The Vector or Cross Product


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1 The Vector or ross Product 1 ppendix The Vector or ross Product We saw in ppendix that the dot product of two vectors is a scalar quantity that is a maximum when the two vectors are parallel and is zero if the two vectors are normal or perpendicular to each other. We now discuss another kind of vector multiplication called the vector or cross product, which is a vector quantity that is a maximum when the two vectors are normal to each other and is zero if they are parallel. Prerequisite knowledge: ppendix The Scalar or Dot Product.1 Definition of the ross Product The vector or cross product of two vectors is written as and reads " cross." It is defined to be a third vector such that =, where the magnitude of is = = sinφ (.1) and the direction of is perpendicular to both and in a righthanded sense as shown in Fig..1. φ is the smaller angle between and and the direction of is found by the following rule. Extend the fingers of your right hand along and then curl them toward as if you were rotating through φ. Your thumb will then point in the direction of. The vector product has a magnitude sinφ but its direction, found by rotating into through φ, is opposite to that of. Therefore, = = ( ) (.2) and the commutative law does not hold for the cross product. φ Figure.1 =
2 2 ppendix.2 Distributive Law for the ross Product The distributive law ( + ) = ( ) + ( ) holds in general for the cross product and is illustrated for the special case shown in Fig..2 where,, and all lie in the xy plane and D = +. The +z direction is out of the paper so from the righthand rule D is into the paper or in the k direction. Similarly is in the +k direction and is in the k direction. We can then write ut since ( + ) = D = Dcosφ and = Dsinφ = k Dsin( θ φ) = k D sinθ cosφ cosθsinφ ( ) ( ) sinθ cos = ( ) + ( ) + = k + k π since cosθ = sin θ 2 and the angle between and is π θ 2. θ y θ φ Figure.2 ( + ) = ( ) + ( ) D x.3 ross Product and Vector omponents We now wish to find the components of = in the rectangular coordinate system shown in Fig..3 if we know the components of and. From the definition of the cross product, if two vectors are parallel, then φ = 0, sinφ = 0, and their cross product is zero. In particular, the cross product of a vector with itself is always zero. Therefore i i = j j= k k = 0. If two vectors are perpendicular, then φ = π 2, sinφ = 1, and the magnitude of their cross product is equal to the product of the magnitudes of the two vectors and the
3 The Vector or ross Product 3 direction of the cross product is given by the righthand rule. In particular i j is in the direction of k (rotate i into j with the fingers of your right hand and watch your thumb) and has a magnitude of unity. ut this is just the unit vector k. Thus, in a similar way, i j = k j i = k k i = j i k = j j k = i k j = i z i k j y x Figure.3 = Let us write the cross product = = i + j+ k i+ j+ k ( ) ( ) ( x xi i) ( x yi j) ( x zi k ) ( y xj i) ( y yj j) ( y zj k ) ( k i) ( k j) ( k k ) = z x z y z z Using the above results to evaluate the cross products of the unit vectors, we can write = = ( y z z y) i+ ( z x x z) j+ ( x y y x) k = i + j+ k Therefore, if =, the components of are given in terms of the components of and by x = yz zy y = zx xz = z x y y x
4 4 ppendix useful way to remember the components of = is to recognize that can be written as the determinant i j k = = (.3) If you evaluate this determinant, you obtain the same result ( y z z y) ( z x x z) ( x y y x) = i+ j+ k (.4).4 ssociative Law The associative law does not in general hold for the vector product. Thus ( ) ( ) as can be seen by the simple example shown in Fig..4. Since and are parallel, = is a vector directed along the +z axis (out of the paper), ( ) 0. ( ) however, so that ( ) is a nonzero vector directed along the y axis. y x Figure.4 ( ) ( ).5 ross Product Geometric Properties onsider the parallelogram with its sides formed by the vectors and as shown in Fig..5. The area of the parallelogram is h where h = sinφ. Recall that the magnitude of the cross product is given by = sinφ
5 The Vector or ross Product 5 Therefore in terms of and the area of the parallelogram is given by. φ h Figure.5 The area of the parallelogram is onsider the parallelepiped with its sides formed by the vectors,, and as shown in Fig..6. The volume of the parallelepiped is (area of parallelogram formed by = n, where n is a unit vector parallel to. Since and ) (height h) ( )( ) = n, then in terms of,, and the volume of the parallelepiped is given by ( ). h n Figure.6 The area of the parallelepiped is ( ).6 Scalar Triple Product The expression D ( ) is called the scalar triple product. It can be evaluated in terms of the rectangular components of,, and D by letting = as in Example 5c. Then D ( ) = D ( ) Dx( yz zy) Dy( zx xz) Dz( xy yx) D = + + Note that the scalar triple product can be written as the determinant D D D D = (.5)
6 6 ppendix.7 Example 1 Given the vectors = i 2j+ 4k and = 3i+ j 2k find and. nswer 1: i j k = = i(4 4) + j(12+ 2) + k (1+ 6) = 14j + 7k = = 245 = 15.7 nswer 2: In Matlab the cross product of vectors and can be written as cross(,) as shown in Matlab Example 1. Matlab Example 1 >> = [12 4] = >> = [3 12] = >> = cross(,) = >> mag = norm() mag = >>
7 The Vector or ross Product 7.8 Example 2 Use = sinφ to find the angle φ between and in Example 5g, and compare with the result of Example 4e. nswer 1: sinφ = 245 = = φ = = 114 nswer 2: In Matlab the solution can be found by writing the single Matlab equation shown in Matlab Example 2. Matlab Example 2 >> = [12 4] = >> = [3 12] = >> phi = (asin(norm(cross(,))/(norm()*norm())))*180/pi phi = >> Note carefully the need to use parentheses in the equation for phi. The Matlab function asin for the arc sine gives the answer in radians. Thus, that result must be multiplied by 180/π to give the answer in degrees. lso note that the angle phi is greater than 90 degrees (as can be determined by plotting the vectors) and therefore the arc sine result must be subtracted from 180 degrees.
8 8 ppendix Problems Where appropriate use Matlab to find the answers to the following problems. 1 If = 5i j 2k and = 2i+ 3j k, find (a) and (b) (c) sin φ and φ where φ is the smaller angle between and. 2 If = 5i j 2k and = 2i+ 3j k, find 3 If = , ( ), and ( ) i j k, = 2i 4j+ 5k, and = i+ j 2k, find (a) ( ) (b) ( ) 4 Evaluate (a) 2i ( 3j 4k ) (b) ( i + 2j) k 2i 4j i+ k (c ) ( ) ( ).
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