5.3 The Cross Product in R 3


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1 53 The Cross Product in R 3 Definition 531 Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] Then the vector given by [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] is called the cross product (or vector product) of u and v and denoted u v Example 532 Suppose u = [1, 2, 1] and v = [2, 1, 0] Find u v and show it is orthogonal to both u and v Solution: u v = [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] = [2(0) ( 1)(1), 1(2) (1)(0), 1(1) 2(2)] = [1, 2, 3] u( u v) = [1, 2, 1][1, 2, 3] = = 0 v( u v) = [2, 1, 0][1, 2, 3] = = 0 Thus u u v and v u v This is a general feature of the cross product : it is often used to produce a vector which is orthogonal to two given ones Note: If u and v have the same initial point, there is exactly one plane in R 3 which contains the line segments representing both of them Since u v is orthogonal to both u and v, it is normal to this plane Hence the cross product provides us with the means to finish Example 626 u v v u 77
2 Back to Example 526: P contains the points A(1, 2, 1), B(2, 4, 1) and C( 1, 0, 3), so the vectors AB = [1, 2, 0] and orthogonal to both AC = [ 2, 2, 2] lie within P A vector n which is normal to P is AB and AC Let n = AB AC = [1, 2, 0] [ 2, 2, 2] u v = [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] = [1, 2, 0] [ 2, 2, 2] = [2(2) 0( 2), 0( 2) 1(2), 1( 2) 2( 2)] (Check this is orthogonal to both So n = [4, 2, 2] is normal to P : = [4, 2, 2] AB and AC) P : 4x 2y + 2z = d A(1, 2, 1) belongs to P = 4(1) 2(2) + 2(1) = d : d = 2; P : 4x 2y + 2z = 2 Equation of P : 2x y + z = 1 Check that the coordinates of the three original points A, B and C do indeed satisfy this equation Properties of the Cross Product Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] be vectors in R 3 Then : 1 u( u v) = 0 and v( u v) = 0 Proof : u( u v) = [u 1, u 2, u 3 ][u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] = u 1 (u 2 v 3 u 3 v 2 ) + u 2 (u 3 v 1 u 1 v 3 ) + u 3 (u 1 v 2 u 2 v 1 ) = u 1 u 2 v 3 u 1 u 3 v 2 + u 2 u 3 v 1 u 2 u 1 v 3 ) + u 3 u 1 v 2 u 3 u 2 v 1 = 0 Similarly for v( u v) 2 The cross product is not commutative In fact v u = ( u v) So u v and v u have opposite directions This is easily checked from the formula (Exercise) 78
3 3 u u = 0 (= [0, 0, 0]) The cross product of any vector with itself is the zero vector Again this is easily seen from the formula 4 If k is a scalar, then u (k v) = k( u v); eg u (2 v) = 2( u v), etc Note: In particular this means : if k is a scalar then u (k u) = k( u u) = k 0 by Property 3 : ie, if two vectors in R 3 have the same (or opposite) direction, their cross product is the zero vector 5 If u and v are nonzero vectors in R 3, and v is not a scalar multiple of u, then u v is a nonzero vector orthogonal to both u and v 6 Distributivity of the Cross Product over Vector Addition : Let u, v and w be vectors in R 3 Then (a) u ( v + w) = ( u v) + ( u w) (b) ( u + v) w = ( u w) + ( v w) The Standard Basis Vectors in R 3 Let v = [2, 3, 4] Then we could write v = [2, 0, 0] + [0, 3, 0] + [0, 0, 4] v = 2[1, 0, 0] + 3[0, 1, 0] + 4[0, 0, 1] Note that [1, 0, 0], [0, 1, 0] and [0, 0, 1] are unit vectors pointing along the positive X, Y and Z axes respectively The above example indicates that any vector in R 3 can be written as the sum of three vectors, each a scalar multiple of [1, 0, 0], [0, 1, 0] or [0, 0, 1] Definition 533 The vectors [1, 0, 0], [0, 1, 0] and [0, 0, 1] are called the standard basis vectors in R 3 and are denoted by ı, j and k respectively ı = [1, 0, 0], j = [0, 1, 0], k = [0, 0, 1] 79
4 Z ı k j X Y Then for example [2, 3, 4] = 2 ı+3 j+4 k and in general the vector [a, b, c] may also be written a ı + b j + c k The ı, j, k notation will be convenient for computing cross products Computing Cross Products using 3 3 Determinants Another method for computing u v Take u = [2, 1, 1] and v = [1, 0, 2] Step 1 Write a 3 3 matrix whose rows consist of 1 ı j k ı j k 2 Components of u Components of v Step 2 u v is the determinant of this matrix, and can be computed for example using the basketweave method (or cofactor expansion along the first row): ı j k ı j u v is given by ( ı(1)(2) + j(1)(1) + k(2)(0)) } {{ } products of terms along top right bottom left diagonals ( k(1)(1) + ı(1)(0) + k(1)(1)) } {{ } products of terms along top left bottom right diagonals u v = 2 ı + j k 4 j = 2 ı 3 j k = [2, 3, 1] Example 534 Find [3, 1, 1] [3, 1, 1] 80
