Vectors and Vector Algebra

Chapter 4
Vectors and Vector Algebra
b. Find the components and the magnitude of 2.00A − B.
EXERCISES
Exercise 4.1. Draw vector diagrams and convince yourself that the two schemes presented for the construction of
D = A − B give the same result.
Exercise 4.2. Find A − B if A = (2.50,1.50) and
B = (1.00,−7.50)
A − B = (1.50,9.00) = 1.50i + 9.00j
Exercise 4.3. Let |A| = 4.00,|B| = 2.00, and let the angle
between them equal 45.0◦ . Find A · B.
A · B = (4.00)(2.00) cos (45◦ ) = 8.00 × 0.70711 = 5.66.
Exercise 4.4. If A = (3.00)i − (4.00)j and B =
(1.00)i + (2.00)j.
a. Draw a vector diagram of the two vectors.
b. Find A · B and (2.00A) · (3.00B).
A · B = 3.00 × 1.00
+(−4.00)(2.00) = −5.00
(2A) · (3B) = 6(−5.00) = −30.00
2.00A − B = i(4.00 + 1.00)+j(−6.00 + 4.00)
= i(5.00)+j(−2.00)
√
|2.00A − B| = 25.00 + 4.00 = 5.385
c. Find A · B.
A · B = (2.00)(−1.00) + (−3.00)(4.00) = −14.00
d. Find the angle between A and B.
A·B
−14.00
=
= −0.94176
AB
(3.606)(4.123)
α = arccos (−0.94176) = 2.799 rad = 160.3◦
cos (α) =
Exercise 4.6. Find the magnitude of the vector
A = (−3.00, 4.00, −5.00).
√
√
A = 9.00 + 16.00 + 25.00 = 50.00 = 7.07
Exercise 4.7. a. Find the Cartesian components of the
position vector whose spherical polar coordinates are
r = 2.00, θ = 90◦ , φ = 0◦ . Call this vector A.
x = 2.00
y = 0.00
Exercise 4.5. If A = 2.00i − 3.00 j and B = −1.00i +
4.00 j
a. Find |A| and |B|.
|A| = A =
|B| = B =
√
4.00 + 9.00 = 3.606
√
1.00 + 16.00 = 4.123
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00050-1
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z = r cos (θ ) = 0.00
A = (2.00)i
b. Find the scalar product of the vector A from part a
and the vector B whose Cartesian components are
(1.00, 2.00, 3.00).
A · B = 2.00 + 0 + 0 = 2.00
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Mathematics for Physical Chemistry
c. Find the angle between these two vectors. We must first
find the magnitude of B.
√
|B| = B = 1.00 + 4.00 + 9.00
√
= 14.00 = 3.742
2.00
= 0.26726
cos (α) =
(2.00)(3.742)
α = arccos (0.26726) = 74.5◦ = 1.300 rad
Exercise 4.8. From the definition, show that
Exercise 4.9. Show that the vector C is perpendicular to B.
We do this by showing that B · C = 0.
B · C = B x C x + B y C y + Bz C z
Exercise 4.10. The magnitude of the earth’s magnetic
field ranges from 0.25 to 0.65 G (gauss). Assume that the
average magnitude is equal to 0.45 G, which is equivalent to
0.000045 T. Find the magnitude of the force on the electron
in the previous example due to the earth’s magnetic field,
assuming that the velocity is perpendicular to the magnetic
field.
×(1.000 × 105 m s−1 )(0.000045 T)
= 7.210 × 10−19 A s m s−1 kg s−2 A−1
= 7.210 × 10
−19
N
Exercise 4.11. A boy is swinging a weight around his
head on a rope. Assume that the weight has a mass of
0.650 kg, that the rope plus the effective length of the boy’s
arm has a length of 1.45 m and that the weight makes
a complete circuit in 1.34 s. Find the magnitude of the
angular momentum, excluding the mass of the rope and that
of the boy’s arm. If the mass is moving counterclockwise
in a horizontal circle, what is the direction of the angular
momentum?
2π(1.45 m)
= 6.80 m s−1
1.34 s
L = mvr = (0.650 kg)(6.80 m s−1 )
v =
×(1.45 m) = 6.41 kg m2 s−1
5. Find A · B if A = (1.00)i + (2.00) j + (3.00)k and
B = (1.00)i + (3.00) j − (2.00)k.
