Actividad de aprendizaje 2. Vectores en Rn

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Ejercicios resueltos de Vectores en Rn para algebra lineal

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1- Bases geométricas.docx

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Actividad 2, Operaciones Con

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ALUMNO: JUAN GUILLERMO MARTÍNEZ GÓMEZ

MATRÍCULA 83931

GRUPO: K040 MATERIA: ALGEBRA LINEAL

Master your semester with Scribd Free Foron 30this Days Sign to vote title DOCENTE: MTRO. MANUEL DERead LAupROSA LÓPEZ & The New York Times  Useful  Not useful Special offer for students: Only $4.99/month.

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1.- Sean u=(-1,1,2), v=(2,0,3) y w=(-1,3,9). Hallar en forma de coordenadas resultado de las siguientes operaciones: a) u + v - w b) 6u + 2v - 2w

c)⅓ (v + 3u - 4w) d) 4(u + v) - (v – w + u)

a) u+v-w (-1,1,2)+(2,0,3)-(-1,3,9) (-1,1,2)+(2,0,3)= (1,1,5) (1,1,5) – (-1,3,9) = (2,-2,4) You're Reading a Preview

b) 6u+2v-2w 6(-1,1,2)+2(2,0,3)-2(-1,3,9)

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6(-1,1,2)= (-6,6,12) 2(2,0,3)= (4,0,6) Master your semester with Scribd 2(-1,3,9)= (-2,6,18) & The New York Times = (-6,6,12)+(4,0,6)-(-2,6,18) Special offer for students: Only $4.99/month.

(-6,6,12)+(4,0,6)=(-2,6,18)

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4(-1,3,9)= (-4,12,36) = 1/3 [(2,0,3)+(-3,3,6)-(-4,12,36) (2,0,3) + (-3,3,6) = (-1,3,9) =1/3 [(-1,3,9) – (-4,12,36)] (-1,3,9) – (-4,12,36) = (3,-9,-27) = 1/3 (3,-9,27) = (1,-3,-9)

d) 4(u + v) - (v – w + u) 4[(-1,1,2) + (2,0,3)] – [(2,0,3) – (-1,3,9) + (-1,1,2)] =4[(-1,1,2) + (2,0,3)] – [(2,0,3) – (-1,3,9) + (-1,1,2)] = (2,6,24) =4[(-1,1,2) + (2,0,3)] – [(2,0,3) – (-1,3,9) + (-1,1,2)] =4(1,1,5) – [(2,0,3) – (-1,3,9) + (-1,1,2)] You're Reading a Preview =4(1,1,5) – [(3,-3,-6) + (-1,1,2)] =4(1,1,5) – (2,-2,4) = (4,4,20) – (2,-2,-4)



= (2,6,24)

Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  2.- Hallar ecuaciones vectoriales y paramétricas lasNot siguientes rectas & The New Yorklas Times useful  Usefulde  Special offer for students: $4.99/month. a) LaOnly recta paralela

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a (2,-1,0) que pasa por P(1,-1,3)

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x = l t + x1 y = m t + y1 z = n t + z1 donde las:  

{l; m; n} - coordenadas del vector director: AB; (x1, y1, z1) - coordenadas de un punto situado en la recta (coordenadas de un punto A).

AB = {xb - xa; yb - ya; zb - za} = { 1 - 2 ; -1 - (-1) ; 3 - 0 } = { -1 ; 0 ; 3 }

Ecuación paramétrica de la recta: x=-t+2 y = -1 z = 3t

You're Reading a Preview Unlock full access with a free trial.

