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Unformatted text preview: 352 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) 13.4 The Cross Product (ET Section 12.4) Preliminary Questions 1. What is the ( 1 , 3 ) minor of the matrix 3 4 2 5 1 1 4 3 ? SOLUTION The ( 1 , 3 ) minor is obtained by crossing out the first row and third column of the matrix. That is, 3 4 2 5 1 1 4 3  5 1 4 2. The angle between two unit vectors e and f is 6 . What is the length of e f ? SOLUTION We use the Formula for the Length of the Cross Product: e f = e f sin Since e and f are unit vectors, e = f = 1. Also = 6 , therefore, e f = 1 1 sin 6 = 1 2 The length of e f is 1 2 . 3. What is u w , assuming that w u = 2 , 2 , 1 ? SOLUTION By anticommutativity of the cross product, we have u w = w u = 2 , 2 , 1 = 2 , 2 , 1 4. Find the cross product without using the formula: (a) 4 , 8 , 2 4 , 8 , 2 (b) 4 , 8 , 2 2 , 4 , 1 SOLUTION By properties of the cross product, the cross product of parallel vectors is the zero vector. In particular, the cross product of a vector with itself is the zero vector. Since 4 , 8 , 2 = 2 2 , 4 , 1 , the vectors 4 , 8 , 2 and 2 , 4 , 1 are parallel. We conclude that 4 , 8 , 2 4 , 8 , 2 = and 4 , 8 , 2 2 , 4 , 1 = . 5. What are i j and i k ? SOLUTION The cross product i j and i k are determined by the righthand rule. We can also use the following figure to determine these crossproducts: j i k We get i j = k and i k = j 6. When is the cross product v w equal to zero? SOLUTION The cross product v w is equal to zero if one of the vectors v or w (or both) is the zero vector, or if v and w are parallel vectors. S E C T I O N 13.4 The Cross Product (ET Section 12.4) 353 Exercises In Exercises 14, calculate the 2 2determinant. 1. 1 2 4 3 SOLUTION Using the definition of 2 2 determinant we get 1 2 4 3 = 1 3 2 4 = 5 2. 2 3 1 6 5 2 SOLUTION Using the definition we get 2 3 1 6 5 2 = 2 3 2 1 6 ( 5 ) = 4 3 + 5 6 = 13 6 3. 6 9 1 1 SOLUTION We evaluate the determinant to obtain 6 9 1 1 = 6 1 9 1 = 15 4. 9 25 5 14 SOLUTION The value of the 2 2 determinant is 9 25 5 14 = 9 14 5 25 = 1 In Exercises 510, calculate the 3 3determinant. 5. 1 2 1 4 3 1 1 SOLUTION Using the definition of 3 3 determinant we obtain 1 2 1 4 3 1 1 = 1 3 1 2 4 1 1 + 1 4 3 1 = 1 ( 3 1 ) 2 ( 4 1 1 ) + 1 ( 4  ( 3 ) 1 ) = 3 8 + 3 = 8 6. 1 1 2 3 1 3 1 SOLUTION We evaluate the 3 3 determinant to obtain 1 1 2 3 1 3 1 = 1 3 3 1 2 3 1 1 + 1 2 1 3 = 1 ( ( 1 ) 3 3 ) + 1 ( 2 3 1 ) = 9 6 = 15 7. 1 1 1 3 1 2 3 354 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) SOLUTION We use the definition to write 1 1 1 3 1 2 3 = 1 3 1 3 1 1 2 3 + 1 1 3 2 = ( 3 3 1 ) + 1 ( 1  3 ( 2 )) = 9 + 6 = 15 8....
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This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Hubscher

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