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Math Algebra Elementary Linear Algebra: Applications Version

(a) Find Euclidean distance and cosine between vectors v and u. Also determine whether the cosine is acute, obtuse or 90 ° . Distance between given vectors d = 14 ≈ 3.74 . Cosine between given vectors cos θ = 5 51 51 ≈ 0.7 . Angle between given vectors is acute. Given: The vectors are equivalent to u = ( 3 , 3 , 3 ) and, v = ( 1 , 0 , 4 ) . Concept used: For vectors x = ( x 1 , x 2 , x 3 ) and y = ( y 1 , y 2 , y 3 ) , Euclidean distance between them is given by d = ( y 1 − x 1 ) 2 + ( y 2 − x 2 ) 2 + ( y 3 − x 3 ) 2 . Angle θ between two vectors x and y is given by: cos θ = x → ⋅ y → | x → | | y → | If two vectors are ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) , then their dot product is x 1 x 2 + y 1 y 2 + z 1 z 2 . For a vector x = ( x 1 , x 2 , x 3 ) , | x | = ( x 1 ) 2 + ( x 2 ) 2 + ( x 3 ) 2 . Calculation: Euclidean distance between u and v is given by: d = ( 1 − 3 ) 2 + ( 0 − 3 ) 2 + ( 4 − 3 ) 2 ⇒ d = ( − 2 ) 2 + ( − 3 ) 2 + 1 2 ⇒ d = 4 + 9 + 1 ⇒ d = 14 ≈ 3.74 For cosine of the angle between u and v , we have: cos θ = u → ⋅ v → | u → | | v → | ⇒ cos θ = ( 3 , 3 , 3 ) ⋅ ( 1 , 0 , 4 ) | ( 3 , 3 , 3 ) | | ( 1 , 0 , 4 ) | ⇒ cos θ = 3 × 1 + 3 × 0 + 3 × 4 3 2 + 3 2 + 3 2 1 2 + 0 2 + 4 2 ⇒ cos θ = 3 + 12 3 ⋅ 3 2 1 + 16 ⇒ cos θ = 15 3 3 17 ⇒ cos θ = 15 3 51 ⇒ cos θ = 5 51 ......................... [ cancelling out 3 from the numerator and denominator] ⇒ cos θ = 5 51 51 ....................... [ rationalize the denominator] ⇒ cos θ ≈ 0.7 Now, θ = cos − 1 ( 0.7 ) ⇒ θ = ± 45 .572996 ° So, the angle between the vectors u and v is acute. Conclusion: So, Distance between given vectors d = 14 ≈ 3.74 . Cosine between given vectors cos θ = 5 51 51 ≈ 0.7 . Angle between given vectors is acute. (b) Find Euclidean distance and cosine between vectors v and u . Also determine whether the cosine is acute, obtuse or 90 ° . Distance between given vectors d = 59 ≈ 7.68 . Cosine between given vectors ⇒ cos θ ≈ − 0.123 . Angle between given vectors is obtuse. Given: The vectors are equivalent to u = ( 0 , − 2 , − 1 , 1 ) and v = ( − 3 , 2 , 4 , 4 ) Concept used: For vectors x = ( x 1 , x 2 , x 3 ) and y = ( y 1 , y 2 , y 3 ) , Euclidean distance between them is given by d = ( y 1 − x 1 ) 2 + ( y 2 − x 2 ) 2 + ( y 3 − x 3 ) 2 . Angle θ between two vectors x and y is given by: cos θ = x → ⋅ y → | x → | | y → | If two vectors are ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) , then their dot product is x 1 x 2 + y 1 y 2 + z 1 z 2 . For a vector x = ( x 1 , x 2 , x 3 ) , | x | = ( x 1 ) 2 + ( x 2 ) 2 + ( x 3 ) 2 . Calculation: Euclidean distance between u and v is given by : d = ( − 3 − 0 ) 2 + ( 2 − ( − 2 ) ) 2 + ( 4 − ( − 1 ) ) 2 + ( 4 − 1 ) 2 ⇒ d = ( − 3 ) 2 + ( 4 ) 2 + 5 2 + 3 2 ⇒ d = 9 + 16 + 25 + 9 ⇒ d = 59 ≈ 7.68 For cosine of the angle between u and v , we have : cos θ = u → ⋅ v → | u → | | v → | ⇒ cos θ = ( 0 , − 2 , − 1 , 1 ) ⋅ ( − 3 , 2 , 4 , 4 ) | ( 0 , − 2 , − 1 , 1 ) | | ( − 3 , 2 , 4 , 4 ) | ⇒ cos θ = 0 × ( − 3 ) + ( − 2 ) × 2 + ( − 1 ) × 4 + 1 × 4 0 2 + ( − 2 ) 2 + ( − 1 ) 2 + 1 2 ( − 3 ) 2 + 2 2 + 4 2 + 4 2 ⇒ cos θ = − 2 − 4 + 4 4 + 1 + 1 9 + 4 + 16 + 16 ⇒ cos θ = − 2 6 ⋅ 3 5 ⇒ cos θ = − 2 3 30 ⇒ cos θ = − 2 30 3 ⋅ 30 ....................... [ rationalize the denominator] ⇒ cos θ = − 30 3 ⋅ 15 ⇒ cos θ = − 30 45 ⇒ cos θ ≈ − 0.123 Now, θ = cos − 1 ( − 0.123 ) ⇒ θ = ± 97 .065273 ° So, the angle between the vectors u and v is obtuse. Conclusion: So, Distance between given vectors d = 59 ≈ 7.68 . Cosine between given vectors ⇒ cos θ ≈ − 0.123 . Angle between given vectors is obtuse.

