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Unformatted text preview: 360 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) We now verify the equality: k v k 2 k w k 2 − ( v · w ) 2 = 17 · 21 − 18 2 = 33 = k v × w k 2 40. Find the volume of the parallelepiped spanned by u , v , and w in Figure 17. SOLUTION Using u = h 1 , , 4 i , v = h 1 , 3 , 1 i and w = h− 4 , 2 , 6 i , the scalar triple product is the following 3 × 3 determinant: u · ( v × w ) = ¯ ¯ ¯ ¯ ¯ ¯ 1 4 1 3 1 − 4 2 6 ¯ ¯ ¯ ¯ ¯ ¯ = 1 · ( 18 − 2 ) − + 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 = h 1 , 3 , 1 i and w = h− 4 , 2 , 6 i . 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 p 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 = h 2 , 2 , 1 i , v = h 1 , , 3 i , w = h , − 4 , i SOLUTION Using u = h 2 , 2 , 1 i , v = h 1 , , 3 i , and w = h , − 4 , i , the volume is given by the following scalar triple product: u · ( v × w ) = ¯ ¯ ¯ ¯ ¯ ¯ 2 2 1 1 3 − 4 ¯ ¯ ¯ ¯ ¯ ¯ = 2 ( + 12 ) − 2 ( − ) + 1 ( − 4 − ) = 24 − 4 = 20 . 43. Sketch and compute the volume of the parallelepiped spanned by u = h 1 , , i , v = h , 2 , i , w = h 1 , 1 , 2 i SOLUTION Using u = h 1 , , i , v = h , 2 , i , and w = h 1 , 1 , 2 i , the volume is given by the following scalar triple product: u · ( v × w ) = ¯ ¯ ¯ ¯ ¯ ¯ 1 2 1 1 2 ¯ ¯ ¯ ¯ ¯ ¯ = 1 ( 4 − ) − + = 4 . u v w y x z S E C T I O N 13.4 The Cross Product (ET Section 12.4) 361 44. Sketch the parallelogram spanned by u = h 1 , 1 , 1 i and v = h , , 4 i , and compute its area. SOLUTION The parallelogram spanned by u = h 1 , 1 , 1 i and v = h , , 4 i 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 = k u × v k 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 + = 4 i − 4 j = 4 h 1 , − 1 , i Hence, A = k 4 h 1 , − 1 , ik = 4 kh 1 , − 1 , ik = 4 q 1 2 + ( − 1 ) 2 + 2 = 4 √ 2 45. Calculate the area of the parallelogram spanned by u = h 1 , , 3 i and v = h 2 , 1 , 1 i . 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 3 2 1 1 ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 3 1 1 ¯ ¯ ¯ ¯ i − ¯ ¯ ¯ ¯ 1 3 2 1 ¯ ¯ ¯ ¯ j + ¯ ¯ ¯ ¯ 1 2 1 ¯ ¯ ¯ ¯ k = − 3 i − ( 1 − 6 ) j + k = − 3 i + 5 j + k The area A is the length A = k u × v k = q ( − 3 ) 2 + 5 2 + 1 2 = √ 35 ≈ 5 . 92 ....
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This note was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.
 Winter '08
 GANGliu

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