44 Section 10.7 The Cross Product Objective #0: Evaluating Determinants. Recall the following definition for determinants: Determinants a The determinant for matrix 1 b 1 is denoted as a 1 b 1 a b a b and is defined as: a 1 b 1 a b a 1 b a b 1. 3 3 Determinants The determinant for 3 3 matrix is denoted as defined as: An easy way to remember this: A a b c b c B and is a b c a 1 c 1 a c + C a 1 b 1 a b a b c Subtract a b c a b c Cross out the row and column that the number occurs. This will give us the correct matrix. Also, remember to subtract B.
45 Evaluate the following: 5 3 3 Ex. 1a Ex. 1b 4 1 3 5 3 4 1 5(1) 4( 3) 3 3( 3) () 3 5 + 1 17 9 4 13 Ex. 1 1 1 1 3 4 3 1 1 1 1 3 4 3 1 4 3 1 1 ( 1) 3 3 + 1 1 3 4 [( 3) 4( 1)] + 1[(1)( 3) 3( 1)] + 1[(1)(4) 3()] [ 6 + 4] + 1[ 3 + 3] + 1[4 6] [ ] + 1[0] + 1[ ] 4 + 0 6. Objective #1 Find the Cross Product of Two Vectors. In two dimensions and three dimensions, we saw that the dot product was defined as finding the sum of the products of the corresponding components. A second type of vector product exists for vectors in threedimensional space (and only in three dimensional space) called the Cross Product. Cross Product Definition Let v a 1 i + b 1 j + c 1 and w a i + b j + c be non-zero two vectors in space. The Cross Product v w is defined as the vector: v w a b c b c i a 1 c 1 a c j + a 1 b 1 a b (b 1 c b c 1 )i (a 1 c a c 1 )j + (a 1 b a b 1 )
46 Unlie the dot product that produces a scalar for the answer, the cross product produces a vector that is orthogonal to both v and w. Given v 3i + j + 4 and w i 7j 3, find: Ex. 3a v w Ex. 3b w v Ex. 3c v v Ex. 3d w w a) v w i j + 17 3 4 7 3 4 7 3 i 3 4 3 j + 3 7 b) w v 7 3 3 4 i + j 17 7 3 4 i 3 3 4 j + 7 3 c) v v 3 4 3 4 0i 0j + 0 0 4 4 i 3 4 3 4 j + 3 3 d) w w 7 3 7 3 0i 0j + 0 0 7 3 7 3 i 3 3 j + 7 7 If we compare the answers to part a and part b, we notice that v w (w v), so cross product is not a commutative operation. This means how you find the cross product of two vectors will determine the direction. Objective #: Algebraic Properties of Cross Products. We will now list some Algebraic properties of cross products.
Algebraic Properites of Cross Products Let u, v, and w be non-zero vectors in space and let α be a scalar. Then: 1) u u 0 ) u v (v u) 3) α(u v) (αu) v u (αv) 4) u (v + w) (u v) + (u w) Proof: 1) Let u a 1 i + b 1 j + c 1. Then u u 0i 0j + 0 0 Hence, u u 0 i a 1 c 1 a 1 c 1 j + a 1 b 1 a 1 b 1 ) Let u a 1 i + b 1 j + c 1 and v a i + b j + c. b Then u v 1 c 1 i a 1 c 1 j + a 1 b 1 b a b c c a c a b (b 1 c b c 1 )i (a 1 c a c 1 )j + (a 1 b a b 1 ) and (v u) a b c [ [(b c 1 b 1 c )i (a c 1 a 1 c )j + (a b 1 a 1 b )] (b c 1 b 1 c )i (a c 1 a 1 c )j (a b 1 a 1 b ) (b 1 c b c 1 )i (a 1 c a c 1 )j + (a 1 b a b 1 ) Hence, u v (v u). 3) Let u a 1 i + b 1 j + c 1 and v a i + b j + c. Then α(u v) α α a b c 47 b c i a c a 1 c 1 j + a b a 1 b 1 ] b c i α a 1 c 1 a c j + α a 1 b 1 a b α(b 1 c b c 1 )i α(a 1 c a c 1 )j + α(a 1 b a b 1 ) But
