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Math Algebra Elementary Linear Algebra: Applications Version

(a) Find u ⋅ v u ⋅ u , u ⋅ v and v ⋅ v The dot products are u ⋅ u = 26 v ⋅ v = 24 u ⋅ v = − 8 Given: The vectors are equivalent to u = ( 3 , 1 , 4 ) and v = ( 2 , 2 , − 4 ) Concept used: 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 Calculation: For u ⋅ u , Put u = ( x 1 , y 1 , z 1 ) = ( 3 , 1 , 4 ) and u = ( x 2 , y 2 , z 2 ) = ( 3 , 1 , 4 ) u ⋅ u = ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) ⇒ u ⋅ u = ( 3 , 1 , 4 ) ⋅ ( 3 , 1 , 4 ) ⇒ u ⋅ u = ( 3 × 3 ) + ( 1 × 1 ) + ( 4 × 4 ) ............. [ ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ u = 9 + 1 + 16 ⇒ u ⋅ u = 26 For v ⋅ v , Put v = ( x 1 , y 1 , z 1 ) = ( 2 , 2 , − 4 ) and v = ( x 2 , y 2 , z 2 ) = ( 2 , 2 , − 4 ) v ⋅ v = ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) ⇒ v ⋅ v = ( 2 , 2 , − 4 ) ⋅ ( 2 , 2 , − 4 ) ⇒ v ⋅ v = ( 2 × 2 ) + ( 2 × 2 ) + ( − 4 × − 4 ) ............. [ ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ v ⋅ v = 4 + 4 + 16........................................ [ ( − x ) × ( − y ) = + x y ] ⇒ v ⋅ v = 24 For u ⋅ v , Put u = ( x 1 , y 1 , z 1 ) = ( 3 , 1 , 4 ) and v = ( x 2 , y 2 , z 2 ) = ( 2 , 2 , − 4 ) u ⋅ v = ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) ⇒ u ⋅ v = ( 3 , 1 , 4 ) ⋅ ( 2 , 2 , − 4 ) ⇒ u ⋅ v = ( 3 × 2 ) + ( 1 × 2 ) + ( 4 × − 4 ) ............. [ ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ v = 6 + 2 + ( − 16 ) ........................................ [ ( − x ) × ( + y ) = − x y ] ⇒ u ⋅ v = − 8 Conclusion: Therefore, the dot products of the given vectors are u ⋅ u = 26 v ⋅ v = 24 u ⋅ v = − 8 (b) Find u ⋅ v u ⋅ u , u ⋅ v and v ⋅ v The dot products are u ⋅ u = 54 v ⋅ v = 21 u ⋅ v = 0 Given: The vectors are equivalent to u = ( 1 , 1 , 4 , 6 ) and v = ( 2 , − 2 , 3 , − 2 ) Concept used: If two vectors are ( w 1 , x 1 , y 1 , z 1 ) and ( w 2 , x 2 , y 2 , z 2 ) , then their dot product is w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 Calculation: For u ⋅ u , Put u = ( w 1 , x 1 , y 1 , z 1 ) = ( 1 , 1 , 4 , 6 ) and u = ( w 2 , x 2 , y 2 , z 2 ) = ( 1 , 1 , 4 , 6 ) u ⋅ u = ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) ⇒ u ⋅ u = ( 1 , 1 , 4 , 6 ) ⋅ ( 1 , 1 , 4 , 6 ) ⇒ u ⋅ u = ( 1 × 1 ) + ( 1 × 1 ) + ( 4 × 4 ) + ( 6 × 6 ) ............. [ ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) = w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ u = 1 + 1 + 16 + 36 ⇒ u ⋅ u = 54 For v ⋅ v , Put v = ( w 1 , x 1 , y 1 , z 1 ) = ( 2 , − 2 , 3 , − 2 ) and v = ( w 2 , x 2 , y 2 , z 2 ) = ( 2 , − 2 , 3 , − 2 ) v ⋅ v = ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) ⇒ v ⋅ v = ( 2 , − 2 , 3 , − 2 ) ⋅ ( 2 , − 2 , 3 , − 2 ) ⇒ v ⋅ v = ( 2 × 2 ) + ( − 2 × − 2 ) + ( 3 × 3 ) + ( − 2 × − 2 ) ............. [ ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) = w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ v ⋅ v = 4 + 4 + 9 + 4...................................................... [ ( − x ) × ( − y ) = x y ] ⇒ v ⋅ v = 21 For u ⋅ v , Put u = ( w 1 , x 1 , y 1 , z 1 ) = ( 1 , 1 , 4 , 6 ) and v = ( w 2 , x 2 , y 2 , z 2 ) = ( 2 , − 2 , 3 , − 2 ) u ⋅ v = ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) ⇒ u ⋅ v = ( 1 , 1 , 4 , 6 ) ⋅ ( 2 , − 2 , 3 , − 2 ) ⇒ u ⋅ v = ( 1 × 2 ) + ( 1 × − 2 ) + ( 4 × 3 ) + ( 6 × − 2 ) ............. [ ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) = w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ v = 2 − 2 + 12 − 12...................................................... [ ( + x ) × ( − y ) = − x y ] ⇒ u ⋅ v = 0 Conclusion: Therefore, the dot products of the given vectors are u ⋅ u = 54 v ⋅ v = 21 u ⋅ v = 0

