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Introductory Linear Algebra por Bernard Kolman 8/E

Module by: Julian Daza. E-mail the author

Summary: Algunos ejercicios resueltos sobre algebra lineal de Beernard Kolman

kol2005pgn245ejc21.tm por Julian Camilo Daza

21. Find de cosine of the angle betwen each pair of vectors u and v .

(a) u=(2,3,1)v=(3,-2,0)u=(2,3,1)v=(3,-2,0)

cos θ = | u • v 〚 u 〛 〚 v 〛 | cos θ = | u • v 〚 u 〛 〚 v 〛 |
(1)

 sage: u=vector([2,3,1])

 sage: v=vector([3,-2,0])

 sage: u.dot_product(v)

       0

 sage: u.dot_product(u)

       14

 sage: v.dot_product(v)

       13

cos θ = | 0 〚 u 〛 〚 v 〛 | cos θ = | 0 〚 u 〛 〚 v 〛 |
(2)

El angulo es 0.

(b) u=(1,2,-1,3)v=(0,0,-1,-2)u=(1,2,-1,3)v=(0,0,-1,-2)

 sage: u=vector([1,2,-1,3])

 sage: v=vector([0,0,-1,-2])

 sage: u.dot_product(v)

       -5

 sage: u.dot_product(u)

       15

 sage: v.dot_product(v)

       5

cosθ=|-553|cosθ=|-553|=|-13||-13|

θ = cos - 1 | - 1 3 | = 125 , 26 θ = cos - 1 | - 1 3 | = 125 , 26

(c) u=(2,0,1)v=(2,2,0)u=(2,0,1)v=(2,2,0)

 sage: u=vector([2,0,1])

 sage: v=vector([2,2,-1])

 sage: u.dot_product(v)

       3

 sage: u.dot_product(u)

       5

 sage: v.dot_product(v)

       9

cosθ=|335|cosθ=|335|=|15||15|

θ = cos - 1 | 1 5 | = 63 . 43 θ = cos - 1 | 1 5 | = 63 . 43

(d) u=(0,4,2,3)v=(0,-1,2,0)u=(0,4,2,3)v=(0,-1,2,0)

 sage: u=vector([0,4,2,3])

 sage: v=vector([0,-1,2,0])

 sage: u.dot_product(v)

       0

 sage: u.dot_product(u)

       29

 sage: v.dot_product(v)

       5

cos θ = | 0 〚 u 〛 〚 v 〛 | cos θ = | 0 〚 u 〛 〚 v 〛 |
(3)

El angulo es 0

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