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13.4.Ex40-62 - 360 C H A P T E R 13 V E C T O R G E O M E T...

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360 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) We now verify the equality: v 2 w 2 ( v · w ) 2 = 17 · 21 18 2 = 33 = v × w 2 40. Find the volume of the parallelepiped spanned by u , v , and w in Figure 17. SOLUTION Using u = 1 , 0 , 4 , v = 1 , 3 , 1 and w = 4 , 2 , 6 , the scalar triple product is the following 3 × 3 determinant: u · ( v × w ) = 1 0 4 1 3 1 4 2 6 = 1 · ( 18 2 ) 0 + 4 ( 2 + 12 ) = 16 + 56 = 72 41. Find the area of the parallelogram spanned by v and w in Figure 17. y x z u = ⟨ 1, 0, 4 w = ⟨− 4, 2, 6 v = ⟨ 1, 3, 1 FIGURE 17 SOLUTION The area of the parallelogram equals the length of the cross product of the two vectors v = 1 , 3 , 1 and w = − 4 , 2 , 6 . We calculate the cross product as follows: v × w = i j k 1 3 1 4 2 6 = ( 18 2 ) i ( 6 + 4 ) j + ( 2 + 12 ) k = 16 i 10 j + 14 k The length of this vector 16 i 10 j + 14 k is 16 2 + 10 2 + 14 2 = 2 138. Thus, the area of the parallelogram is 2 138. 42. Calculate the volume of the parallelepiped spanned by u = 2 , 2 , 1 , v = 1 , 0 , 3 , w = 0 , 4 , 0 SOLUTION Using u = 2 , 2 , 1 , v = 1 , 0 , 3 , and w = 0 , 4 , 0 , the volume is given by the following scalar triple product: u · ( v × w ) = 2 2 1 1 0 3 0 4 0 = 2 ( 0 + 12 ) 2 ( 0 0 ) + 1 ( 4 0 ) = 24 4 = 20 . 43. Sketch and compute the volume of the parallelepiped spanned by u = 1 , 0 , 0 , v = 0 , 2 , 0 , w = 1 , 1 , 2 SOLUTION Using u = 1 , 0 , 0 , v = 0 , 2 , 0 , and w = 1 , 1 , 2 , the volume is given by the following scalar triple product: u · ( v × w ) = 1 0 0 0 2 0 1 1 2 = 1 ( 4 0 ) 0 + 0 = 4 . u v w y x z
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S E C T I O N 13.4 The Cross Product (ET Section 12.4) 361 44. Sketch the parallelogram spanned by u = 1 , 1 , 1 and v = 0 , 0 , 4 , and compute its area. SOLUTION The parallelogram spanned by u = 1 , 1 , 1 and v = 0 , 0 , 4 is shown in the figure. y x z v u 5 4 3 2 1 1 1 We find its area A using the formula for the area of a parallelogram: A = u × v We first find the cross product vector u × v : u × v = ( i + j + k ) × 4 k = 4 i × k + 4 j × k + 4 k × k = − 4 j + 4 i + 0 = 4 i 4 j = 4 1 , 1 , 0 Hence, A = 4 1 , 1 , 0 = 4 1 , 1 , 0 = 4 1 2 + ( 1 ) 2 + 0 2 = 4 2 45. Calculate the area of the parallelogram spanned by u = 1 , 0 , 3 and v = 2 , 1 , 1 . SOLUTION The area of the parallelogram is the length of the vector u × v . We first compute this vector: u × v = i j k 1 0 3 2 1 1 = 0 3 1 1 i 1 3 2 1 j + 1 0 2 1 k = − 3 i ( 1 6 ) j + k = − 3 i + 5 j + k The area A is the length A = u × v = ( 3 ) 2 + 5 2 + 1 2 = 35 5 . 92 . 46. Find the area of the parallelogram determined by the vectors a , 0 , 0 and 0 , b , c . SOLUTION The area A is the length of the cross product of the two vectors. We first compute the cross product: a , 0 , 0 × 0 , b , c = a i × ( b j + c k ) = ab i × j + ac i × k = ab k ac j = 0 , ac , ab j i k The area of the parallelogram is therefore A = 0 , ac , ab = 0 2 + ( ac ) 2 + ( ab ) 2 = a 2 c 2 + a 2 b 2 = | a | b 2 + c 2 47. Sketch the triangle with vertices O , P = ( 3 , 3 , 0 ) , and Q = ( 0 , 3 , 3 ) , and compute its area using cross products. SOLUTION The triangle O P Q is shown in the following figure. y x O Q = (0, 3, 3) P = (3, 3, 0) z
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362 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) The area S of the triangle is half of the area of the parallelogram determined by the vectors −→ O P = 3 , 3 , 0 and −→ OQ = 0 , 3 , 3 . Thus, S = 1 2 −→ O P × −→ OQ (1) We compute the cross product: −→ O P × −→ OQ = i j k 3 3 0 0 3 3 = 3 0 3 3 i 3 0 0 3 j + 3 3 0 3 k = 9 i 9 j + 9 k = 9 1 , 1 , 1
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