(a) Find u ⋅ v u ⋅ u , u ⋅ v and v ⋅ v The dot products are u ⋅ u = 26 v ⋅ v = 24 u ⋅ v = − 8 Given: The vectors are equivalent to u = ( 3 , 1 , 4 ) and v = ( 2 , 2 , − 4 ) Concept used: 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 Calculation: For u ⋅ u , Put u = ( x 1 , y 1 , z 1 ) = ( 3 , 1 , 4 ) and u = ( x 2 , y 2 , z 2 ) = ( 3 , 1 , 4 ) u ⋅ u = ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) ⇒ u ⋅ u = ( 3 , 1 , 4 ) ⋅ ( 3 , 1 , 4 ) ⇒ u ⋅ u = ( 3 × 3 ) + ( 1 × 1 ) + ( 4 × 4 ) ............. [ ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ u = 9 + 1 + 16 ⇒ u ⋅ u = 26 For v ⋅ v , Put v = ( x 1 , y 1 , z 1 ) = ( 2 , 2 , − 4 ) and v = ( x 2 , y 2 , z 2 ) = ( 2 , 2 , − 4 ) v ⋅ v = ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) ⇒ v ⋅ v = ( 2 , 2 , − 4 ) ⋅ ( 2 , 2 , − 4 ) ⇒ v ⋅ v = ( 2 × 2 ) + ( 2 × 2 ) + ( − 4 × − 4 ) ............. [ ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ v ⋅ v = 4 + 4 + 16........................................ [ ( − x ) × ( − y ) = + x y ] ⇒ v ⋅ v = 24 For u ⋅ v , Put u = ( x 1 , y 1 , z 1 ) = ( 3 , 1 , 4 ) and v = ( x 2 , y 2 , z 2 ) = ( 2 , 2 , − 4 ) u ⋅ v = ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) ⇒ u ⋅ v = ( 3 , 1 , 4 ) ⋅ ( 2 , 2 , − 4 ) ⇒ u ⋅ v = ( 3 × 2 ) + ( 1 × 2 ) + ( 4 × − 4 ) ............. [ ( x 1 , y 1 , z 1 ) ⋅ ( x 2 , y 2 , z 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ v = 6 + 2 + ( − 16 ) ........................................ [ ( − x ) × ( + y ) = − x y ] ⇒ u ⋅ v = − 8 Conclusion: Therefore, the dot products of the given vectors are u ⋅ u = 26 v ⋅ v = 24 u ⋅ v = − 8 (b) Find u ⋅ v u ⋅ u , u ⋅ v and v ⋅ v The dot products are u ⋅ u = 54 v ⋅ v = 21 u ⋅ v = 0 Given: The vectors are equivalent to u = ( 1 , 1 , 4 , 6 ) and v = ( 2 , − 2 , 3 , − 2 ) Concept used: If two vectors are ( w 1 , x 1 , y 1 , z 1 ) and ( w 2 , x 2 , y 2 , z 2 ) , then their dot product is w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 Calculation: For u ⋅ u , Put u = ( w 1 , x 1 , y 1 , z 1 ) = ( 1 , 1 , 4 , 6 ) and u = ( w 2 , x 2 , y 2 , z 2 ) = ( 1 , 1 , 4 , 6 ) u ⋅ u = ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) ⇒ u ⋅ u = ( 1 , 1 , 4 , 6 ) ⋅ ( 1 , 1 , 4 , 6 ) ⇒ u ⋅ u = ( 1 × 1 ) + ( 1 × 1 ) + ( 4 × 4 ) + ( 6 × 6 ) ............. [ ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) = w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ u = 1 + 1 + 16 + 36 ⇒ u ⋅ u = 54 For v ⋅ v , Put v = ( w 1 , x 1 , y 1 , z 1 ) = ( 2 , − 2 , 3 , − 2 ) and v = ( w 2 , x 2 , y 2 , z 2 ) = ( 2 , − 2 , 3 , − 2 ) v ⋅ v = ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) ⇒ v ⋅ v = ( 2 , − 2 , 3 , − 2 ) ⋅ ( 2 , − 2 , 3 , − 2 ) ⇒ v ⋅ v = ( 2 × 2 ) + ( − 2 × − 2 ) + ( 3 × 3 ) + ( − 2 × − 2 ) ............. [ ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) = w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ v ⋅ v = 4 + 4 + 9 + 4...................................................... [ ( − x ) × ( − y ) = x y ] ⇒ v ⋅ v = 21 For u ⋅ v , Put u = ( w 1 , x 1 , y 1 , z 1 ) = ( 1 , 1 , 4 , 6 ) and v = ( w 2 , x 2 , y 2 , z 2 ) = ( 2 , − 2 , 3 , − 2 ) u ⋅ v = ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) ⇒ u ⋅ v = ( 1 , 1 , 4 , 6 ) ⋅ ( 2 , − 2 , 3 , − 2 ) ⇒ u ⋅ v = ( 1 × 2 ) + ( 1 × − 2 ) + ( 4 × 3 ) + ( 6 × − 2 ) ............. [ ( w 1 , x 1 , y 1 , z 1 ) ⋅ ( w 2 , x 2 , y 2 , z 2 ) = w 1 w 2 + x 1 x 2 + y 1 y 2 + z 1 z 2 ] ⇒ u ⋅ v = 2 − 2 + 12 − 12...................................................... [ ( + x ) × ( − y ) = − x y ] ⇒ u ⋅ v = 0 Conclusion: Therefore, the dot products of the given vectors are u ⋅ u = 54 v ⋅ v = 21 u ⋅ v = 0

Solution Summary: The author explains that if two vectors are (3,1,4) and

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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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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