Problem 2.125 Two vectors UD3iC2jand VD2i
C4j.
Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
Problem 2.126 The two segments of the L-shaped bar
are parallel to the xand zaxes. The rope AB exerts
a force of magnitude jFjD500 lb on the bar at A.
Determine the cross product rCA ðF, where rCA is the
position vector form point Cto point A.
y
4 ft
5 ft
C
Solution: We need to determine the force Fin terms of its
66
Problem 2.127 The two segments of the L-shaped bar
are parallel to the xand zaxes. The rope AB exerts
a force of magnitude jFjD500 lb on the bar at A.
Determine the cross product rCB ðF, where rCB is the
position vector form point Cto point B. Compare your
answers to the answer to Problem 2.126.
y
4 ft
4 ft
5 ft
C
Solution: We need to determine the force Fin terms of its compo-
nents. The vector from Ato Bis used to define F.
Problem 2.128 Suppose that the cross product of two
vectors Uand Vis UðVD0.IfjUj 6D 0, what do you
know about the vector V?
Problem 2.129 The cross product of two vectors U
and Vis UðVD30iC40k. The vector VD4i
2jC3k. The vector UD⊲4iCUyjCUzk⊳. Determine
Solution: From the given information we have
ijk
Problem 2.130 The magnitudes jUjD10 and jVjD
20.
(a) Use the definition of the cross product to determine
UðV.
(b) Use the definition of the cross product to determine
VðU.
(c) Use Eq. (2.34) to determine UðV.
(d) Use Eq. (2.34) to determine VðU.
U
V
x
y
45°
30°
Problem 2.131 The force FD10i4j(N). Deter-
mine the cross product rAB ðF.
y
x
B
A
rAB
(6, 3, 0) m
(6, 0, 4) m
F
z
Solution: The position vector is
y
A
(6, 3, 0)
68
Problem 2.132 By evaluating the cross product
UðV, prove the identity sin⊲12⊳Dsin 1cos 2
cos 1sin 2.
x
y
V
θ
1
θ
2
U
Solution: Assume that both Uand Vlie in the x–yplane. The
The vectors are:
The cross product is
ijk
y
Problem 2.133 In Example 2.15, what is the minimum
distance from point Bto the line OA?
Solution: Let be the angle between rOA and rOB. Then the
minimum distance is
Problem 2.134 (a) What is the cross product rOA ð
rOB? (b) Determine a unit vector ethat is perpendicular
to rOA and rOB.
( 4, 4, –4) mB
y
rOB
Solution: The two radius vectors are
Problem 2.135 For the points O,A, and Bin Pro-
blem 2.134, use the cross product to determine the length
of the shortest straight line from point Bto the straight
line that passes through points Oand A.
Solution:
(The magnitude of Cis 338.3)
70
Problem 2.136 The cable BC exerts a 1000-lb force F
on the hook at B. Determine rAB ðF.
rAB
B
y
F
6 ft
Solution: The coordinates of points A,B, and Care A(16, 0, 12),
B(4, 6, 0), C(4, 0, 8). The position vectors are
rAB ðFD
12 6 12
0600 800
D2400iC9600jC7200k(ft-lb)
Problem 2.137 The force vector Fpoints along the
straight line from point Ato point B. Its magnitude
is jFjD20 N. The coordinates of points Aand B
are xAD6m,y
AD8m,z
AD4 m and xBD8m,y
BD
1m,z
BD2m.
(a) Express the vector Fin terms of its components.
(b) Use Eq. (2.34) to determine the cross products
rAðFand rBðF.x
y
z
rB
rA
A
B
F
Solution: We have rAD⊲6iC8jC4k⊳m,rBD⊲8iCj2k⊳m,
FD⊲20 N⊳⊲86⊳miC⊲18⊳mjC⊲24⊳mk
⊲2m⊳2C⊲7m⊳2C⊲6m⊳2
Note that both cross products give the same result (as they must).
Problem 2.138 The rope AB exerts a 50-N force T
on the collar at A. Let rCA be the position vector from
point Cto point A. Determine the cross product rCA ðT.
0.4 m
0.15 m
y
BC
T
Solution: We define the appropriate vectors.
Problem 2.139 In Example 2.16, suppose that the
attachment point Eis moved to the location (0.3, 0.3,
y
(0.2, 0.4, ⫺0.1) m
Solution: We first develop the force T.
From Example 2.16 we know that the unit vector perpendicular to the
door is
72
Problem 2.140 The bar AB is 6 m long and is perpen-
dicular to the bars AC and AD. Use the cross product to
determine the coordinates xB,yB,zBof point B.
C
A
B
y
(xB, yB, zB)
Solution: The strategy is to determine the unit vector perpendic-
ular to both AC and AD, and then determine the coordinates that will
agree with the magnitude of AB. The position vectors are:
rOA D0iC3jC0k,rOD D0iC0jC3k,and
rOC D4iC0jC0k. The vectors collinear with the bars are:
rAD D⊲00⊳iC⊲03⊳jC⊲30⊳kD0i3jC3k,
rAC D⊲40⊳iC⊲03⊳jC⊲00⊳kD4i3jC0k.
jRjD0.4685iC0.6247jC0.6247k.
Problem 2.141* Determine the minimum distance
from point Pto the plane defined by the three points
A,B, and C.
(0, 5, 0) m
B
y
P
Solution: The strategy is to find the unit vector perpendicular to
0.7212iC0.4327jC0.5409k. The distance of point Pfrom the plane
is dDrOP ÐerOA ÐeD11.792 2.164 D9.63 m. The second term
is the distance of the plane from the origin; the vectors rOB,orrOC
z
A[3,0,0]
Problem 2.142* The force vector Fpoints along
the straight line from point Ato point B. Use
Eqs. (2.28)– (2.31) to prove that
rBðFDrAðF.
Strategy: Let rAB be the position vector from point A
to point B. Express rBin terms of of rAand rAB . Notice
that the vectors rAB and Fare parallel. x
y
z
rB
rA
A
B
F
Solution: We have
Problem 2.143 For the vectors UD6iC2j4k,
VD2iC7j, and WD3iC2k, evaluate the following
mixed triple products: (a) UÐ⊲VðW⊳; (b) WÐ⊲Vð
U⊳; (c) VÐ⊲WðU⊳.
Solution: Use Eq. (2.36).
624
74
Problem 2.144 Use the mixed triple product to calcu-
late the volume of the parallelepiped.
x
y
z
(140, 90, 30) mm
(200, 0, 0) mm
(160, 0, 100) mm
rOA Ð⊲rOC ðrOD⊳D
60 90 30
40 0 100
200 0 0
D60⊲0⊳C90⊲200⊳⊲100⊳C⊲30⊳⊲0⊳mm3
D1,800,000 mm3
z
Problem 2.145 By using Eqs. (2.23) and (2.34), show
that
UÐ⊲VðW⊳D
UxUyUz
VxVyVz
WxWyWz
.
Problem 2.146 The vectors UDiCUYjC4k,VD
2iCj2k, and WD3iCj2kare coplanar (they
lie in the same plane). What is the component Uy?
Problem 2.147 The magnitude of Fis 8 kN. Express
F
x
(7, 2) m
Solution: The unit vector collinear with the force Fis developed
eDr
jrjD0.6247i0.7809j. The force vector is
Problem 2.148 The magnitude of the vertical force W
is 600 lb, and the magnitude of the force Bis 1500 lb.
Given that ACBCWD0, determine the magnitude of
the force Aand the angle ˛.
BW
Solution: The strategy is to use the condition of force balance to
76
Problem 2.149 The magnitude of the vertical force
vector Ais 200 lb. If ACBCCD0, what are the mag-
nitudes of the force vectors Band C?
70 in. 100 in.
Solution: The strategy is to express the forces in terms of scalar
Problem 2.150 The magnitude of the horizontal force
vector Din Problem 2.149 is 280 lb. If DCECFD0,
what are the magnitudes of the force vectors Eand F?
Solution: The strategy is to express the force vectors in terms of
scalar components, and then solve the force balance equation for the
unknowns. The force vectors are:
Problem 2.151 What are the direction cosines of F?
Refer to this diagram when solving Problems 2.151–
2.157.
y
A
(4, 4, 2) ft
F ⫽ 20i ⫹ 10j ⫺ 10k (lb)
u
Solution: Use the definition of the direction cosines and the
Problem 2.152 Determine the scalar components of
a unit vector parallel to line AB that points from A
Solution: Use the definition of the unit vector, we get
Problem 2.153 What is the angle between the line
AB and the force F?
