¿Alguien que sepa desarrollar estos ejercicios sobre la unidad de vectores, en geometría y trigonometría?
Respuestas
Al efectuar las operaciones se obtiene:
1. (26, 46)
2. u = (√34, 30.96°); y = (8, 70°); v = (√85, 77.47°); w = (5√2, 81.87°)
3. x = (4.69, 1.71) ; y = (2.73, 7.51)
4. u·v = 37
5. x·y = 25.71
6. uₙ = (5, 3)/√34 ; vₙ = (2, 9)/√85
7. Θ_uv = 46.51° ; Θ_vw = 4.4° ; Θ_xy = 50°
8. u×v = 39k ; w×y = -11.6k
9. 5u+4x-7y = (51.65, 30.73)
10. 9u+8w-12y = (20.24, -7.12)
Explicación:
Datos;
u = (5, 3)
v = (2, 9)
w = (1, 7)
x = (5, 20°)
y = (8, 70°)
1. 3u - 5w + 8v
3u = 3(5, 3) = (15,9)
5w = 5(1, 7) = (5, 35)
8v = 8(2, 9) = (16, 72)
Sustituir;
3u - 5w + 8v = (15,9) - (5, 35) + (16, 72)
3u - 5w + 8v = (15-5+16, 9-35+72)
3u - 5w + 8v = (26, 46)
2. Forma polar u, v, y, w;
(modulo, ángulo) = (√[(x)²+(y)²] , Θ = tan⁻¹(y/x))
|u| =√[(5)²+(3)²] = √34
Θ = tan⁻¹(3/5) = 30.96°
u = (√34, 30.96°)
|v| =√[(2)²+(9)²] = √85
Θ = tan⁻¹(9/2) = 77.47°
v = (√85, 77.47°)
|w| =√[(1)²+(7)²] = 5√2
Θ = tan⁻¹(7) = 81.87°
w = (5√2, 81.87°)
3. Forma cartesiana (rcos(Θ), rsen(Θ));
x = (5, 20°) = (5cos(20°), 5sen(20°)) = (4.69, 1.71)
y = (8, 70°) = (8cos(70°), 8sen(70°)) = (2.73, 7.51)
4. u·v
Producto escalar;
u·v = (5, 3) · (2, 9)
u·v = (5×2)+(3×9)
u·v = 10 + 27 = 37
5. x·y
Producto escalar;
x·y = (5cos(20°), 5sen(20°)) · (8cos(70°), 8sen(70°))
x·y = (5cos(20°)×8cos(70°))+(5sen(20°)×8sen(70°))
x·y = 25.71
6. Normalizar los vectores;
A = r/|r|
Sustituir;
uₙ = (5, 3)/√34
vₙ = (2, 9)/√85
7. Ángulos de separación;
Θ_uv = Θ_v - Θ_u
Sustituir;
Θ_uv = 77.47° - 30.96°
Θ_uv = 46.51°
Θ_vw = 81.87° - 77.47°
Θ_vw = 4.4°
Θ_xy = 70° - 20°
Θ_xy = 50°
8. u×v ; w×y
Producto vectorial;
u×v = (45-6)k
u×v = 39k
w×y = (7.51-19.11)k
w×y = -11.6k
9. 5u+4x-7y
= 5(5, 3)+ 4(4.69, 1.71) - 7(2.73, 7.51)
= (25+18.76-19.11, 15+6.84-52.57)
= (51.65, 30.73)
10. 9u+8w-12y
= 9(5, 3)+8(1, 7)-12(2.73, 7.51)
= (45+8-32.76, 27+56-90.12)
= (20.24, -7.12)