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Matemática básica
TEMA:
vectores
: 02Q Grupo Horario : INTEGRANTES:
Jhon jamer Vallejos Gallardo Calisaya Choque Franz Vertiz del Águila Carolina Milla cueva Diana Rivera Cano Melissa
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Useful Not useful SEMESTRE ACADEMICO: 2013-A
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VECTORES
⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ( (⃗ ⃗ ) ( (⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ) 1.- Si
,
; Determinar
2.-Sean los vectores
⃗ ⃗
Entonces:
Como:
⃗
⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗
determinar los vector
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Remplazando
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⃗
Entonces:
⃗ ⃗
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3.- Sea ABCD un rectángulo, una de cuyas diagonales tiene por extremos los puntos A= (1,5); C= (9 Si los lados de mayor longitud son paralelos al vector
. (2,3) = 0
⃗
. Calcular los puntos B y D
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9.- Encontrar los vectores a y b, tales que paralelo a (1,-1)
= (-1,5);
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es ortogonal a (-5,3) y
⃗ ⃗
es
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:
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10.- Sea el rectángulo ABCD, A =(-1,6); B=(2,3). B y D son vértices opuestos,
⃗
(-3,1). Halla los vértices C y D.
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⃗
es paralelo a (3,1)
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Download With Free Trial 20. Si A= (-1,0,1) y C=(3,4,-5) son los extremos de una diagonal del rombo ABCD. Hallar vectorialmente los vértices B y D si el lado AB es paralelo al vector a=(4,5,3) up to voteyon this title Tomo a M como punto medio de las diagonales: AC y BD Sign y a B(a,b,c) a D(d,e,f).
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Hallo las coordenadas de M, como es punto medio de AC, entonces:
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a + 1=4
a=3 b=5 c+1=3
c=2 B = (3,5,2)
como M es punto medio de BD y ya tengo las coordenadas de B, entonces D= ( -1,-1
20. Si A=(3,1,1) , B= (0,-2,1) , C= (4,0,3) y D son puntos coplanares, hallar C,tal que con B sean vértices de un triángulo equilátero.
Siendo coplanares A,B,C entonces D será coplanar con A y B así que tengo la igualdad: D(AB)=D(DB)=D(DA)
Represento las coordenadas de D como (dx,dy,dz) You're Reading a Preview
D(DB)=D(DA) 2
2
2
2
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2
(dx-0) +(dy +2) +(dz-1) =(dx-3) +(dy -1) +(dz-1) 6dy + 6 dx =6
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dy + dx =1…(a)
Me dicen que A,B y C son coplanares, entonces : AB=B-A=(-3,-3,2) AC=C-A=(1,-1,4) AD=D-A=( dx -3,dy-1,dz-1)
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3
21. En R , hallar 3 vectores unitarios de componentes no negativas, de modo que la diferencia entre ellos sean también vectores unitarios. El primero es ortogonal a vectores j y k, y el segundo es perpendicular al vector K.
Sean los tres vectores a,b y c los vectores. Teniendo en cuenta q a es ortogonal a los vectores j y K,entonces ,según :
No tamos que a =(1,0,0) El segundo es perpendicular al vector k,entonces: b.k= 0 bx.0+0.by+bz=0 bz =0
You're Reading a Preview
Unlock full access a free trial. Ahora nos dicen que la diferencia entre loswith vectores es un vector unitario,por lo mis
a-b = (1-bx,0-by,0-bz) = a un vector unitario (módulo de todo vector unitario es 1) 2 2 2 Download With Free Trial (1-b ) +(b ) +(b ) =1 x
y
2
z
2
1+ bx -2 bx+ by =1 2
2
bx + by =2 bx……(a) Sign up to vote on this title
Ahora recordando que el modulo de un vector unitario es igual a 1.
De b(bx,by,0)
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+ b
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2 y =1
2
by =
by =
√
√ 0), hallo C:
Tengo : a =(1,0,0) ; b==( ,
Teniendo en cuenta que el vector c es unitario,entonces su modulo es 1. 2
2
2
cx +cy +cz
=1…(c)
Despejo: |a-c|=1 2
2
2
(1-cx) +(0-cy) +(0-cz) =1 2
2
2
1+cx -2cx+cy +cz
=1….(d)
Reeemplazo (c) en (d): 2
2
2
1+cx -2cx+cy +cz =1 1-2cx+1=1
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-2cx=-1
cx =
Despejo para hallar cy:
+c +c =1 c +c = …(e) 2 y
2
2 z
2
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√ +√ c =1 c= √ y
y
Finalemte reemplazo el valor anterior en (e):
) +c = √
(
2
2 z
25.-Si
√ tengo: a = (1, 0,0); b= (, √ 0) , c=( √ , √ ) √ √
cz =
€ a R
3
.Demostra que:
= -||
Como supuestamente se sabe que: axbxc=(a.c)b-(a.b)c
You're Reading a Preview (PROPIEDAD) Unlock full access with a free trial.
