Espacios Con  Producto  Interno  Resumen
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Espacios Con Producto Interno Resumen

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Explicación del Producto Interno en vectores, así como de sus propiedades

Explicación del Producto Interno en vectores, así como de sus propiedades

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  • 1. Producto Interno y sus propiedades
  • 2. Producto Interno
    • En matemáticas el producto escalar, también conocido como producto interno, interior o punto, es una operación definida sobre un espacio vectorial cuyo resultado es una magnitud escalar.
  • 3. Propiedades
    • (u • u)>0
    u = 4i + 6j ( 4i + 6j ) • ( 4i + 6j ) > 0 si: u Є V u u (4)(4) + (6)(6) > 0 i j 16 + 36 > 0 52 > 0
  • 4.
    • (u)(v + w)=(u •v)+(u•w)
    u = 4i + 6j {u,v,w Є V} v = 3i - 7j w = i + j (4i + 6j) • [ (3i - 7j) + (i + j) ]=[ (4i + 6j) • (3i - 7j) ]+[ (4i + 6j) • (i + j) ] (4i + 6j) • [ 3i - 7j + i + j ]=[ (4i + 6j) • (3i - 7j) ]+[ (4i + 6j) • (i + j) ] (4i + 6j) • [ 4i - 6j ]=[ (4i + 6j) • (3i - 7j) ]+[ (4i + 6j) • (i + j) ] (4) (4) + (6) (-6) =[ (4) (3) + (6) (-7) ]+[ (4) (1) + (6) (1) ] 16 + (-36)=[12 + (-42)]+[4 + 6] -20 = -30 + 10 -20 = -20 i j i i j j
  • 5.
    • (u)(v+w)
    u = 4i + 6j v = 3i - 7j w = i + j i j i j -j -j (u•v)+(u•w)
  • 6. (u+v)(w)=(u •w)+(v•w) u = 4i + 6j {u,v,w Є V} v = 3i - 7j w = i + j [ (4i + 6j) + (3i - 7j) ] • (i + j) =[ (4i + 6j) • (i + j) ]+[ (3i - 7j) • (i + j) ] (7) (1) + (-1) (1) =10 + (-4) 7 + (-1)=10 - 4 7-1 = 6 6 = 6 [ 4i + 6j + 3i - 7j ] • (i + j) =[ (4) (1) + (6) (1) ]+[ (3) (1) + (-7) (1) ] [ 7i – j ] • (i + j) =[4 + 6]+[3 + (-7)] i j i i j j
  • 7. u = 4i + 6j v = 3i - 7j w = i + j i j -j i j -j (u+v)(w) (u •w)+(v•w)
  • 8.
    • (u • v)=(v • u)
    u = 4i + 6j {u,v Є V} v = 3i - 7j (4i + 6j) • (3i - 7j) = (4i + 6j) • (3i - 7j) (4i + 6j) • (3i - 7j) = (4i - 6j) • (3i + 7j) (4) (3) + (6) (-7) = (4) (3) + (-6) (7) 12 + (-42) = 12 + (-42) -30 = -30 i j i j 12 - 42 = 12 - 42
  • 9. u = 4i + 6j v = 3i - 7j u = 4i - 6j v = 3i + 7j i j -j i j -j
  • 10. u = 4i + 6j v = 3i - 7j u •v = -30 = u •u = (4i + 6j)(4i + 6j) = (4)(4)+(6)(6) = 16+36 = 52 = (3i - 7j)(3i - 7j) = (3)(3)+(-7)(-7) = 9+49 = 58 cos θ = -30 = -30 52 58 3016 θ = cos -1 (-0.54) θ = 123.11° cos θ = u • v u v | | | | | | | | u | | | | u | | | | u | | | | u | | | | u | | | | v | | | | v | | | | v | | | | v | | | |
  • 11. cos θ = -30 = -30 52 58 3016 θ = cos -1 (-0.54) θ = 123.11°
  • 12.
    • (u • α v)= α (u • v)
    u = 4i + 6j {u,v Є V} v = 3i - 7j (4i + 6j) • [ ( ½ ) (3i - 7j) ] = ( ½ )[ (4i + 6j) • (3i - 7j) ] (4i + 6j) • (3/2i – 7/2j) = ( ½ ) [ (4i - 6j) • (3i + 7j) ] (4) (3/2) + (6) (-7/2) = ( ½ ) [ (4) (3) + (-6) (7) ] 12/6 + (-42/2) = ( ½ ) (12 + (-42)) -15 = -15 i j i j 6 - 21 = ( ½ ) (12 – 42) α = 1/2
  • 13. (u • v) = 0 u = 3i - 1j {u,v Є V} v = 2i + 6j (3i - 1j) • (2i + 6j) = 0 (3) (2) + (-1) (6) = 0 6 + (-6) = 0 Resultado = 0 ES UN VECTOR ORTOGONAL i j 6 - 6 = 0
  • 14. = (3i - j)(3i – j) = (3)(3)+(1)(1) = 9+1 = 10 = 40 d = 10 + 40 d = 50 Teorema de Pitágoras u | | | | u | | | | u | | | | u | | | | v | | | |