13.4.Ex15-28 - S E C T I O N 13.4 SOLUTION The Cross...

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S E C T I O N 13.4 The Cross Product (ET Section 12.4) 355 SOLUTION We use the definition of the cross product to write v × w = i j k 0 1 1 1 1 0 = 1 1 1 0 i 0 1 1 0 j + 0 1 1 1 k = ( 0 1 ) i ( 0 + 1 ) j + ( 0 1 ) k = − i j k 15. v = 1 3 , 1 , 1 3 , w = − 1 , 1 , 2 SOLUTION The cross product is the following vector: v × w = i j k 1 3 1 1 3 1 1 2 = 1 1 3 1 2 i 1 3 1 3 1 2 j + 1 3 1 1 1 k = 2 + 1 3 i 2 3 + 1 3 j + 1 3 + 1 k = 7 3 i j + 2 3 k 16. v = 1 , 1 , 0 , w = 0 , 1 , 1 SOLUTION The cross product v × w is the following vector: v × w = i j k 1 1 0 0 1 1 = 1 0 1 1 i 1 0 0 1 j + 1 1 0 1 k = ( 1 0 ) i ( 1 0 ) j + ( 1 0 ) k = i j + k In Exercises 17–20, calculate the cross product. 17. ( i + j ) × k SOLUTION We use basic properties of the cross product to obtain ( i + j ) × k = i × k + j × k = − j + i j i k i × k = − j j × k = i 18. ( j k ) × ( j + k ) SOLUTION Using properties of the cross product we get ( j k ) × ( j + k ) = ( j k ) × j + ( j k ) × k = j × j k × j + j × k k × k = 0 + i + i 0 = 2 i 19. ( i + 2 k ) × ( j k ) SOLUTION Using the distributive law we obtain ( i + 2 k ) × ( j k ) = ( i + 2 k ) × j + ( i + 2 k ) × ( k ) = i × j + 2 k × j i × k 2 k × k = k 2 i + j 0 = − 2 i + j + k 20. ( 2 i 3 j + 4 k ) × ( i + j 7
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