A · B = 1.00 + 6.00 − 6.00 = 1.00
A × B = A × B = i(A y Bz − A z B y )
+ j(A z Bx − A x Bz ) + k(A x B y − A y Bx )
= i(0.00 − 2.00) + j(4.00 − 0.00)
+ k(0.00 − 2.00)
= −2.00i + 4.00j − 2.00k
9. Find the angle between A and B if A = 1.00i +2.00 j +
1.00k and B = 1.00i − 1.00k.
|F| = F = (1.602 × 10−19 C)
kg m s
A·B=0+0=0
7. Find A × B if A = (0.00,1.00,2.00) and B =
(2.00,1.00,0.00).
= −1 + 2 − 1 = 0.
= 7.210 × 10
1. Find A − B if A = 2.00i + 3.00 j and B = 1.00i +
3.00 j − 1.00k.
3. Find A · B if A = (0,2) and B = (2,0).
This result follows immediately from the screw-thread
rule or the right-hand rule, since reversing the order of the
factors reverses the roles of the thumb and the index finger.
−2
PROBLEMS
A − B = −1.00i + 2.00k.
A × B = −(B × A) .
−19
By the right-hand rule, the angular momentum is vertically
upward.
A · B = 1.00 + 0 − 2.00 = −1.00
√
√
A = 1.00 + 4.00 + 1.00 = 6.00 = 2.4495
√
√
B = 1.00 + 1.00 = 2.00 = 1.4142
−1.00
= −0.28868
cos (α) =
(2.4495)(1.4142)
α = arccos (−0.28868) = 107◦ = 1.86 rad
11. A spherical object falling in a fluid has three forces
acting on it: (1) The gravitational force, whose
magnitude is Fg = mg, where m is the mass of the
object and g is the acceleration due to gravity, equal to
9.80 m s−2 ; (2) The buoyant force, whose magnitude
is Fb = m f g, where m f is the mass of the displaced
fluid, and whose direction is upward; (3) The frictional
force, which is given by Ff = −6π ηr v, where r
is the radius of the object, v its velocity, and η the
coefficient of viscosity of the fluid. This formula for
CHAPTER | 4 Vectors and Vector Algebra
the frictional forces applies only if the flow around
the object is laminar (flow in layers). The object is
falling at a constant speed in glycerol, which has a
viscosity of 1490 kg m−1 s−1 . The object has a mass
of 0.00381 kg, has a radius of 0.00432 m, a mass of
0.00381 kg, and displaces a mass of fluid equal to
0.000337 kg. Find the speed of the object. Assume
that the object has attained a steady speed, so that the
net force vanishes.
Fz,total = 0 = −(0.00381 kg)(9.80 m s−2 )
+ (0.000337 kg)(9.80 m s−2 )
+ 6π(1490 kg m−1 s−1 )(0.00432 m)v
−(0.00381 kg)(9.80 m s−2 )+(0.000337 kg)(9.80 m s−2 )
v=
6π(1490 kg m−1 s−1 )(0.00432 m)
= 0.18 m s−1
13. An object of mass 12.000 kg is moving in the x
direction. It has a gravitational force acting on it equal
to −mgk, where m is the mass of the object and g is
the acceleration due to gravity, equal to 9.80 m s−1 .
There is a frictional force equal to (0.240 N)i. What
is the magnitude and direction of the resultant force
(the vector sum of the forces on the object)?
Ftotal = −(12.000 kg)(9.80 m s−1 )k + (0.240 N)i
Ftotal
= −(117.60 N)k + (0.240 N)i
= (117.6 N)2 + (0.240 N)2 = 118 N
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The angle between this vector and the negative z axis is
0.240
= 0.117◦ = 0.00294 rad
α = arctan
117.6
15. According to the Bohr theory of the hydrogen atom,
the electron in the atom moves around the nucleus
in one of various circular orbits with radius r = a0 n 2
where a0 is a distance equal to 0.529×10−10 m, called
the Bohr radius and n is a positive integer. The mass
of the electron is 9.109 × 10−31 kg. According to the
theory, L = nh/2π , where h is Planck’s constant,
equal to 6.626 × 10−34 J s. Find the speed of the
electron for n = 1 and for n = 2.
Since the orbit is circular, the position vector and the
velocity are perpendicular to each other, and L = mvr .
For n = 1:
v =
L
(6.626 × 10−34 J s)
=
mr
2π(9.109 × 10−31 kg)(0.529 × 10−10 m)
= 2.188 × 106 m s−1
For n = 2
v =
2(6.626 × 10−34 Js)
L
=
mr
2π(9.109 × 10−31 kg)22 (0.529 × 10−10 m)
= 1.094 × 106 m s−1
Notice that the speed for n = 1 is nearly 1% of the
speed of light.
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