(αu) v αa 1 αb 1 αc 1 a b c 48 αb 1 αc 1 b c i αa 1 αc 1 a c j + αa 1 αb 1 a b (αb 1 c αb c 1 )i (αa 1 c αa c 1 )j + (αa 1 b αa b 1 ) α(b 1 c b c 1 )i α(a 1 c a c 1 )j + α(a 1 b a b 1 ) By a similar argument, it can be shown that: u (αv) α(b 1 c b c 1 )i α(a 1 c a c 1 )j + α(a 1 b a b 1 ) Hence, α(u v) (αu) v u (αv). 4) Let u a 1 i + b 1 j + c 1, v a i + b j + c, and w a 3 i + b 3 j + c 3. v + w (a + a 3 )i + (b + b 3 )j + (c + c 3 ) So, u (v + w) a + a 3 b +b 3 c +c 3 b 1 c 1 a i 1 c 1 a j + 1 b 1 b +b 3 c +c 3 a + a 3 c +c 3 a + a 3 b +b 3 (b 1 c + b 1 c 3 b c 1 b 3 c 1 )i (a 1 c + a 1 c 3 a c 1 a 3 c 1 )j + (a 1 b + a 1 b 3 a b 1 a 3 b 1 ) (b 1 c b c 1 )i (a 1 c a c 1 )j + (a 1 b a b 1 ) + (b 1 c 3 b 3 c 1 )i (a 1 c 3 a 3 c 1 )j + (a 1 b 3 a 3 b 1 ) (u v) + (u w) Thus, u (v + w) (u v) + (u w). Objective #3 Geometric Properties of the Cross Product. We will state some important geometric properties of Cross Products. Geometric Properites of Cross Products Let u and v be non-zero vectors in space. Then: 1) u v is orthogonal to both u and v. ) u v u v (u v) 3) u v u v sin(θ) where θ is the angle between u & v. 4) u v is the area of the parallelogram have u and v as adjacent sides. 5) u v 0 if and only if u and v are parallel. Proof:
1) Let u a 1 i + b 1 j + c 1 and v a i + b j + c. Then u v (b 1 c b c 1 )i (a 1 c a c 1 )j + (a 1 b a b 1 ) Now, consider u (u v): (a 1 i + b 1 j + c 1 ) ((b 1 c b c 1 )i (a 1 c a c 1 )j + (a 1 b a b 1 )) a 1 (b 1 c b c 1 ) b 1 (a 1 c a c 1 ) + c 1 (a 1 b a b 1 ) a 1 b 1 c a 1 b c 1 a 1 b 1 c + a + a 1 b c 1 a 0. If the dot product of two vectors is zero, then the vectors are orthogonal. Thus, u v is orthogonal to u. By a similar argument, it can be shown that u v is orthogonal to v. Hence, u v is orthogonal to both u and v. ) Let u a 1 i + b 1 j + c 1 and v a i + b j + c. u v 49 (b 1 c b c 1 )i (a 1 c a c 1 )j + (a 1 b a b 1 ) a b c ( (b 1 c b c 1 ) +(-[a 1 c a c 1 ]) +(a 1 b a b 1 ) ) (b 1 c b c 1 ) + (a 1 c a c 1 ) + (a 1 b a b 1 ) b 1 b1 b c 1 c + b + a1 a1 a c 1 c + a + a1 a1 a b 1 b + a b1 Now, u v ( a 1 +b1 + c1 ) ( a +b + c ) ( a 1 +b 1 +c1 )( a +b +c ) a 1 a + a1 + a1 + a b1 + b1 + b1 + a + b + c1 and (u v) (a 1 a + b 1 b + c 1 c ) (a 1 a + b 1 b + c 1 c )(a 1 a + b 1 b + c 1 c ) ( a 1 a + a1 a b 1 b + a 1 a c 1 c + a 1 a b 1 b + b 1 + b 1 b c 1 c + a 1 a c 1 c + b 1 b c 1 c + c 1 ) ( a 1 a + a1 a b 1 b + a 1 a c 1 c + b 1 + b1 b c 1 c + c 1 ) a 1 a a1 a b 1 b a 1 a c 1 c b 1 b1 b c 1 c c 1 So, u v (u v) a 1 a + a1 + a1 + a b1 + b1 + b1 + a + b + c1 a 1 a a1 a b 1 b a 1 a c 1 c b 1 b1 b c 1 c c 1 a 1 + a1 + a b1 + b1 + a + b a1 a b 1 b a 1 a c 1 c b 1 b c 1 c b 1 b1 b c 1 c + b + a1 a1 a c 1 c + a + a1 a1 a b 1 b + a b1
