{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

13.4.Ex40-62

# 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...

This preview shows pages 1–4. Sign up to view the full content.

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

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

View Full Document
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
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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern