Taller 1 (1)

PROBLEMAS ALGEBRA VECTORIAL
L. Rasch, J. Vega, S. Ospina, J. Torres
April 2022
1.1 Identify which of the following quantities is not a vector:
(a) Force
(b) Momentum
(c) Acceleration
(d) Work
(e) Weight
1.2 Which of the following is not a scalar field:
(a) Displacement of a mosquito in space
(b) Light intensity in a drawing room
(c) Temperature distribution in your classroom
(d) Atmospheric pressure in a given region
(e) Humidity of a city
1.3 The rectangular coordinate systems shown in the next
Figure are right-handed except:
Respuesta: La coordenadas rectangulares que no cumplen con la mano derecha son b y e.
1.4 Which of these is correct?
2
(a) A × A = |A|
(b) A × B + B × A = 0
(c) A · B · C = B · C · A
(d) ax · ay = az
(e) ak = ax − ay
1
Where ak is a unit vector.
Esta respuesta se puede comprobar conociendo que una de las propiedades del producto cruz es
ser anticonmutativo, es decir:
A × B = −B × A
1.5 Which of the following identities is not valid:
(a) a(b + c) = ab + bc
(b) a×(b + c) = a × b + a × c
(c) a · b = b · a
(d) c·(a × b) = −b·(a × c)
(e) aA · aB = cosθAB
1.6 Which of the following statements are meaningless:
(a) A · B + 2A = 0
(b) A · B + 5 = 2A
(c) A(A + B) + 2 = 0
(d) A · A + B · B = 0
1.7 Let F = 2ax − 6ay + 10az and G = ax + Gy ay + 5az : If F and
G have the same unit vector, Gy is
(a) 6
(b) −3
(d) 0
(e) 6
Sabemos que -3 es la respuesta correcta porque si suponemos que F y G tienen el mismo vector
unitario podemos encontrar otro vector con igual dirección y sentido, pero con diferente magnitud
a partir de uno de estos. Por definición le vector unitario:
⃗ua es el vector unitaria del vector ⃗a y fácilmente observamos que la magnitud de nuestro F es:
⃗|=
|F
p
√
22 + (−6)2 + 102 = 2 35
Y ahora divido entre el F
⃗
F
1 1
1
= √ (2, −6, 10) = √ (1, −3, 5)
⃗
2
35
35
|F |
1.8 Given that A = ax + αay + az and B = αax + ay + az , if A
and B are normal to each other, α is
(a) −2
(b) − 12
(c) 0
(d) 1
(e) 2
2
Esta respuesta se obtiene conociendo que si dos vectores A y B son normales es necesario que
A · B = 0. Luego se plantea la equación:
A · B = (1, α, 1) · (α, 1, 1) = 0
De donde se obtiene 2α = −1 y finalmente, α = − 21
1.9 The component of 6ax + 2ay − 3az along 3ax − 4ay is:
(A · B)B
(18 − 8)(3, −4, 0)
=
2
|B|
(9 + 16)
10
=
(3, −4, 0)
25
2
= (3, −4, 0)
5
1.10 Given A = −6ax + 3ay + 2az the projection of A along
ay is:
A · ay = (−6, 3, 2) · (0, 1, 0) = 3
1.1 A bug starts at a point and travels 1m northward, 21 m
eastward, 41 m southward, 18 m westward, and so on, making a
90 turn to the right and halving the distance each time.(a)
What is the total distance traveled by the bug? (b) Find the
final position of the bug relative to its starting location (c)
Find the straight-line distance from the starting location to
the final position.
Para el movimiento en el eje Y tenemos la siguiente serie:
1−
1
1
1
+
−
+ ...
4 16 64
Lo cual ponemos resumir en una sumatoria dada por:
n
∞ X
−1
4
n=0
Y por evaluacion de series geometricas infinitas tenemos que como
n
∞ X
−1
n=0
4
=
1
1−
−1
4
=
−1
4
< 1, entonces:
4
= 0,8
5
Para el movimiento en el eje X tenemos una serie similar:
1 1
1
1
− +
−
+ ...
2 8 32 128
Simplificada en:
n
n
∞
∞ X
1 −1
1 X −1
1 4
2
= ·
= · = = 0,4
2
4
2 n=0 4
2 5
5
n=0
3
Para hallar la distancia total recorrida realizamos la misma serie con el valor absoluto de cada
trayecto:
1+
1 1 1
1
1
1
1
+ + +
+
+
+
+ ...
2 4 8 16 32 64 128
∞ n
X
1
1
=
1 =2
2
1
−
2
n=0
Tomando el punto de origen como el (0,0) tenemos que la posición final es ( 25 , 45 ), ası́ que la distancia
en linea recta hasta el punto final es:
p
p
p
x2 + y 2 = 0,42 + 0,82 = 0,8 = 0,8944
1.2 Solve the following equations for A, B, and C:
⃗+B
⃗ +C
⃗ = 2aˆ1 + 3aˆ2 + 2aˆ3
A
⃗+B
⃗ −C
⃗ = aˆ1 + 3aˆ2
2A

1


2

1
1
1
1
−1
−2

3
..
. 2
..
. 1
..
. 4
1

···
0

0
0
−2
1
3
0
11
⃗ − 2B
⃗ + 3C
⃗ = 4aˆ1 + 5aˆ2 + aˆ3
A


.
1 ..
3 2
1 1


.


3 0 =⇒ 0 −1 −3 ..


.
0 −3 2 ..
5 1


..
. −1 0 −2
1 0


..
 =⇒ 
. 3
3
4
0 1


..
0 0
. 11 11 11

2
3
−3
−3
2
2
0
0
1
..
.
..
.
..
.
2


−4 =⇒

−1

1
2
0
0
1
1
0


1

1
⃗ = 1aˆ1 + 2aˆ2
A
⃗ = aˆ3
B
⃗ = aˆ1 + aˆ2 + aˆ3
C
1.3 Show that (A + B) · (A − B) = A2 − B 2 and that (A + B) ×
(A − B) = 2B × A. Verify the above for A = 3a1 − 5a2 + 4a3 y
B = a1 + a2 − 2a3
(A + B) · (A − B) = A2 − B 2
(A + B) · (A − B) = A2 − B 2
A · (A − B) + B · (A − B) =
A2 − A · B + B · A − B 2 =
A2 − A · B + A · B − B 2 =
A2 − B 2 =
p
2
32 + (−5)2 + 42 −
√ 2 √ 2
(4, −4, 2) · (2, −6, 6) = 50 − 6
(3 + 1, −5 + 1, 4 − 2) · (3 − 1, −5 − 1, 4 − (−2)) =
8 + 24 + 12 = 50 − 6
44 = 44
4
p
12 + 12 + (−2)2
2
(A + B) × (A − B) = 2B × A
(A + B) × (A − B) = 2B × A
A × (A − B) + B × (A − B) =
A×A−A×B+B×A−B×B =
A×A+B×A+B×A−B×B =
0+B×A+B×A−0=
2B × A =
(4, −4, 2) × (2, −6, 6) = (2, 2, −4) × (3, −5, 4)
(−12, −20, −16) = (−12, −20, −16)
1.4 Given A = −2a1 + a2 , B = a1 − 2a2 + a3 and C = 3a1 + 2a2 + a3
find A × (B × C) + B × (C × A) + C × (A × B)
A × (B × C) + B × (C × A) + C × (A × B)
= (−2, 1, 0) × [(1, −2, 1) × (3, 2, 1)] + (1, −2, 1) × [(3, 2, 1) × (−2, 1, 0)] + (3, 2, 1) × [(−2, 1, 0) × (1, −2, 1)]
= (−2, 1, 0) × (−4, 2, 8) + (1, −2, 1) × (−1, −2, 7) + (3, 2, 1) × (1, 2, 3)
= (8, 16, 0) + (−12, −8, 4) + (4, −8, 4)
= (0, 0, 0)
1.1 Find de unit vector along the line joining point (2,4,4) to
point (-3,2,2)
Primero hallamos el vector r:
r = (−3, 2, 2) − (2, 4, 4) = (−5, −2, −2)
Y su magnitud:
|r| =
p
√
(−5)2 + (−2)2 + (−2)2 = 33
Ahora hallamos el vector unitario que pasa por estos:
ar =
r
(−5, −2, −2)
√
=
= −0,87ax − 0,34ay − 0,34az
|r|
33
1.2 Let A = 2ax + 5ay − 3az , B = 3ax − 4ay and C = ax + ay + az .
1.2.1 Determine A + 2B
Primero se realiza:
2B = 2(3ax ) − 2(4ay )
2B = 6ax − 8ay
Luego se suma A
A + 2B = (2ax + 5ay − 3az ) + (6ax − 8ay )
A + 2B = (2ax + 6ax ) + (5ay − 8ay ) + (−3az + 0)
A + 2B = 8ax − 3ay − 3az
5
1.2.2 Calculate |A − 5C|
Se halla 5C
5C = 5(ax ) + 5(ay ) + 5(az )
5C = 5ax + 5ay + 5az
Con este resultado hallamos la resta para posteriormente hallar la magnitud del vector resultante
A − 5C = (2ax + 5ay − 3az ) + (5ax + 5ay + 5az )
A − 5C = (2ax − 5ax ) + (5ay − 5ay ) + (−3az − 5az )
A − 5C = −3ax − 8az
Luego:
p
(−3ax )2 + (−8az )2
√
|A − 5C| = 73 ≈ 8,544
|A − 5C| =
1.2.3 For what values of k is |kB| = 2?
Hallamos kB
kB = k(3ax ) − k(4ay )
kB = k3ax − k4ay
Se tiene según propiedades que la magnitud de un vector escalado es igual a la magnitud del vector
por el valor absoluto del escalar, es decir,
|⃗a ∗ a| = |⃗a|a
Por tanto, se desarrolla
q
(3ax )2 + (−4ay )2
√
|B| = 25 = 5
|B| =
Obteniendo
|k||B| = 2
|k|5 = 2
2
|k| =
5
Por lo tanto los numeros con los que se cumple la condición son ± 25
1.2.4 Find
A×B
A·B
Se realiza A × B

ax
A×B=2
3
ay
5
−4

az
−3
0
De esta matriz se obtiene
0ax − 9ay − 8az − (0ay + 12ax + 15az )
−9ay − 8az − 12ax − 15az
A × B = −12ax − 9ay − 23az
Ahora se halla producto punto A· B
A· B = (2ax + 5ay − 3az )· (3ax − 4ay )
A· B = (2 ∗ 3) + (5 ∗ −4) + (−3 ∗ 0)
A· B = 6 − 20 + 0 = −14
A· B = −14
Y se remplaza finalmente
A×B
−12ax − 9ay − 23az
=
A· B
−14
A×B
6
9
23
= ax + ay + az
A· B
7
14
14
6
1.3 If
A = 2ax + ay − 3az
B = ay − az
C = 3ax + 5ay + 7az
determine:
(a) A − 2B + C
A − 2B + C = (2 − 0 + 3, 1 − 2 + 5, −3 − 2 + 7)
= (5, 4, 2)
= 5ax + 4ay + 2az
(b) C − 4(A + B)
C − 4(A + B) = (3, 5, 7) − 4(2, 2, −4)
= (3, 5, 7) − (8, 8, −16)
= (−5, −3, 23)
= −5ax − 3ay + 23az
(c)
2A−3B
|C|
(4, 2, −6) − (0, 3, −3)
2A − 3B
√
=
|C|
32 + 52 + 72
(4, −1, −3)
√
=
83
(d) A · C − |B|2
A · C − |B|2 = 6 + 5 − 21 −
p
2
12 + 1 2
= −10 − 2
= −12
(e) 21 B ×
1
3A
+ 14 C
1
B×
2
1
1
A+ C
3
4
!
1
2 1
3 5 7
= (0, 1, −1) ×
, , −1 +
, ,
2
3 2
4 4 4
1 1
17 19 3
= 0, , −
×
, ,
2 2
12 12 4
28 17 17
=
, ,−
24 24 24
7
1.4 If the position vectors of points T and S are 3ax − 2ay + az
and 4ax + 6ay + 2az respectively, find (a) the coordinates of
T and S, (b) the distance vector from T to S, (c) the distance
between T and S.
(a)
T = (3, −2, 1)
S = (4, 6, 2)
(b)
S − T = (4, 6, 2) − (3, −2, 1)
= (1, 8, 1)
(c)
|S − T | =
=
p
√
12 + 82 + 12
66
1.5 If
A = 5ax + 3ay + 2az
B = −ax + 4ay + 6az
C = 8ax + 2ay
find the values of α and β such that αA + βB + C is parallel to the y-axis.
Comenzamos reescribiendo nuestros vectores de la forma de la ecuación pedida
(5α + 3β + 2)ax (−α + 4β + 6)ay + (8α + 2β)az
Ahora para encontrar α y β paralelos al eje y utilizamos las ax y az igualadas a 0.
5α + 3β + 2 = 0 (1) y 8α + 2β = 0 de donde α = 4β (2)
Reemplazando en la primera ecuación (1):
5(4β) + 3β + 2 = 0
20β + 3β + 2 = 0
23β + 2 = 0
β=
−2
23
Y ahora usamos el valore β para hallar α reemplazando en (2)
α=
8
−8
23
1.6 Given vectors
A = αax − ay + 4az
B = 3ax + βay − 6az
C = 5ax − 2ay + γaz
Determine α,β,and γ such that the vectors are mutually orthogonal.
Se sabe que para que dos vectores sean ortogonales se necesita que su producto punto de igual a
cero (A · B = 0). De esta forma, se puede crear el siguiente sistema de tres ecuaciones:
A · B = 3α + β − 24 = 0
A · C = 52 + 4γ = 0
A · B = 15 − 2β − 6γ = 0
El cual simplemente se resuelve obteniendo α = 38 ,β = 16,γ = − 17
6 .
1.7
(a) Show that (A · B)2 + (A × B)2 = (AB)2
(A · B)2 + (A × B)2 = (AB)2
(ABcosθAB )2 + (ABsinθAB )2 =
(AB)2 (cos2 θAB + sin2 θAB ) =
(AB)2 =
(b) Show that
ax =
ay ×az
ax ·ay ×az
ay × az
ax · ay × az
ax
=
ax · ax
ax
=
1
= ax
ax =
ay =
az ×ax
ax ·ay ×az
az × ax
ax · ay × az
ay
=
ax · ax
ay
=
1
= ay
ay =
9
az =
ax ×ay
ax ·ay ×az
ax × ay
ax · ay × az
az
=
ax · ax
az
=
1
= az
az =
1.8 given that:
P = 2ax − ay − 2az
Q = 4ax + 3ay + 2az
R) = −ax + ay + 2az
find: (a) |P + Q − R|, (b) P · Q × R, (c) Q × P · R, (d) (P × Q) · (Q × R), (e) (P × Q)(Q × R),
(f) cosθ P R , (g) sinθ P Q .
(a)
|P + Q − R| = |(2, −1, −2) + (4, 3, 2) − (−1, 1, 2)|
= |(7, 1, −2)|
√
=3 6
(b)
P · Q × R = (2, −1, −2) · (4, 3, 2) × (−1, 1, 2)
= 1 × (−1, 1, 2)
= (−1, 1, 2)
(c)
Q × P · R = (4, 3, 2) × (2, −1, −2) · (−1, 1, 2)
= (−4, 12, −10) · (−1, 1, 2)
= −4
(d)
(P × Q) · (Q × R) = [(2, −1, −2) × (4, 3, 2)] · [(4, 3, 2) × (−1, 1, 2)]
= (4, −12, 10) · (4, −10, 7)
= 206
(e)
(P × Q)(Q × R) = [(2, −1, −2) × (4, 3, 2)] × [(4, 3, 2) × (−1, 1, 2)]
= (4, −12, 10) × (4, −10, 7)
= (16, 12, 8)
10
(f )
P ·R
|P ||R|
−7
= √
3 6
cosθ P R =
(g)
P ×Q
|P ||Q|
√
260
= √
3 29
sinθ P Q =
1.9 Given vectors T = 2ax − 6ay + 3az and S = ax + 2ay + az .
find:
(a) the scalar projection of T on S
TS =
T ·S
(2, −6. − 3) · (1, 2, 1)
2 − 12 − 3
−13
√
=
=
=
2
2
2
|S|
4
4
1 +2 +1
(b) the vector projection of S on T
ST =
(S · T )T
−13(2, −6, −3)
−13(2, −6, −3)
−13
=
=
=
(2, −6, −3)
T2
(2, −6, −3) · (2, −6, −3)
4 + 36 + 9
49
(c) the smaller angle between T and S
Usando los puntos anteriores
cos θ =
T ·S
−13
−13
−13
p
=
=
=
= −0,46
2
2
2
|T ||S|
4
·
7
28
4 · 2 + (−6) + (−3)
θ = cos−1 (−0,46) = 117, 66◦
1.10 If A = −ax + 6ay + 5az and B = ax + 2ay3 az , find:
1.10.1 The scalar proyections of A on B
Esta proyección escalar se puede obtener como:
|AB | =
A·B
(−1, 6, 5) · (1, 2, 3)
26
√
=
= √ = 6,95
|B|
1+4+9
14
1.10.2 The vector proyection of B on A
La proyección de B en A se obtiene como:
B·A
(−1, 6, 5) · (1, 2, 3)
(A) = √
(−1, 6, 5) = 0,42(−1, 6, 5) = (−0,42, 2,52, 2,1)
|A|2
1 + 36 + 25
Ası́ se obtiene finalmente: −0,42ax + 2,52ay + 2,1az
11
1.10.3 The unit vector perpendicular to the plane containing A and B
Para encontrar el vector perpendicular únicamente es necesario realizar el producto cruz de los
vectores dados y dividir por su magnitud:
A×B=
−1
1
6 5
= 8ax + 8ay − 8az
2 3
Y luego, el vector unitario es:
8ax + 8ay − 8az
aA×B = √
= 0,58ax + 0,58ay − 0,58az
64 + 64 + 64
1.11 Calculate the angles that vector H = 3ax + 5ay − 8az
makes with the x, y and z axes.
Eje x. X=(1,0,0)
H ·X
|H||X|
3
=√
98
= 0,3030
cosθHX =
cos−1 (0, 3030) = 72,36◦
Eje y. Y=(0,1,0)
H ·Y
|H||Y |
5
=√
98
= 0,5050
cosθHY =
cos−1 (0, 5050) = 59,66◦
Eje z. Z=(0,0,1)
H ·Z
|H||Z|
−8
=√
98
= −0,8081
cosθHZ =
cos−1 (−0, 8081) = 143,91◦
1.12 Find the triple scalar product of P, Q and R given that:
P = 2ax − ay + az Q = ax + ay + az R = 2ax + 3az
triple scalar product:
P · (Q × R) = (2, −1, 1)[(1, 1, 1) × (2, 0, 3)]
= (2, −1, 1) · (3, −1, −2)
= (6, 1, −2)
12
1.13 Symplify the following expressions:
Usamos:
(A × B) × C = (A · C)B − (B · C)A
(a) A X (A X B)
A × (A × B) = −(A × B) × A = (B · A)A − (A · A)B
(a) A X [A X (A X B)]
A × [A × (A × B)] = A × [(B · A)A − (A · A)B] = (A · B)(A × A) − (A · A)(A × B)
1.14 Show that the dot and the cross in the triple scalar
product may be interchanged, i.e., A · (B × C) = (A × B) · C
Esto puede ser comprobado conociendo la propiedad del triple producto escalar dada por:
A · (B × C) = B · (C × A) = C · (A × B)
La cual se obtuvo del siguiente procedimiento:
C · (A × B) = (Ay Bz − Az By , −Ax Bz + Az Bx , Ax By − Ay Bx ) · (Cx , Cy , Cz )
C · (A × B) = Ax By Cz − Ax Bz Cy − Ay Bx Cz + Ay Bz Cx + Az Bx Cy − Az By Cx
Del cual se obtiene el mismo resultado con A · (B × C):
A · (B × C) = (Ax , Ay , Az ) · (By Cz − Bz Cy , −Bx Cz + Bz Cx , Bx Cy − Bz Cx )
A · (B × C) = Ax By Cz − Ax Bz Cy − Ay Bx Cz + Ay Bz Cx + Az Bx Cy − Az By Cx
1.15 Points P1 = (1, 2, 3), P2 = (−5, 2, 0) and P3 = (2, 7, −3) form
a triangle in space. Calculate the area of the triangle.
−−−→
−−−→
P1 P2 = (−6, 0, −3)
P1 P3 = (1, 5, −6)
−−−→ −−−→
P1 P2 × P1 P3 = (15, −39, −30)
1 −−−→ −−−→
A = |P1 P2 × P1 P3 | = 25,72
2
13
1.16 The vertices of the triangle are located at p1 = (4, 1, −3),
p2 = (−2, 5, 4) and p3 = (0, 1, 6). Find the three angles of the
triangle.
A = p2 − p1 = (−2, 5, 4) − (4, 1 − 3) = (−6, 4, 7)
B = p3 − p2 = (0, 1, 6) − (−2, 5, 4) = (2, −4, 2)
C = p1 − p3 = (4, 1 − 3) − (0, 1, 6) = (4, 0, −9)
A·B
∠AB = acos(
)
|A||B|
(−6, 4, 7) · (2, −4, 2)
p
)
= acos( p
2
(−6) + 42 + 72 ∗ 22 + (−4)2 + 22
14
= acos( √
)
2424
= 73,47◦
B·C
∠BC = acos(
)
|B||C|
(2, −4, 2) · (4, 0, −9)
p
= acos( p
)
2
2 + (−4)2 + 22 ∗ 42 + 02 + (−9)2
10
)
= acos( √
2328
= 78,04◦
A·C
)
∠AC = acos(
|A||C|
(−6, 4, 7) · (4, 0, −9)
p
= acos( p
)
(−6)2 + 42 + 72 ∗ 42 + 02 + (−9)2
87
)
= acos( √
9797
= 28,48◦
1.17 Points P , Q, and R are located at (-1,4,8), (2,-1,3), and
(-1,2,3), respectively. Determine:
(a) the distance between P and Q
rP Q = P − Q = (−3, 5, 5)
rP Q =
p
(−3)2 + (5)2 + (5)2 = 7,68
(b) the distance vector from P to R
rP R = R − P = (0, −2, −5)
(c) the angele between QP and QR
Usamos rP Q pero falta calcular rRQ
rRQ = R − Q = (−3, 3, 0)
14
rRQ =
p
(−3)2 + (3)2 + (0)2 = 4,24
Ahora si utilizamos nuestros datos para hallar el ángulo.
cos θ =
rP Q · rRQ
(−3, 5, 5) · (−3, 3, 0)
=
= 0,74
|rP Q ||rRQ |
7, 68 · 4, 24
θ = cos−1 (0,74) = 42, 52◦
(d) the area of triangle P QR

