + (x 1 y x y 1 ). Cross Product in Component Form: We (some of us) may already now that 3 3 determinants can be computed through the expansion: a b c x y z = a y1 z 1 y z b x 1 z 1 x z + c x 1 y 1 x y. For instance, 1 3 1 3 4 7 5 = 1 3 4 7 5 1 4 5 + 3 1 3 7 = ( 15 8) ( 5 8) + 3 (7 6) = 14. We nd the cross product formula may be memorized by using the determinant expansion formula: i j u v = x y z = y1 z 1 y z i x1 z 1 x z j + x1 y 1 x y. Example 3.4. Find u v if (a) u =< 1, 3, 4 >, v =<, 7, 5 >, (b) u =< 1,, 1 >, v =< 1,, 0 >. Solution: (a) Note u and v are the same as in Example 3.3. i j u v = x y z = i j 1 3 4 7 5 = 3 4 7 5 i 1 4 5 j + 1 3 7 = ( 15 8) i ( 5 8) j + (7 6) = 43 i + 13 j +. 5
(b) u v = = i j 1 1 1 0 1 0 i 1 1 1 0 j + 1 1 = i + j. Example 3.5. Consider the triangle with vertices P (1, 4, 6), Q (, 5, 1), R (1, 1, 1). Q P R (a) Find a vector that is perpendicular to this triangle. (b) Find the area of the triangle. Solution: (a) Let u =! P R =! OR! OP = h1, 1, 1i h1, 4, 6i = h0, 5, 5i v =! P Q =! OQ! OP = h, 5, 1i h1, 4, 6i = h 3, 1, 7i Then the vector u v is a vector perpendicular to both u and v, and consequently it is perpendicular to the triangle. Now, i j u v = 0 5 5 3 1 7 = 5 5 1 7 i 0 5 3 7 j + 0 5 3 1 = 40 i + 15 j 15 = 5 8 i + 3 j 3. 6
Ans: h8, 3, 3i is perpendicular to the triangle. (b) Area = 1 j u vj = 1 5 8 i + 3 j 3 = 5 8 i + 3 j 3 = 5 p 5 p 8 64 + 9 + 9 = =. 638. Triple products. There are at least two ways to de ne product for 3 vectors u, v and w. (i) (scalar) triple product: ( u v) w is a scalar. x W h θ Consider the parallelepiped formed by three vectors u, v and w.the bottom face is a parallelogram by u and v, whose area is j u vj. The cross product ( u v) is perpendicular to the bottom face, and j( u v) wj = j u vj (j w cos θj) = j u vj h = volume of parallelepiped 7
If u =< x 1, y 1, z 1 >, v =< x, y, z >, w =< x 3, y 3, z 3 >, then i j ( u v) w = x y z w y = 1 z 1 µ y z i x1 z 1 x z j + x1 y 1 x y < x 3, y 3, z 3 > = y1 z 1 y z x3 x1 z 1 x z y3 + x1 y 1 x y z3 x 3 y 3 z 3 = x y z = x 3 y 3 z 3 x y z = x y z x 3 y 3 z 3. Example 3.6. Show that the following three vectors u =< 1, 4, 7 >, v =<, 1, 4 >, w =< 0, 9, 18 > are on the same plane. Solution: 1 4 7 ( u v) w = 1 4 0 9 18 = 1 ( 1) 18 + 4 4 0 + ( 7) ( 9) ( 7) ( 1) 0 4 18 1 4 ( 9) = 18 + 16 144 + 36 = 36 = 0. Since the volume of the parallelepiped is zero, they are on one plane. (ii) Another way to de ne triple product is called vector triple product: ( u v) w is a vector. One can show the following formula a b c = ( a c) b a b c Homewor: 1. Which of the following expressions are meaningful? (a) a b c 8
(b) a b c (c) a b + c (d) a b c (e) (f) a b a b c d c d. ector a and b are shown below. Given j aj = 3, b =. (a) Find a b (b) a b = hx, y, zi has three components x, y, z. For each component, determine whether it is positive, or zero, or negative. Z b Y X a (c) Find a b, where = h0, 0, 1i. 3. Find a unit vector that is orthogonal to the plane passing through three points P, Q, and R. Find also the triangle whose vertices are P, Q, and R. (a) P (0, 1, 0), Q (3, 1, ), R (4, 3, 1) 9
(b) P (, 1, 1), Q (1,, 3), R (1, 3, ) 4. Determine whether four points P, Q, R, S lie on a plane. If not, nd the volume of the parallelepiped with edges P Q, P R, P S. (a) P (0, 1, 0), Q (3, 1, ), R (4, 3, 1), S (,, 1) (b) P (1, 1, 1), Q (, 6, 1), R (4, 0, 1), S (6, 10, 3) 5. Find the magnitude of the torque about P if a 36 lb force is applied as shown. 4 ft P 4 ft 30 o 36 lb 6. Determine whether the following formulas hold: (a) a b a + b = a b (b) a b a b = a b (c) a b a + b = 0 10