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12.4-sm.dvi - 352 C H A P T E R 13 V EC T OR G E OME TRY(ET...

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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 0 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 0 3 - 5 - 1 4 0 2. The angle between two unit vectors e and f is π 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 θ Since e and f are unit vectors, e = f = 1. Also θ = π 6 , therefore, e × f = 1 · 1 · sin π 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 anti-commutativity 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 = 0 and 4 , 8 , 2 × 2 , 4 , 1 = 0 . 5. What are i × j and i × k ? SOLUTION The cross product i × j and i × k are determined by the right-hand rule. We can also use the following figure to determine these cross-products: 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.
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S E C T I O N 13.4 The Cross Product (ET Section 12.4) 353 Exercises In Exercises 1–4, calculate the 2 × 2 -determinant. 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 5–10, calculate the 3 × 3 -determinant. 5. 1 2 1 4 - 3 0 1 0 1 SOLUTION Using the definition of 3 × 3 determinant we obtain 1 2 1 4 - 3 0 1 0 1 = 1 - 3 0 0 1 - 2 4 0 1 1 + 1 4 - 3 1 0 = 1 · ( - 3 · 1 - 0 · 0 ) - 2 · ( 4 · 1 - 0 · 1 ) + 1 · ( 4 · 0 - ( - 3 ) · 1 ) = - 3 - 8 + 3 = - 8 6. 1 0 1 - 2 0 3 1 3 - 1 SOLUTION We evaluate the 3 × 3 determinant to obtain 1 0 1 - 2 0 3 1 3 - 1 = 1 0 3 3 - 1 - 0 - 2 3 1 - 1 + 1 - 2 0 1 3 = 1 · ( 0 · ( - 1 ) - 3 · 3 ) - 0 + 1 · ( - 2 · 3 - 0 · 1 ) = - 9 - 6 = - 15 7. 1 0 1 1 3 1 - 2 0 3
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354 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) SOLUTION We use the definition to write 1 0 1 1 3 1 - 2 0 3 = 1 3 1 0 3 - 0 1 1 - 2 3 + 1 1 3 - 2 0 = ( 3 · 3 - 1 · 0 ) - 0 + 1 · ( 1 · 0 - 3 ( - 2 )) = 9 + 6 = 15 8. 0 1 0 5 93 6 4 78 5 SOLUTION We have 0 1 0 5 93 6 4 78 5 = 0 93 6 78 5 - 1 5 6 4 5 + 0 5 93 4 78 = 0 - ( 5 · 5 - 6 · 4 ) + 0 = - 1 9. 1 0 0 0 1 0 0 0 1 SOLUTION Using the definition of 3 × 3 determinant yields 1 0 0 0 1 0 0 0 1 = 1 1 0 0 1 - 0 0 0 0 1 + 0 0 1 0 0 = 1 ( 1 · 1 - 0 · 0 ) - 0 + 0 = 1 10. 1 2 3 2 4 6 - 3 - 4 2 SOLUTION We have 1 2 3 2 4 6 - 3 - 4 2 = 1 4 6 - 4 2 - 2 2 6 - 3 2 + 3 2 4 - 3 - 4 = 1 ( 4 · 2 - 6 · ( - 4 )) - 2 ( 2 · 2 - 6 · ( - 3 )) + 3 ( 2 · ( - 4 ) - 4 · ( - 3 )) = 32 - 44 + 12 = 0 In Exercises 11–16, calculate v × w .
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