50 u v Hence, u v u v (u v) 3) Recall that u v u v cos(θ), so (u v) u v cos (θ). Thus u v u v u v cos (θ) Factor out u v : u v u v [1 cos (θ)] But, sin (θ) 1 cos (θ), so u v u v [sin (θ)] u v ± u v [sin(θ)] But u > 0, v > 0 and for 0 θ π, sin(θ) > 0, so we reject the negative root. Hence, u v u v sin(θ) 4) Let u and v be adjacent sides of a parallelogram: u is the length of the side of u. v is the length of the side of v. v θ is the angle between u and v. The area of a parallelogram is θ the base times the height. The height of the parallelogram u is v sin(θ). Thus, the area u v sin(θ)] u v. 5) We will need to prove both directions of this theorem. ( ) Suppose u v 0. Then u v 0 which means u v sin(θ) 0. Since u 0 and v 0, then sin(θ) 0 which implies that θ 0 or π. In either case, u and v are parallel. ( ) Suppose u and v are parallel. Then v αu for some scalar α. So, u v u (αu) α(u u) α(0) 0. Thus, u v 0 if and only if u and v are parallel. Objective #4: Find a Vector Orthogonal to Two Given Vectors. We now that u v is orthogonal to both u and v. If u and v are not parallel, then the two vectors create a plane. Thus, u v will be normal
51 (another word for perpendicular) to this plane. There are two possible vectors that are normal to the plane. One will be u v and the other will be v u. u u v or v u? v u v or v u? To determine which is the correct direction, we can use the right-hand rule. We place our right hand so that the index finger points in the same direction as the first vector in the cross product and have our thumb sticing out. Then we curl our remaining three fingers towards the second vector in the cross product. Our thumb then will tell us which is the correct direction. If we loo at the figure above, we can determine which vector normal to the plane is u v and v u. u v v u Place the index finger in the Place the index finger in the same direction as as the vector same direction as as the vector u. Curl the other three fingers v. Curl the other three fingers towards v (counterclocwise). towards u (clocwise). Since the thumb points up, then Since the thumb points down, then the top vector is u v. the bottom vector is v u. u u v v v u
5 Find a vector orthogonal to the two given vectors: Ex. 4 v i 3j + 8 and w i + 4j 5 To find a vector that is orthogonal to both v and w, we need to simply find the cross product: v w 3 8 3 8 4 5 i 8 1 5 j + 3 1 4 1 4 5 17i + j + 5 Find the area of the parallelogram with the given vertices: Ex. 5 P 1 (, 1, 4), P (1, 3, 8), P 3 ( 4, 4, 3) and P 4 ( 5, 6, 7) Two adjacent sides of a parallelogram share a common endpoint. u P 1 P (1 )i + (3 1)j + (8 4) i + j + 4 v P 1 P 3 ( 4 )i + (4 1)j + (3 4) 6i + 3j Now, find u v: u v 1 4 6 3 1 14i + 6j + 9 4 3 1 i 1 4 6 1 j + 1 6 3 Area u v ( 14) +(6) +(9) 196+ 676+ 81 953 30.87 units