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Math Algebra Elementary Linear Algebra: Applications Version

(a) Whether the given vectors are orthogonal vectors or not. The given vectors are orthogonal. Given: The two vectors are u = ( 6 , 1 , 4 ) and v = ( 2 , 0 , − 3 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 〉 and b = 〈 b 1 , b 2 , b 3 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 = 0 Calculation: The dot product of two vectors u = ( 6 , 1 , 4 ) and v = ( 2 , 0 , − 3 ) is: u ⋅ v = ( 6 ) ( 2 ) + ( 1 ) ( 0 ) + ( 4 ) ( − 3 ) u ⋅ v = 12 + 0 − 12 u ⋅ v = 0 Since u ⋅ v = 0 , therefore the given vectors are orthogonal. Conclusion: The given vectors are orthogonal. (b) Whether the given vectors are orthogonal vectors or not. The given vectors are not orthogonal. Given: The two vectors are u = ( 0 , 0 , − 1 ) and v = ( 1 , 1 , 1 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 〉 and b = 〈 b 1 , b 2 , b 3 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 = 0 Calculation: The dot product of two vectors u = ( 0 , 0 , − 1 ) and v = ( 1 , 1 , 1 ) is: u ⋅ v = ( 0 ) ( 1 ) + ( 0 ) ( 1 ) + ( − 1 ) ( 1 ) u ⋅ v = 0 + 0 − 1 u ⋅ v = − 1 Since u ⋅ v ≠ 0 , therefore the given vectors are not orthogonal. Conclusion: The given vectors are not orthogonal. (c) Whether the given vectors are orthogonal vectors or not. The given vectors are not orthogonal. Given: The two vectors are u = ( 3 , − 2 , 1 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 , a 4 〉 and b = 〈 b 1 , b 2 , b 3 , b 4 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 = 0 Calculation: The dot product of two vectors u = ( 3 , − 2 , 1 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) is: u ⋅ v = ( 3 ) ( − 4 ) + ( − 2 ) ( 1 ) + ( 1 ) ( − 3 ) + ( 3 ) ( 7 ) u ⋅ v = − 12 − 2 − 3 + 21 u ⋅ v = − 17 + 21 u ⋅ v = 4 Since u ⋅ v ≠ 0 , therefore the given vectors are not orthogonal. Conclusion: The given vectors are not orthogonal. (d) Whether the given vectors are orthogonal vectors or not. The given vectors are not orthogonal. Given: The two vectors are u = ( 5 , − 4 , 0 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 , a 4 〉 and b = 〈 b 1 , b 2 , b 3 , b 4 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 = 0 Calculation: The dot product of two vectors u = ( 5 , − 4 , 0 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) is: u ⋅ v = ( 5 ) ( − 4 ) + ( − 4 ) ( 1 ) + ( 0 ) ( − 3 ) + ( 3 ) ( 7 ) u ⋅ v = − 20 − 4 + 0 + 21 u ⋅ v = − 24 + 21 u ⋅ v = − 3 Since u ⋅ v ≠ 0 , therefore the given vectors are not orthogonal. Conclusion: The given vectors are not orthogonal.