5 Solution: ı j k ı j [3, 1, 1] [3, 1, 1] = ı(1)(1) = j( 1)(3) + k(3)( 1) k(1)(3) ı( 1)( 1) j(3)(1) = ı 3 j 3 k 3 k ı 3 j = 6 j 6 k [3, 1, 1] [3, 1, 1] = [0, 6, 6] Remark: The cross product is not associative : ie, if u, v and w are vectors, ( u v) w need not be equal to u ( v w) For example ı ( ı j) = ı k = j (Check) ( ı ı) j = 0 j = 0 j Length of the Cross Product So far our discussion of the cross product has focussed on its direction The length of u v also has significance, relating to the angle θ between u and v Fact 535 (Lagrange s Identity) For any vectors u and v in R 3 u v 2 = u 2 v 2 ( u v) 2 This can be proved by writing each term in terms of components of u and v Replacing u v by u v cos θ, Lagrange s Identity becomes : u v 2 = u 2 v 2 ( u v cos θ) 2 = u 2 v 2 u 2 v 2 cos 2 θ = u 2 v 2 (1 cos 2 θ) = u 2 v 2 (sin 2 θ) = u v = u v sin θ (Note that sin θ 0 since θ is between 0 and π (180 )) Application: Area of a Parallelogram 81
6 Suppose u and v are vectors representing adjacent sides of a parallelogram P The area of P is u h, where u is regarded as the base, and h denotes the perpendicular height of P above u Then v θ u h sin θ = Area of P = u h = u v sin θ = u v h = h = v sin θ v Example 536 (Summer 1999 Q6) Find the area of the parallelogram in R 3 having the vectors [1, 2, 3] and [7, 6, 7] as adjacent sides Solution: Area is [1, 2, 3] [7, 6, 7] ı j k ı j [1, 2, 3] [7, 6, 7] = ı(2)( 7) + j(3)(7) + k(1)(6) k(2)(7) ı(3)(6) j(1)( 7) = 32 ı + 28 j 8 k = [ 32, 28, 8] Area of parallelogram = [ 32, 28, 8] = 4 [ 8, 7, 2] = = = Example 537 (Summer 2001 Q3) Find the area of the triangle in R 3 with vertices A(1, 2, 1), B(2, 4, 1) and C( 1, 0, 3) Solution: AB = [1, 2, 0], AC = [ 2, 2, 2] AB AC 82
7 From the diagram, we want half of the area of the parallelogram having adjacent sides; ie AB and Area of triangle = 1 2 AB AC = 1 2 [4, 2, 2] (from Problem 626 ) = = 6 Summary: Area of parallelogram with adjacent sides u and v : u v 1 Area of triangle with adjacent sides u and v : 2 u v Another Application: Volume of a Parallelepiped AC as A parallelepiped in R 3 is a sixfaced object, in which pairs of opposite faces consist of similar parallelograms A Parallelpiped If u, v and w are vectors in R 3 having different directions and initial points at O, they form three adjacent sides of a unique parallelepiped Example 538 (Summer 1999 Q6 (b)) Find the volume of the parallelepiped P having u = [1, 2, 3], v = [7, 6, 7] and w = [4, 5, 3] as adjacent sides Solution: Suppose the parallelogram with u and v as sides forms the base of P v w u v u h 83
8 Then V = Volume of P = A h where A is the area of the base and h is the (perpendicular) height of P above this base A = u v The vector u v is perpendicular to the base of P and from the diagram we see that h = proj u v w = h = w( u v) u v Then w( u v) V = A h = u v u v Volume of P = w( u v) From Example 536 u v = [1, 2, 3] [7, 6, 7] = [ 32, 28, 8] Then Volume of P = [4, 5, 3][ 32, 28, 8] = 4( 32) + 5(28) 3( 8) = 36 84
9 Remarks: 1 If w, u and v are adjacent sides of a parallelepiped P, the volume of P is given by w( u v) 2 w( u v) is called the scalar triple product of w, u and v Although this definition looks nonsymmetric, it turns out that w( u v) does not depend on the order in which u, v and w are written; ie w( u v) = u( v w) = v( w u) = w( v u) = v( u w) = u( w v), for any vectors u, v, w in R 3 (Typically three of the six expressions inside the absolute value signs above will be negative and three positive, but all will have the same absolute value) In Example 538 there was no particular reason to choose the parallelogram defined by u and v as the base : choosing a different face, for example the one defined by u and w would have resulted in v( u w) as the volume formula Exercise: Check that v( u w) = w( u v) for this example 3 Scalar Triple Products and the 3 3 Determinant In Example 538 suppose we want to compute u( w v) = [1, 2, 3]([4, 5, 3] [7, 6, 7]) This is just the determinant of the matrix Components of u Components of w Components of v This determinant can be computed for example by the basketweave method: u( w v) = 1(5)( 7) + 2( 3)(7) + 3(4)(6) 3(5)(7) 1( 3)(6) 2(4)( 7) = = 36 85
10 So u( w v) = 36 (As expected u( w v) = 36 = Volume of P ) This completes our discussion of the cross product Summary of Section 53 : 1 Definition of Cross Product 2 Computation of cross product using, formula, standard basis vectors or basketweave method 3 Use of cross product to find the equation of a plane given three points 4 Area of parallelograms and triangles 5 Volume of a parallelepiped and the scalar triple product 86
v 1 v 3 u v = (( 1)4 (3)2, [1(4) ( 2)2], 1(3) ( 2)( 1)) = ( 10, 8, 1) (d) u (v w) = (u w)v (u v)w (Relationship between dot and cross product)
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