DE ARRIBA
axaxb==(a.b)a - (a.a)b
Download With Free Trial a.a= a
2
a x a x a x b =a x ((a.a)a - (a.a)b) Multiplicamos al parêntesis : a x a x a x b = (a.a)(axa) -
se elimina axa=0
a
2
(a x b)
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y lo vuelvo a cambiar como a.(bxc) resultándonos
̅ ̅ = -||
⃗ ⃗ ⃗ ⃗ ‖⃗⃗⃗⃗
29. Sea ABC un triángulo y “O” un pinto cualquiera y sean Demostrar que el área del triángulo ABC, está dado por:
B
,
,
O
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C
Download With Free Trial A
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⃗ ⃗ ⃗
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‖ ⃗ ⃗⃗ ⃗
S=
⃗ ‖
⃗ ⃗ ⃗ ⃗⃗ S=
‖⃗⃗⃗ ⃗⃗ ⃗ ‖
30. hallar los vectores unitarios que son perpendiculares al plano determinado por lo puntos: A= (3,-6,4); B= (2, 1,1); C= (5, 0,-2) SOLUCION: D= (x,y,z) A.D=0….3x-6y+4z=0---- ① You're Reading a Preview ② B.D=0….2x+y+z=0---Unlock full access with a free trial.
C.D=0….5x+0y-2z=0----3 Download With Trial De ①, ② y 3 Free se tiene: X=2z/5: y=-2z C= (2z/5, -2z, z)=Z/5(2,-10,5) up to vote on this title Sign R:U= √ Useful Not useful
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√ √ √ √ (
5
=10
..
Por formar ángulo obtuso: Z=16
⃗ ⃗ ⃗ ⃗ ‖‖ √
32. halle el vector “X” perpendicular a los vectores ecuación:
⃗
=6
y
y satisfac
X=-z; y=z
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33. Los vértices de un triángulo son A=(-2,3,-1); B=(1,1,5); C=(-1,5,-3). Hallar el vector “x” en dirección de la bisectriz del ángulo BAC si
B
=2
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⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ (⃗) ⃗ ① ⃗⃗ ⃗⃗ ② ⃗ ⃗
DE
① ② ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗⃗ ⃗ ⃗ Y
…..
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34. El vector x es perpendicular a los vectores a=(3,2,2) y b=(18,-22,-5) y forma con eje OY un ángulo obtuso. Hallar sus componentes si llxll=14
①
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‖‖
35. demostrar que si entonces
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⃗
es el ángulo que forman los vectores no ortogonales a y b
‖‖⃗ (⃗) ⃗ ⃗⃗ ‖‖ ⃗ ⃗ ⃗⃗ 36.
Solución:
⃗ ⃗ ‖‖⃗ ⃗ [ ][ ] …①
Y el producto interno:
….②
You're Reading a Preview
Efectuando
① y ② comprobamos quefullson idénticas, Unlock access with a freepor trial.tanto:
⃗⃗ ‖⃗‖⃗ (⃗⃗) ⃗ Download With Free Trial
37. Demostrar que el área de triangulo cuyos vértices son los extremos de los vecto a, b, c es: A
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PRODUCTO VECTORIAL 1) Sean
A(-3,4,0) B(3,6,3) y C(-1,2,1) los tres vértices de un triangulo. Calcular el area del
triangulo.
⃗ ⃗ √ √ √ ⃗ √ √ ⃗ ⃗ ⃗ ⃗ =
= (3,6,3) – (-3,4,0) = (6,2,3)
=
= (-1,2,1) – (-3,4,0) = (2,-2,1)
II x II =
= (8,0,-16)
II(8,0,-16)II =
=
=8
Entonces el area del triangulo es : A = II x II A = 8 A=4
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2) Dados los vectores
= (2,1,3)
= (1,2,3)
= (-1,-1,0) y hallar el volumen del paralelepíp
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que tiene por aristas los vectores dados.
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V = I x x I
=6
V=6
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4) Dados los puntos A(1,0,1) B(1,1,1) C(1,6,k). Determinar : a) Para que valores del parámetro k están alineados.
