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Get Full Access to University Physics - 13 Edition - Chapter 1 - Problem 47e
Get Full Access to University Physics - 13 Edition - Chapter 1 - Problem 47e

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# Find the angle between each of these pairs of vectors:

ISBN: 9780321675460 31

## Solution for problem 47E Chapter 1

University Physics | 13th Edition

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Problem 47E

Find the angle between each of these pairs of vectors:

Step-by-Step Solution:
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Solution 47E a) Two vectors are given as, A = 2 i + 6 j B = 2 i 3 j We know the dot product of two vectors can be written as, A. B = AxB x A y -y------------------(1) Where A xB axd A ,y aye the components of the two vectors. And it can also be shown as, A. B = | || |s ---------------(2) Where is the angle between two vectors. The magnitude of a vector we can get by the following rule, | |= A + A 2 = ( 2) + 6 = 4 + 36 = 40 = 6.32 . x y | |= Bx2+ B y2= (2) + 3 = 4 + 9 = 13 = 3.60 From equation (1) and (2), A. B = A B cos = A B + A B | || | x x y y 6.32 × 3.60 × cos = ( 2) × 2 + 6 × ( 3) 22.78 × cos = 4 18 = 22 cos = 22 / 22.78 = 0.965 1 0 = cos ( 0.965) = 165 So the actual angle between these two vectors is 180 165 = 15 . 0 b) Two vectors are given as, A = 3 i + 5 j B = 10 i + 6 j We know the dot product of two vectors can be written as, A. B = A x +xA B y--y----------------(1) Where A ,x anx A ,B yre yhe components of the two vectors. And it can also be shown as, A. B = A B cos ---------------(2) | || | Where is the angle between two vectors. The magnitude of a vector we can get by the following rule, | |= Ax+ A y2 = + 5 = 9 + 25 = 34 = 5.830 . | |= B x + By2 = 0 + 6 = 100 + 36 = 136 = 11.66 From equation (1) and (2), A. B = | || |s = A Bx+xA B y y 5.830 × 11.66 × cos = 3 × 10 + 5 × 6 67.98 × cos = 30 + 30 = 60 cos = 60 / 67.98 = 0.882 1 0 = cos (0.882) = 28.04 = 28 (approximately) 0 So the angle between these two vectors is 28 . c) Two vectors are given as, A = 4 i + 2 j B = 7 i + 14 j We know the dot product of two vectors can be written as, A. B = AxB x A y -y------------------(1) Where A ,x axd A ,Byary the components of the two vectors. And it can also be shown as, A. B = | || |s ---------------(2) Where is the angle between two vectors. The magnitude of a vector we can get by the following rule, | | A + A 2 = ( 4) + 2 = 16 + 4 = 20 = 4.472 . x y | |= Bx2 + By2= + 14 = 49 + 196 = 245 = 15.652 From equation (1) and (2), A. B = | || |s = A x +xA By y 4.472 × 15.652 × cos = 4 × 7 + 2 × 14 70 × cos = 28 + 28 = 0 cos = 0 / 70 = 0 1 0 = cos (0) = 90 (approximately) 0 So the angle between these two vectors is 90 .

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##### ISBN: 9780321675460

University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: angle, Find, pairs, these, vectors. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 47E from chapter: 1 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 47E from 1 chapter was answered, more than 361 students have viewed the full step-by-step answer. The answer to “Find the angle between each of these pairs of vectors:” is broken down into a number of easy to follow steps, and 10 words.

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