Free aTrial P (1,0,1) yWith b) Las rectas que pasan porDownload cortan la recta de ecuación vectorial p= (1,2,0) + t (2,-1,2) en los puntos situados a 3 unidades de distancia del punto Pₒ (1,2,0) Master your with Scribd x =semester t + 1  Read Free Foron 30this Days Sign up to vote title  = - t Times & The NewyYork Not useful  Useful  Cancel anytime. Special offer for students: Only $4.99/month. z = t + 1 L=[Pₒ + ta / t € R

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3.- Hallar el punto de intersección (si existe) del siguiente par de rectas: x=4+2s y=6+3s z=1+s x=3+t y=1−2t

z=3+3t x= 4+2s = 3+t

despejamos

y= 6+3s = 1-2t

despejamos

z= 1 + s = 3+3t despejamos

s = t – 1 2 s = 1/3 (-2t-5) s = 3t + 2

tomamos dos ecuaciones anteriores y despejamos 3t + 2 = 1/3 (-2t-5)

t = -1 en la ecuación

s = 3(-1) + 2 y obtenemos You're Reading a Preview Sustituimos en las ecuaciones (x, y, z,)

s= -1


x = 4 + 2(-1) = 3 + (-1) 2 = 2 y = 6 + 3(-1) = 1 – 2(-1)


El punto de intersección es (2, 3, 0)

Master your semester 3 = 3 with Scribd z =York 1 + (-1)Times = 3 + 3(-1) & The New Special offer for students: Only $4.99/month.

0 = 0


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· 2 -1 1 -1 2 1 = 2 · - 1 + 1 · = 0 2 1 2 1 0

= -2 Por lo tanto el volumen del paralelepípedo es 2 5.- Hallar el área del triángulo definido por los siguientes vértices: b. A(3,-1,1), B(4,1,0) y C(2,-3,0)

CA = (1,2,1)

CB = (2,4, 0) You're Reading a Preview

CA × CB = det

i

j

k

1

2

1

2

4

0


= 2j – 4i Download With Free Trial

Area del triángulo = √ (2)² - (4)² = √4 – 16 = √-12

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d) a(x, y ,z) = (0,0,0) (ax, ay, az)= (0,0,0) Si a=0 (0,0,0)=(0,0,0)

7.- Subespacios de espacios vectoriales

En cada caso determinar si U es un subespacio de R ᵌ, justificar la respue a. ={[1

] | ,

}

b. ={[ 0 ] | 2+ 2=0, ,

}

Independencia lineal 8. ¿Cuáles de los siguientes subconjuntos son independientes? Justificar respuesta. a Preview a. {[1−1 0] ,[3 2−1] ,[3 5−2] }You're de Reading 3 No son independientes

{ e1 – e2, e2 – e3, e3 – e1}


Download With Free Trial donde e1; e2; e3; son tres vectores independientes.

b) {(1,-1,0),(3,2,-1)(3,5,2)}

Master your semester with Scribd cualesquiera que sean (x; y; z) & The New York Times

Serán independientes Special offer for students: Only $4.99/month. si los unicos coeficientes poner en la combinación lineal:


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ʎ1, ʎ2, ʎ3, , que podemos

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{ e1 – e2, e2 – e3, e3 – e1} donde e1; e2; e3; son tres vectores independientes. b) {(1,-1,1, -1),(2,0,1,0)(0,-2,1,-2)} cualesquiera que sean (x; y; z) Serán independientes si los unicos coeficientes ʎ1, ʎ2, ʎ3, , que podemos poner en la combinación lineal:

ʎ1(e1 - e2) + ʎ2(e2 - e3) + ʎ3(e3 - e1) para que salga cero son todos cero. Por lo tanto det

1 -1 1 -1 2 0 1 0 0 -2 1 -2

=0

9. Demostrar que cada uno de los siguientes conjuntos de vectores es linealmente independiente. You're Reading a Preview Unlock full access with a free trial.

a. [1+ ,1− , + 2] en P2 1+x=0 1- x=0 x + x² = 0

det

1 1 0 Download With Free Trial 1 -1 0 =0 si en linealmente independiente 1 1 0

Master your semester with Scribd & The New York Times b. 1Only 1$4.99/month. 0 Special offer for students:

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