Los puntos son colineales :
Si
⃗ =
= (1,1,1) = (1,0,1) = (0,1,0)
=
= (1,6,K) – (1,1,1) = (0,5,K-1)
=
= (1,0,1) – (1,6,K) = (0,-6,1-K)
//
Si
//
(0,1,0) = (0,5,k-1) 1 = 5t
,
=t
(0,1,0) = r (0,-6,1-k)
0 = t(k-1)
1 = -6r
,
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k=1
r=-
0 = r(1-k) k=1
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b) Valores de k si existen tal que A,B y C sean tres vértices de un paralelogramo de area II
x
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II = 3
II (0,1,0)x(o,6,k-1) II = 3
II (K-1,0,0) II =
=
II (K-1,0,0) II
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=3
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⃗ ⃗ ⃗
V=5
V = II V = II
II = 5
II = 30
=
= (3,0,1) - (2,1,-1) = (1,-1,2)
=
= (2,-1,3) – (3,0,1) = (-1,-1,2)
=
= (D) – (2,1,-1) = (m,n,p)
= (m,n,p)
= 1(-p-2m) + 1(-p-2n) + 2(-m+n)
-2p = 30 P = -15
= (m,n,p) = (0,0,-15)
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D – (2,1,-1) = (0,0,-15) D = (0,0,-15) – (2,1,-1) Sign up to vote on this title
D = (2,1,-16)
Useful Not useful 6) Si A(-1,3,2) C(3,-5,0) son los extremos de una diagonal del rectángulo ABCD hallar B y
lado
es paralelo al vector
= (1,1,1).
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Q=2
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s = -1
D = (q,r,s) = D (2,-6,-1)
=
= D – B = (2,-6,-1) – (m,n,p) = (2-m, -6-n, -1-p)
=
= (3,-5,0) – (-1,3,2) = (4,-8,-2)
=
=
(2-m, -6-n, -1-p) = (4,-8,-2) 2 – m = 4
-6 – n = -8
,
m = -2
,
-1 – p = -2
n=2
p=1
B (m,n,p) = B (-2,2,1)
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7. La fuerza
Download With Free Trial =(2,-3,6) esta aplicada al punto P(1,-2,5) . Determinar el Momento del la fuerza F
respecto al punto A(3,4,-5). SOLUCION
⃗
Sea =
= P – A = (1,-2,5) - (3,4,-5) = (-2,-6,10)
Momento de la fuerza =
= (-6,32,18)
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⃗ ⃗ ⃗
(
⃗ ⃗ ⃗ ). (
) = (
). (
) + (
). (
) - (
) + (
). (
). (
)
) =0
9. un paralelepípedo esta determinado por los vectores =(3,1,1) , Determinar su volumen
02_Componentes
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(
). (
Banco de Preguntas -
⃗
=(1,-2,-3) y =(-1,2,-3) .
SOLUCION
V=
V= 3(6+2) -1 (-3-1) + 1 (2-2) V= 28
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Download With Free Trial 10. Halla los vectores unitarios que son perpendiculares al plano determinado por los puntos A=(3,-6,4) , B=(2,1,1) y C =(5,0,-2) SOLUCION SEA M EL VECTOR (X1,X2,X3)
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Banco de Preguntas -
02_Componentes
Search document
(2, 6,-6). (X1, X2, X3)= 0 (3,-1,-3) . (X1, X2, X3)= 0 DESAROLLANDO -X1+7X2-3X3 = 0 2X1+6X2-6X3 = 0 3X1-X2-3X3 = 0
X1= 6
X2=3
X3=5
⃗ √
=
⃗
=
√
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12. Halllar el volumen de un tetraedro cuyos vértices son los puntos A(2,-3,5) B(0,2,1) C (-2,-2,3) y d (3,2,4) Sign up to vote on this title
Sean los puntos:
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⃗ V=
Geometria Analitica-
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Banco de Preguntas -
02_Componentes
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= (5, 4,1)
V= 1(-12) -5 (3-15) - 1 (12) V= 6
13. Los vectores a=(0,1,2) , b=(0,1,k) y c =(k,1,1) forman un paralelepípedo de 3 los valore de k
3=
V= 0 -1 (K= 3 ,
de volumen cal
) +2 (-k)
k = -1
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14. dado el tetraedro ABCD, A (2, 3,1), B (4, 1,-2)
C (6, 3,7) y D (-5,-4,8) hallar la longitud
altura bajada desde el vértice D SOLUCION
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⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ℎ⃗ ℎ⃗ ℎ⃗ ⃗⃗ ⃗⃗ | | =
Banco de Preguntas -
02_Componentes
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= (2, -2,3)
=
= (4, 0,6)
=
= (-7, -7,7)
=
= (-12,-24,8)
Sea
la altura bajada. Luego
=
=
de donde
=
= = √
11
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Download With Free Trial 15. si
⃗ √ ⃗ √ ⃗ =
,
=
y
= 1 hallar los módulos de los vectores
ángulo que forman estos vectores SOLUCION
‖ ‖ ‖ ‖ ⃗ ⃗ ‖⃗‖ ⃗ =
+
…………..(1)
+2
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⃗
y e
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‖‖ ‖⃗‖ ⃗
=2
=4 =2
‖‖ ‖⃗ ‖ ⃗
of 26
Banco de Preguntas -
02_Componentes
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=1
=1
,
=1
‖ ‖ ⃗ =
=
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