13.4.Ex40-62

13.4.Ex40-62 - 360 C H A P T E R 13 VECTOR GEOMETRY (ET...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

This note was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

Page1 / 9

13.4.Ex40-62 - 360 C H A P T E R 13 VECTOR GEOMETRY (ET...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online