Las componentes de u.v.w son u=(1,2,3) , v=(2,5,-4) , w=(1,1,3) halla
a) (u×v)×w
b) u×(v×w)
Porfavor ayuda
Respuestas a la pregunta
Dadas las componentes de los vectores u, v y w, se obtiene:
a) (u×v)×w = (29, -70, -33)
b) u×(v×w) = (27, -60, -28)
Explicación paso a paso:
Datos;
u=(1,2,3)
v=(2,5,-4)
w=(1,1,3)
a) (u×v)×w
Aplicar producto vectorial o cruz;
u×v = i[(2)(-4)-(5)(3)]-j[(1)(-4)-(2)(3)]+k[(1)(5)-(2)(2)]
u×v = (-23i +10j +k)
u×v = (-23, 10, 1)
Aplicar producto vectorial o cruz;
(u×v)×w = i[(10)(3)-(1)(1)]-j[(-23)(3)-(1)(1)]+k[(-23)(1)-(1)(10)]
(u×v)×w = (29i-70j-33k)
(u×v)×w = (29, -70, -33)
b) u×(v×w)
Aplicar producto vectorial o cruz;
v×w = i[(5)(3)-(1)(-4)]-j[(2)(3)-(1)(-4)]+k[(2)(1)-(1)(5)]
v×w = (19i +10j -3k)
v×w = (19, 10, -3)
Aplicar producto vectorial o cruz;
u×(v×w) = i[(2)(-3)-(10)(3)]-j[(1)(-3)-(19)(3)]+k[(1)(10)-(19)(2)]
u×(v×w) = (-36i+60j-28k)
u×(v×w) = (27, -60, -28)
Dadas las componentes de los vectores u, v y w, se obtiene:
a) (u×v)×w = (29, -70, -33)
b) u×(v×w) = (27, -60, -28)
Explicación paso a paso:
u=(1,2,3)
v=(2,5,-4)
w=(1,1,3)
a) (u×v)×w
Aplicar producto vectorial o cruz;
uxv= i j k
1 2 3
2 5 -4
u×v = i[(2)(-4)-(5)(3)]-j[(1)(-4)-(2)(3)]+k[(1)(5)-(2)(2)]
u×v = (-23i +10j +1k)
u×v = (-23, 10, 1)
Aplicar producto vectorial o cruz;
uxvxw= i j k
-23 10 1
1 1 3
(u×v)×w = i[(10)(3)-(1)(1)]-j[(-23)(3)-(1)(1)]+k[(-23)(1)-(1)(10)]
(u×v)×w = (29i+70j-33k)
(u×v)×w = (29, +70, -33)
b) u×(v×w)
Aplicar producto vectorial o cruz;
vxw= i j k
2 5 -4
1 1 3
v×w = i[(5)(3)-(1)(-4)]-j[(2)(3)-(1)(-4)]+k[(2)(1)-(1)(5)]
v×w = (19i -10j -3k)
v×w = (19, -10, -3)
Aplicar producto vectorial o cruz;
uxvxw= i j k
1 2 3
19 -10 -3
u×(v×w) = i[(2)(-3)-(-10)(3)]-j[(1)(-3)-(19)(3)]+k[(1)(-10)-(19)(2)]
u×(v×w) = (24i+60j-48k)
u×(v×w) = (24, 60, -48)