Solution: Use the definition of the dot product Eq. (2.18), and
Eq. (2.24):
Problem 2.154 Determine the vector component of F
that is parallel to the line AB.
Solution: Use the definition in Eq. (2.26): UPD⊲eÐU⊳e, where e
is parallel to a line L. From Problem 2.152 the unit vector parallel to
Problem 2.155 Determine the vector component of F
that is normal to the line AB.
Problem 2.156 Determine the vector rBA ðF, where
rBA is the position vector from Bto A.
Solution: Use the definition in Eq. (2.34). Noting rBA DrAB,
78
Problem 2.157 (a) Write the position vector rAB from
point Ato point Bin terms of components.
(b) A vector Rhas magnitude jRjD200 lb and is
parallel to the line from Ato B. Write Rin terms of
components.
x
y
A
(4, 4, 2) ft
B (8, 1, ⫺2) ft
F ⫽ 20i ⫹ 10j ⫺ 10k (lb)
u
Solution:
Problem 2.158 The rope exerts a force of magnitude
jFjD200 lb on the top of the pole at B.
(a) Determine the vector rAB ðF, where rAB is the
position vector from Ato B.
(b) Determine the vector rAC ðF, where rAC is the
position vector from Ato C.
x
yB (5, 6, 1) ft
C (3, 0, 4) ft
A
F
Solution: The strategy is to define the unit vector pointing from B
to A, express the force in terms of this unit vector, and take the cross
product of the position vectors with this force. The position vectors
Problem 2.159 The pole supporting the sign is parallel
to the xaxis and is 6 ft long. Point Ais contained in the
y–zplane. (a) Express the vector rin terms of compo-
nents. (b) What are the direction cosines of r?
Ox
z
Solution: The vector ris
rDjrj⊲sin 45°iCcos 45°sin 60°jCcos 45°cos 60°k⊳
Problem 2.160 The zcomponent of the force Fis
80 lb. (a) Express Fin terms of components. (b) what
are the angles x,
y, and zbetween Fand the positive
coordinate axes?
y
x
F
Solution: We can write the force as
80
Problem 2.161 The magnitude of the force vector FB
is 2 kN. Express it in terms of scalar components.
(4, 3, 1) m
y
x
(6, 0, 0) m
(5, 0, 3) m
B
C
D
A
FA
FC
FB
F
z
The magnitude is
jrBDjDp12C32C22D3.74.
The unit vector is
eBD DrBD
jrBDjD0.2673iC0.8018j0.5345k
The force is
FBDjFBjeBD D2eBD (kN) FBD0.5345iC1.6036j1.0693k
D0.53iC1.60j1.07k(kN)
C(6,0,0)
B(5,0,3)
x
z
A
FB
Problem 2.162 The magnitude of the vertical force
vector Fin Problem 2.161 is 6 kN. Determine the vector
components of Fparallel and normal to the line from B
to D.
Problem 2.163 The magnitude of the vertical force
vector Fin Problem 2.161 is 6 kN. Given that FCFAC
FBCFCD0, what are the magnitudes of FA,FB, and
FC?
eAD D0.7845iC0.5883jC0.1961k.
rBD D1iC3j2k,jrBDjDp14 D3.74,
FCFACFBCFCD0,
⊲0.7843jFAj0.2674jFBj0.5348jFCj⊳iD0
Problem 2.164 The magnitude of the vertical force W
is 160 N. The direction cosines of the position vector from
Determine the vector rAG ðW, where rAG is the position
Solution: Express the position vectors in terms of scalar compo-
nents, calculate rAG, and take the cross product. The position vectors
rBG D0.2121iC0.1857j0.1026k.
82
Problem 2.165 The rope CE exerts a 500-N force T
on the hinged door.
(a) Express Tin terms of components.
(b) Determine the vector component of Tparallel to
the line from point Ato point B.
y
A (0.5, 0, 0) m
E
C
D
T
(0, 0.2, 0) m
(0.2, 0.4, ⫺0.1) m
Solution: We have
Problem 2.166 In Problem 2.165, let rBC be the posi-
tion vector from point Bto point C. Determine the cross
product rBC ðT.
y
A (0.5, 0, 0) m
E
C
D
T
(0, 0.2, 0) m
(0.2, 0.4, ⫺0.1) m
Solution: From Problem 2.165 we know that