(a) Whether the given vectors are orthogonal vectors or not. The given vectors are orthogonal. Given: The two vectors are u = ( 6 , 1 , 4 ) and v = ( 2 , 0 , − 3 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 〉 and b = 〈 b 1 , b 2 , b 3 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 = 0 Calculation: The dot product of two vectors u = ( 6 , 1 , 4 ) and v = ( 2 , 0 , − 3 ) is: u ⋅ v = ( 6 ) ( 2 ) + ( 1 ) ( 0 ) + ( 4 ) ( − 3 ) u ⋅ v = 12 + 0 − 12 u ⋅ v = 0 Since u ⋅ v = 0 , therefore the given vectors are orthogonal. Conclusion: The given vectors are orthogonal. (b) Whether the given vectors are orthogonal vectors or not. The given vectors are not orthogonal. Given: The two vectors are u = ( 0 , 0 , − 1 ) and v = ( 1 , 1 , 1 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 〉 and b = 〈 b 1 , b 2 , b 3 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 = 0 Calculation: The dot product of two vectors u = ( 0 , 0 , − 1 ) and v = ( 1 , 1 , 1 ) is: u ⋅ v = ( 0 ) ( 1 ) + ( 0 ) ( 1 ) + ( − 1 ) ( 1 ) u ⋅ v = 0 + 0 − 1 u ⋅ v = − 1 Since u ⋅ v ≠ 0 , therefore the given vectors are not orthogonal. Conclusion: The given vectors are not orthogonal. (c) Whether the given vectors are orthogonal vectors or not. The given vectors are not orthogonal. Given: The two vectors are u = ( 3 , − 2 , 1 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 , a 4 〉 and b = 〈 b 1 , b 2 , b 3 , b 4 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 = 0 Calculation: The dot product of two vectors u = ( 3 , − 2 , 1 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) is: u ⋅ v = ( 3 ) ( − 4 ) + ( − 2 ) ( 1 ) + ( 1 ) ( − 3 ) + ( 3 ) ( 7 ) u ⋅ v = − 12 − 2 − 3 + 21 u ⋅ v = − 17 + 21 u ⋅ v = 4 Since u ⋅ v ≠ 0 , therefore the given vectors are not orthogonal. Conclusion: The given vectors are not orthogonal. (d) Whether the given vectors are orthogonal vectors or not. The given vectors are not orthogonal. Given: The two vectors are u = ( 5 , − 4 , 0 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) . Concept used: Two vectors a = 〈 a 1 , a 2 , a 3 , a 4 〉 and b = 〈 b 1 , b 2 , b 3 , b 4 〉 are orthogonal if a ⋅ b = 0 , i.e., a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 = 0 Calculation: The dot product of two vectors u = ( 5 , − 4 , 0 , 3 ) and v = ( − 4 , 1 , − 3 , 7 ) is: u ⋅ v = ( 5 ) ( − 4 ) + ( − 4 ) ( 1 ) + ( 0 ) ( − 3 ) + ( 3 ) ( 7 ) u ⋅ v = − 20 − 4 + 0 + 21 u ⋅ v = − 24 + 21 u ⋅ v = − 3 Since u ⋅ v ≠ 0 , therefore the given vectors are not orthogonal. Conclusion: The given vectors are not orthogonal.

Solution Summary: The author explains that the given vectors are orthogonal if acdot b=0.

Elementary Linear Algebra: Applications Version

Elementary Linear Algebra: Applications Version

12th Edition

ISBN: 9781119282365

Author: Howard Anton, Chris Rorres, Anton Kaul

Publisher: WILEY

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Chapter 3.2, Problem 2WT

Chapter 3.3, Problem 2E

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In Exercises 1–4, let u = <-3, 4> and v = <2, -5>. Find (a) the component form of the vector and (b) its magnitude. 1. 3u - 4v 2. -2u 3. u + v 4. 5v

Assuming A (2,3, -1 (B (3,1,0), C (4,1,2 Determine the vectors AB, BC, AC.

Chapter 3 Solutions

Elementary Linear Algebra: Applications Version

Ch. 3.1 - In Exercises 12, find the components of the...Ch. 3.1 - In Exercises 12, find the components of the...Ch. 3.1 - In Exercises 34, find the components of the vector...Ch. 3.1 - In Exercises 34, find the components of the vector...Ch. 3.1 - a. Find the terminal point of the vector that is...Ch. 3.1 - Let u = (3, 1, 2), v = (4, 0, 8), and w = (6, 1,...Ch. 3.1 - Show that there do not exist scalars c1, c2, and...Ch. 3.2 - In Exercises 34, evaluate the given expression...Ch. 3.2 - Let v = (2, 3, 0, 6). Find all scalars k such that...Ch. 3.2 - Prob. 9E

Ch. 3.2 - Prob. 11E Ch. 3.3 - Prob. 1E Ch. 3.3 - In Exercises 36, find a point-normal form of the...Ch. 3.3 - In Exercises 36, find a point-normal form of the...Ch. 3.3 - In Exercises 710, determine whether the given...Ch. 3.3 - In Exercises 710, determine whether the given...Ch. 3.3 - In Exercises 1112, determine whether the given...Ch. 3.3 - In Exercises 1314, find projau. a. u = (1, 2), a =...Ch. 3.3 - In Exercises 1520, find the vector component of u...Ch. 3.3 - In Exercises 1520, find the vector component of u...

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Take this test to review the material in Chapters 4and Chapters 5. After you are finished, check your work against the answers in the back of the book. Write w=(7,2,4) as a linear combination of the vectors v1, v2 and v3 if possible. v1=(2,1,0), v2=(1,1,0), v3=(0,0,6)
Determine whether the vectors u=6,10 and v=5,3 are orthogonal.

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