(a) Find Euclidean distance and cosine between vectors v and u. Also determine whether the cosine is acute, obtuse or 90 ° . Distance between given vectors d = 14 ≈ 3.74 . Cosine between given vectors cos θ = 5 51 51 ≈ 0.7 . Angle between given vectors is acute. Given: The vectors are equivalent to u = ( 3 , 3 , 3 ) and, v = ( 1 , 0 , 4 ) . Concept used: For vectors x = ( x 1 , x 2 , x 3 ) and y = ( y 1 , y 2 , y 3 ) , Euclidean distance between them is given by d = ( y 1 − x 1 ) 2 + ( y 2 − x 2 ) 2 + ( y 3 − x 3 ) 2 . Angle θ between two vectors x and y is given by: cos θ = x → ⋅ y → | x → | | y → | If two vectors are ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) , then their dot product is x 1 x 2 + y 1 y 2 + z 1 z 2 . For a vector x = ( x 1 , x 2 , x 3 ) , | x | = ( x 1 ) 2 + ( x 2 ) 2 + ( x 3 ) 2 . Calculation: Euclidean distance between u and v is given by: d = ( 1 − 3 ) 2 + ( 0 − 3 ) 2 + ( 4 − 3 ) 2 ⇒ d = ( − 2 ) 2 + ( − 3 ) 2 + 1 2 ⇒ d = 4 + 9 + 1 ⇒ d = 14 ≈ 3.74 For cosine of the angle between u and v , we have: cos θ = u → ⋅ v → | u → | | v → | ⇒ cos θ = ( 3 , 3 , 3 ) ⋅ ( 1 , 0 , 4 ) | ( 3 , 3 , 3 ) | | ( 1 , 0 , 4 ) | ⇒ cos θ = 3 × 1 + 3 × 0 + 3 × 4 3 2 + 3 2 + 3 2 1 2 + 0 2 + 4 2 ⇒ cos θ = 3 + 12 3 ⋅ 3 2 1 + 16 ⇒ cos θ = 15 3 3 17 ⇒ cos θ = 15 3 51 ⇒ cos θ = 5 51 ......................... [ cancelling out 3 from the numerator and denominator] ⇒ cos θ = 5 51 51 ....................... [ rationalize the denominator] ⇒ cos θ ≈ 0.7 Now, θ = cos − 1 ( 0.7 ) ⇒ θ = ± 45 .572996 ° So, the angle between the vectors u and v is acute. Conclusion: So, Distance between given vectors d = 14 ≈ 3.74 . Cosine between given vectors cos θ = 5 51 51 ≈ 0.7 . Angle between given vectors is acute. (b) Find Euclidean distance and cosine between vectors v and u . Also determine whether the cosine is acute, obtuse or 90 ° . Distance between given vectors d = 59 ≈ 7.68 . Cosine between given vectors ⇒ cos θ ≈ − 0.123 . Angle between given vectors is obtuse. Given: The vectors are equivalent to u = ( 0 , − 2 , − 1 , 1 ) and v = ( − 3 , 2 , 4 , 4 ) Concept used: For vectors x = ( x 1 , x 2 , x 3 ) and y = ( y 1 , y 2 , y 3 ) , Euclidean distance between them is given by d = ( y 1 − x 1 ) 2 + ( y 2 − x 2 ) 2 + ( y 3 − x 3 ) 2 . Angle θ between two vectors x and y is given by: cos θ = x → ⋅ y → | x → | | y → | If two vectors are ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) , then their dot product is x 1 x 2 + y 1 y 2 + z 1 z 2 . For a vector x = ( x 1 , x 2 , x 3 ) , | x | = ( x 1 ) 2 + ( x 2 ) 2 + ( x 3 ) 2 . Calculation: Euclidean distance between u and v is given by : d = ( − 3 − 0 ) 2 + ( 2 − ( − 2 ) ) 2 + ( 4 − ( − 1 ) ) 2 + ( 4 − 1 ) 2 ⇒ d = ( − 3 ) 2 + ( 4 ) 2 + 5 2 + 3 2 ⇒ d = 9 + 16 + 25 + 9 ⇒ d = 59 ≈ 7.68 For cosine of the angle between u and v , we have : cos θ = u → ⋅ v → | u → | | v → | ⇒ cos θ = ( 0 , − 2 , − 1 , 1 ) ⋅ ( − 3 , 2 , 4 , 4 ) | ( 0 , − 2 , − 1 , 1 ) | | ( − 3 , 2 , 4 , 4 ) | ⇒ cos θ = 0 × ( − 3 ) + ( − 2 ) × 2 + ( − 1 ) × 4 + 1 × 4 0 2 + ( − 2 ) 2 + ( − 1 ) 2 + 1 2 ( − 3 ) 2 + 2 2 + 4 2 + 4 2 ⇒ cos θ = − 2 − 4 + 4 4 + 1 + 1 9 + 4 + 16 + 16 ⇒ cos θ = − 2 6 ⋅ 3 5 ⇒ cos θ = − 2 3 30 ⇒ cos θ = − 2 30 3 ⋅ 30 ....................... [ rationalize the denominator] ⇒ cos θ = − 30 3 ⋅ 15 ⇒ cos θ = − 30 45 ⇒ cos θ ≈ − 0.123 Now, θ = cos − 1 ( − 0.123 ) ⇒ θ = ± 97 .065273 ° So, the angle between the vectors u and v is obtuse. Conclusion: So, Distance between given vectors d = 59 ≈ 7.68 . Cosine between given vectors ⇒ cos θ ≈ − 0.123 . Angle between given vectors is obtuse.

Solution Summary: The author analyzes the Euclidean distance and cosine between vectors v and u.

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 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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