rP Q × rRQ
x y
= −3 5
−3 3
 

z
−15
5 = −15
0
6
Área del triangulo:
=
1p
1
rP Q × rRQ =
(−15)2 + (−15)2 + (6)2 = 11,02
2
2
(e) the perimeter of triangle P QR
Para el perimetro hayamos la magnitud del vector restante rRP
|rRP | =
p
(0)2 + (−2)2 + (−5)2 = 5,38
PP QR = |rRP | + rRq + rP Q = 5,38 + 4,24 + 7,68 = 17,3
1.18 If r is the position vector of the point (x, y, z) and A is
a constant vector, show that:
(r − A) · A = 0 is the equation of a constant plane
Se observa que la ecuación es equivalente a: r · A = A · A y si se tiene A = (a, b, c) da como
resultado la ecuación:
ax + by + cz = |A|
2
La cual se sabe que corresponde a la ecuación de un plano
(r − A) · r = 0 is the equation of a sphere
Se observa que la ecuación es equivalente a: r · r = A · r y si se tiene que A = (a, b, c) da como
resultado la ecuación:
x2 + y 2 + z 2 = ax + by + cz
La cual se sabe que corresponde a la ecuación de una esfera
15
1.19
(a) Prove that P = cosθ1 ax + sinθ1 ay and Q = cosθ2 ax + sinθ2 ay are unit vectors in the
xy-plane respectively making angles θ1 and θ2 with the x-axis.
p
cos2 θ1 + sin2 θ1 = 1
p
|Q| = cos2 θ2 + sin2 θ2 = 1
|P | =
(b) By means of dot product, obtain the formula for cos(θ2 − θ1 )
P · Q = |P ||Q|cos(θ2 − θ1 ) = 1 · 1cos(θ2 − θ1 )
Pero
P · Q = (cosθ1 ax + sinθ1 ay ) · (cosθ2 ax + sinθ2 ay )
= cosθ1 cosθ2 + sinθ1 sinθ2
Ası́
cos(θ2 − θ1 ) = cosθ1 cosθ2 + sinθ1 sinθ2
(c) If θ is the angle between P and Q, find 21 |P − Q| in terms of θ.
1
1
|P − Q| = |P 2 + Q2 − 2P Qcosθ|
2
2
1
= |1 + 1 − 2cosθ|
2
= 1 − cosθ
θ
= 2sin2
2
1.20 Consider a rigid body rotating with a constant angular
velocity ω radians per second about a fixed axis through O
as in figure . Let r be the distance vector from O to P , the
position of a particle in the body. The velocity u of the body
at P is —u— = dω =|r|sinθ|ω| or u = ω × r. If the rigid body
is rotating with 3 radians per second about an axis parallel to
ax − 2ay + 2az and passing through point (2,-3,1), determine
the velocity of the body at (1,3,4).
ω(1, −2, 2)
3
= (1, −2, 2)
ω=
r = rp − r0
= (1, 3, 4) − (2, −3, 1)
= (−1, 6, 3)
u=ω×r
= (1, −2, 2) × (−1, 6, 3)
= (−18, −5, 4)
u = −18ax − 5ay + 4az
16
1.21 Given A = x2 yax − yzay + yz 2 az
(a) The magnitude of A at point T (2, −1, 3)
Reemplazando los valores del punto dado en A se obtiene A=(-4,3,-9), de esta forma se saca su
magnitud como:
√
|A| =
16 + 9 + 81 =
√
106 = 10,3
(b) The distance vector from T to S is 5,6 units away from T and in the
same direction A at T
Denominando rT S = B = BaB se obtiene:
B = 5,6
aB = aA =
(−4, 3, −9)
10,3
Luego:
rT S = B =
5,6(−4, 3, 9)
= −2,17ax + 1,63ay − 4,89az
10,3
(c) The position vector of S
rT S = rS − rT → rS = rT + rT S
Reemplazando:
rS = −0, 17ax + 0, 63ay − 1, 89az
1.22 E and F are vector fields given by E = 2xax + ay + yzaz
and F = xyax − y 2 ay + xyzaz . Determine:
1.22.1 |E| at (1, 2, 3)
En el punto (1, 2, 3) se reemplazan los valores y se obtiene que E = (2, 1, 6). Luego únicamente es
necesario sacar la magnitud del vector de donde se obtiene:
|E| =
√
4 + 1 + 36 =
√
41 = 6,4
1.22.2 The component of E along F at (1, 2, 3)
Realizando el mismo procedimiento anterior se reemplazan los valores y se obtiene F = (2, −4, 6).
Luego se realiza la proyección:
EF =
(E · F)F
4 − 4 + 36
=
(2, −4, 6) = 1,29ax − 2,57ay + 3,86az
|F|2
56
1.22.3 A vector perpendicular to both E and F at (0, 1, −3) whose magnitude is unity
En este punto se tienen los valores de E = (0, 1, −3) y F = (0, −1, 0) y para hallar un vector
perpendicular solo se necesita:
17
E×F=
0
0
1
−1
−3
= (−3, 0, 0)
0
Y luego dividir por su magnitud:
aE×F = ±
E×F
= ±ax
|E × F|
18