2,3,4.docx

Ejercicio 2: Resolución de problemas básicos sobre vectores en R2 y R3 Para el desarrollo de los ejercicios 2 y 3, debe revisar los siguientes contenidos ubicados en el entorno de Conocimiento de la Unidad 1. 1.

Descripción del ejercicio 2 a) Dados los vectores representados en el siguiente gráfico, realizar los siguientes pasos:



Nombrar cada uno de los vectores y encontrar la magnitud y dirección de estos. VECTORES:

 ⃗ = 4,1   = 3,5 ;

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|| = =  44 + 1 ≈ 4,1231 || = =  3 + 5 ≈ 5,8309

Luego de obtener la magnitud, se determina el Angulo con la tangente inversa de componente en y sobre su componente en x

 = tan− 41  = 14.036 + 180 = 165.9637  = tan−  53 = 59.0362 

Encontrar el ángulo entre los vectores.

⃗       ∙  −  = cos | | |

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 = cos− 177√ 2 = 106.925 

Sumar los vectores y encontrar la magnitud y dirección del vector resultante.

Magnitud:

Dirección:

 ⃗   =    +  4,4,1 + 3,3, 5 = 1,6   = 1,6 || =  11 + 6 || ≈ 6.0828  = tan−  61 = 80.5377 + 180 = 99.4623

Encontrar el área del paralelogramo formado por los vectores representados, con teoría vectorial. Para determinar el área del paralelogramo se multiplica la magnitud de los dos vectores de siguiente manera: 

  = || ∗ || || ||  == 5,4,81231309   = 4,1231 1231 5,8309 ≈ 24,042



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⃗= 3  4 + 2  = 2 + 5 + 4

b) Dados los vectores  calcular:  para realizar la operación de los vectores es necesario

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



3⃗+2 3 = 333  4 + 2  = 9 + 12  6 2 = 222 +5+ 5 +4+ 4 = 4 +10+ 10 +8+ 8 3⃗ +2 +312⃗ +2 6 +4=+ 45++2210 ++28 = 5 + 22 + 2 + 2 = 9 +12 6⃗.  6⃗.   = 6(3 4 4 +2+ 2 ∙ 22 + 5 +4+ 4) = 66620+8 6 20+8 = 36

⃗ = 3  4 +2 || =  3 + 44 + 2 → || = = √ 2929 ≈ 5.3852  = 3; = 4;4;  = 2  3

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cos = || = 5.32852 ≈ 0.3714  = 2 +5+ 5 +4+ 4 = || = =  2 + 5 + 4 = 3√5 ≈ 6.7082  = 2;2;  = 5;5;  = 4 cos = || = 6.72082 ≈ 0.2981 cos = || = 6.75082 ≈ 0.7454 cos = || = 6.74082 ≈ 0.5963

Para el vector W de igual forma que con el anterior se calcula



Calcular el producto cruz y el producto punto. PRODUCTO CRUZ

    = = 32 45 24 = 〈1610〉 1610〉 〈 〈124〉 1 24〉  +〈+ 〈15+8〉 1 5+8〉 ⃗ = 26 8 +23 . = = 〈33  4 +2+ 2〉 ∙ 〈2 +5+ 5 +4+ 4〉 . = = 32 + 445 + 24 = 6

PRODUCTO PUNTO



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Ejercicio 3: Resolución de problemas básicos sobre vectores en R2 y R3 Descripción del ejercicio 3

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1 = 5,3 2 ==25 4,8 = 29 Vector 1: Vector 2:

∆ ∆  = 21 ∆  = 9,11 ∆∆ = 9,411 ∆∆ = 94 , 11141 



¿Cuánto vale el cambio de velocidad



¿Cuál es la variación de la velocidad por unidad de tiempo?



 . ?

Hallar módulo, dirección, y sentido del siguiente vector. Dados:

⃗



 ⃗



 = (5, 12) y  = (1, k), donde k es un escalar, encuentre (k) tal que la medida en radianes del ángulo  y  sea .

Para hallar el módulo es necesario encontrar la magnitud de los vectores a y b:

|| =  5 + 12 → || = 13 || =  1 +  → || =  1 +  cos− 13√ 13 5 √ +112+  = 3

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2691 +  = 100 + 480 480 +576 + 576 407  + 480  69 = 0  = 0.1295

El valor de k queda entonces.

Ejercicio 4: Resolución de problemas básicos sobre matrices y determinantes. Desarrolla los siguientes ejercicios luego de leer le er detenidamente los conceptos de la unidad 1, referentes a matrices, operaciones con matrices y determinantes. Presentar la solución con editor de ecuaciones.

Descripción del ejercicio 4 Sean las siguientes matrices:

6 1 0 2 3 9 5  2 0 3 0 2 3 5 2 6 3 5 6 1 3   = =  15 02 33 80  = 05 17 53   = 14 30 95 48  =  31 0  +3    ∙∙  ∙∙  6 1 0 2 3 9 5 0 2 3 5 2 6 3 5 6 1 3  =  ∙  ∙  = =  1 0 3 8 ∙ 0 1 3  ∙ 14 30 95 48 Realizar las siguientes operaciones, si es posible:

a)

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b)

c)

d)

4 ∙ 2

6 1 0 2 3 9 5 2 6 5 1 3 6 4 ∙ 05 17 53  ∙ 2  15 02 33 380 →    . 3 ∙7 6 9 5 0 2 3 5 3 ∙∙ 7 7 = 3 14 30 95 48.7 7 105 137 563  0 2 0 6 3 5 9 15 3 = 3 14 30 95 48 = 312 90   2715 1224 6 42 9 5  63 35 6 7 1 3 21 7 = 77 05 17 53  = 350 497 422135  42 35 312 69   159 1512  ∙  63 7 21 42  7 0 21 3 0 27 24 35 49 35  4 8 3  5 4 6 5 8 8   3 ∙∙ 7 7  = 1239  252 777 651 1470 1533    2 0 3   =  =  3  3 

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f)

g)

12 43+ 2 39  6 155 + 18118 8 142  + 3916  ++ 2 9  9 + 3 8 + 8 + 5  ∙   =0  4∙  1  = 235 543 908  0 4  1   ∙  = 23 53 90  ∙ 03 3 2+ 3   +  548 10 191 34 6  13 + 12 620 159 + +54 998+8+89++29  6 9 5  105 137 563 

No es posible hallar el determinante de la matriz porque no es cuadrada.

h)



 2 0 3    3 3  = 9 + 9 1 1 2

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9  9 7    = 330 603 13126 

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