Descripción del ejercicio 2 a) Dados los vectores representados en el siguiente gráfico, realizar los siguientes pasos:
SOLUCIÓN Nombrar cada uno de los vectores y encontrar la magnitud y dirección de los mismos.
•
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⃗ ==3,34,,155 ⃗ |⃗| = 4 + 1 ⃗ | | = 4 , 1 3 ⃗ 11= |⃗| sin |⃗3| = = sisinn 4, 1=3 sin− 3 4, 1 3 = 14,04 | | = 3 + 5 | | =5, 8 3 33= =| | coscos |3| = cos 5, =83 cos− 3 5, 8 3 ==cos59,−05,3383
Sean los vectores dados en la gráfica:
La magnitud del vector es:
La dirección del vector está dada por:
La magnitud del vector es:
La dirección del vector está dada por:
•
• Encontrar el ángulo entre los vectores.
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|⃗| | | cos 4,cos1335,5,=8334cos=s =+112⋅ +35 +5 cos= 24.708 = 73,79
El ángulo entre los dos vectores es: •
• Sumar los vectores y encontrar la magnitud y dirección del
vector resultante.
⃗= ⃗+ = 4+3 4+3 + 1 + 5 ⃗= ⃗+ = 11 + 6 |⃗| = 1 + 6 |⃗| =6,08 1= 1 = |⃗| cos =80,53
Suma de los vectores:
La magnitud del vector suma es:
La dirección del vector suma es:
•
• Encontrar el área del paralelogramo formado por los vectores
representados, con teoría vectorial.
El área del paralelogramo formado por los dos vectores se calcula con el producto cruz:
•
= ⃗× = sin = ⃗× = 4,135,83 sin73,79 = 23,12
• Comprobar y/o graficar los ítems anteriores, según corresponda,
en Geogebra, Matlab, Octave, Scilab, u otro programa similar.
⃗ = 3 4 +2 = 2 + 5 + 4
b) Dados los vectores calcular:
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333 4 +2 +22+5 + 4 99 12 +6 +4+10 + 8
•
•
52 + 14 6⃗ ⋅ 633 4 +2 ∙ 2 + 5 + 4 6 6 20 +8 36 120 +48 ⃗
• Calcular los cosenos directores de cada uno de los vectores.
La magnitud del vector es:
|⃗| = 3 + 4 + 2 |⃗| =5,38 ⃗ cos= |⃗3| =0,55 cos= 4|⃗| = 0,74 2|⃗| =0,37 cos= | | = 2 + 5 + 4 | | =6,71 cos= |2| = 0,0,29 cos= |5| = 0,74 cos= 4 0 59
Los cosenos directores del vector son:
La magnitud del vector es:
Los cosenos directores del vector son:
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⃗∙ = 33 4 +2 ∙ 2 + 5 + 4 ⃗∙ = 6 20 + 8 ⃗∙ = 6 ⃗× = 33 4 +2 × 2 + 5 + 44 ⃗ ⃗ ⃗× = 32 5 4 24 ⃗× = 45 24 ⃗ 32 24⃗ + 32 45 ⃗× = +44325 ⃗ ⃗ ( 3 4 ) ( 2 2 ) 5 42 ⃗× = 1610 1610⃗ 124 1 24⃗ + 15+8 1 5+8 ⃗× = 26⃗ 8⃗ + 23
Producto punto:
Producto cruz:
•
• Comprobar y/o graficar los ítems anteriores, según
corresponda, en Geogebra, Matlab, Octave, Scilab, u otro programa similar.
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Descripción del ejercicio 3 La velocidad de un cuerpo tiene inicialmente inicialmente el valor V1 = (5,-3) m/s, al instante t1 = 25. Después de transcurridos 4 segundos, la velocidad ha cambiado al valor V2 = (-4,8) m/s. • • ¿Cuánto vale el cambio de velocidad .?
∆ ∆ = ∆ =4+853 ∆ =9+11 ∆ = 9 +11 =9,59 ∆ +11 = 0,310,44 = ∆ = =|9|2925 =0,53 ⃗ ⃗ = 60 ⃗ ⃗ | | = 13 = 1 + √ ⃗∙ =5+12 cos= |⃗⃗∙| cos60 = |⃗⃗∙| = 13√ 135+12 √ 1 +
•
• ¿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 π
que la medida en radianes del ángulo y sea 3 La magnitud de es:
.
La magnitud de es:
La llamaremos ecuación 1
El producto punto es:
La fórmula del coseno entre dos vectores es:
Sustituimos la ecuación 1 en la fórmula del coseno entre dos vectores y tenemos:
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1 + = 1013 + 2413 1 + = 100+576 169 169 169100=576 69=407 =2.42
Nos da como resultado que:
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Descripción del ejercicio 4 Sean las siguientes matrices:
Realizar las siguientes operaciones, si es posible:
SOLUCIÓN:
a)
∙ ∙
Recordemos: Para poder multiplicar matrices es necesario que el número de columnas de la primera sea igual al número de filas de la segunda.
9 + 0 + 0 + 1 5 5 + 0 2 + 2 1 6 + 0 + 6 1 5 8+ 5+0+1+155 10+156 0+156+2+211 12+30+ 2+30+11815 815 18+5+0
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4∙36 220 24 2 0 4 6 4=2400 12428 122240 2=1204 1040 1626 1066 4∙ 2 ú ú 3∙7 6 3 3 5 4 2 0 6 9 1 5 3=123 09 1257 2142 7= 7= 7035 72419 324512 4 8 3 5 4 6 5 8 8 3∙3 ∙ 77 = 6152139 2154270 1573737 9 2 3 9 22 = 3 + 3 9 3+3 +36 + 3 + +22 + 2 9 15 + 18 12 16 12 ∙= ∙ = 4 3 3 2 6 95918+ 3 84+ 8+39+ 5 ∙ 0 4 1 b)
c)
d)
e)
f)
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h)
0 3 2 det = 31 0 +3 =9×3 +9×2 +2×2 9×2 × y
9 10 5 = 65 36 31 75 i)
=99
30 363 30 7 6 12 13 9 9 7 = 303 036 16123
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Descripción del ejercicio 5 Uno de los campos de mayor aplicación del algebra lineal es en la Robótica en el Modelado de la Cinemática de Robots. Para representar la posición y la l a orientación de un giro, se utilizan matrices y vectores. Sea el siguiente sistema de coordenadas tridimensional. En él se pueden hacer tres rotaciones: Rotación , Rotación en , Rotación en .
0
,
0
0
Haciendo la rotación, tomando al eje y como eje de giro, la matriz de rotación que se obtiene es:
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0+0+2 0 0 1 1 = ,, ∙ = 0 1 10 00 ∙ 12 = 1+0+0 0+1+0 2 = 11 1 = 23 =45 0 cos45 = √ sin45 = √ √ 22 0 √ 22 1 √ 22 + 0+3 √ 22 = ,, ∙ = √ 02 01 √ 02 ∙ 23 = √ 20+2+0 2 √ + 0+3 [2 2 ] [2 2 ] El vector nos da como resultado:
b) Encontrar el vector
respecto a eje Sabemos que: y
, cuando el punto ,
.
tenemos entonces:
El vector nos da como resultado:
con
, con