1
Solution-Manual-for-Vector-Calculus-Miroslav-Lovric
Vector Calculus Solutions all Chapters Guide (Course Code
MA421)
CHAPTER 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26,10,4) Thus
a
b) the magnitude of 5ax + N − 3M:
(5,0,0) + (8,7,−2) − (−30,12,−24) = (43,−5,22), and |(43,−5,22)| = 48.6. c)
1.2. The three vertices of a triangle are located at A(−1,2,5), B(−4,−2,−3), and C(1,3,−2).
a) Find the length of the perimeter of the triangle: Begin with AB 1),
7). Then the perimeter will be
.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side BC: The
vector from the origin to the midpoint of
The vector from the origin to the midpoint of BC is az). The
vector from midpoint to midpoint is now M a + 7az). The unit
vector is therefore
MAB − MBC (−2ax − ay + 7az)
MM = = = − 0. 27ax − 0. 14ay +0 . 95az
|M AB − M BC | 7. 35
a
where factors of 1/2 have cancelled.
pg. 1
, 2
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the unit
vector is therefore parallel to AC. First we find AC = 2ax +ay −7az, which we recognize as −7.35aMM.
The vectors are thus parallel (but oppositely-directed).
1.3. The vector from the origin to the point A is given as (6, 4), and the unit vector directed from
the origin toward point B is (2,−2,1)/3. If points A and are ten units apart, find the
coordinates of point B.
With
1),
we use the fact that |B − A| = 10, or
Expanding, obtain
or
B2 − 8B − 44 = 0. Thus 2 75 (taking positive
option) and so
B
1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector
in rectangular components that lies in the xy plane, is tangent to the circle at ( 0), and is in the
general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is t = aφ. Its x and y
components are tx = aφ · ax = −sinφ, and ty = aφ · ay =√cosφ. At the point (√3,1), φ = 30◦, and so t.
◦ ◦
= − sin30 a + cos 30 a =0 . 5(− a + 3a )
1.5. A vector field is specified as xG = 24xyax + y12(x2 + 2)ayx+ 18z2azy. Given two points, P(1,2,−1) and Q(−2,1,3),
find:
a) G at P: G(1,2
b) a unit vector in the direction of G at Q: G(−2,1,3) = (−48,72,162), so
(− 48, 72, 162)
G = =( − 0. 26, 0. 39, 0. 88)
|(− 48, 72, 162)|
a
c) a unit vector directed from Q toward P:
P− Q (3, − 1, 4)
QP = = =(0 . 59, 0. 20, 0. 78) a − |
|P
√ −
Q 26
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy,12(x2 + 2),18z2)|, or
pg. 2
,3
10 = |(4xy,2x2 + 4,3z2)|, so the equation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. If a is a unit vector in a given direction, B is a scalar constant, and r = xax + yay + zaz, describe the surface
r · a = B. What is the relation between the the unit vector a and the scalar B to this surface? (HINT:
Consider first a simple example with a = ax and B = 1, and then consider any a and B.):
We could consider a general unit vector, a = A1ax + A2ay + A3az, where
Then r·a = A1x+A2y +A3z = f(x,y,z) = B. This is the equation of a planar surface, where f = B. The
relation of a to the surface becomes clear in the special case in which a = ax. We obtain r · a = f(x)
= x = B, where it is evident that a is a unit normal vector to the surface (as a look ahead (Chapter
4), note that taking the gradient of f gives a).
1.7. Given the vector field E = 4zy2 cos2xax + 2zy sin2xay + y2 sin2xaz for the region |x|, |y|, and |z| less than
2, find:
a) the surfaces on which Ey = 0. With Ey = 2zy sin2x = 0, the surfaces are 1) the plane , with |x z|
< 2; 3) the plane , with |y| < 2,
|z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which : This occurs when 2zy sin2x = y2 sin2x, or on the plane 2z
= y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos2x = zy sin2x = y2 sin2x = 0. This
condition is met on the plane , with |x| < 2, |z| < 2.
1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between two
vectors by finding the angle between A = 3ax − 2ay + 4az and B = 2ax + ay − 2az. Does this ambiguity exist
when the dot product is used?
We use the relation A × B = |A||B|sinθn. With the given vectors we find
A n
±n
where n is identified as shown; we see that n can be positive or negative, as sinθ can be positive
or negative. This apparent sign ambiguity is not the real problem, however, as we really want the
magnitude of the angle anyway. Choosing the positive sign, we are left with
Two values of θ (75.7◦ and 104.3◦) satisfy this equation, and
hence the real ambiguity.
pg. 3
, 4
In using the dot product, we find√ A · B = 6 − 2 − 8 = −4 = |A , or
75.7◦. Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one θ value results when using
the dot product, so no ambiguity.
1.9. A field is given as
G
Find:
a) a unit vector in the direction of G at P(3,4,−2): Have Gp = 25/(9+16)×(3,4,0) = 3ax+4ay, and |Gp| =
5. Thus aG = (0.6,0.8,0).
b) the angle between G and ax at P: The angle is found through aG · ax = cosθ. So cosθ = (0.6,0.8,0) ·
(1,0,0) = 0.6. Thus .
c) the value of the following double integral on the plane y = 7:
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A ) and B = b(ax − ay
+ az). Now use A · B = |A||B|cosθ, or b2(1 − 1 + 1) = (√3b)(√3b)cosθ ⇒ cosθ =
. This result (in magnitude) is the same for any two diagonal vectors.
1.11. Given the points M(0.1,−0.2,−0.1), N(−0.2,0.1,0.3), and P(0.4,0,0.1), find:
a) the vector RMN: RMN = (−0.2,0.1,0.3) − (0.1,−0.2,−0.1) = (−0.3,0.3,0.4).
b) the dot product RMN · RMP: RMP = (0.4,0,0.1) − (0.1,−0.2,−0.1) = (0.3,0.2,0.2). RMN · RMP = (−0.3,0.3,0.4)
· (0.3,0.2,0.2) = −0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RMN on RMP:
R
d) the angle between RMN and RMP:
pg. 4
Solution-Manual-for-Vector-Calculus-Miroslav-Lovric
Vector Calculus Solutions all Chapters Guide (Course Code
MA421)
CHAPTER 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26,10,4) Thus
a
b) the magnitude of 5ax + N − 3M:
(5,0,0) + (8,7,−2) − (−30,12,−24) = (43,−5,22), and |(43,−5,22)| = 48.6. c)
1.2. The three vertices of a triangle are located at A(−1,2,5), B(−4,−2,−3), and C(1,3,−2).
a) Find the length of the perimeter of the triangle: Begin with AB 1),
7). Then the perimeter will be
.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side BC: The
vector from the origin to the midpoint of
The vector from the origin to the midpoint of BC is az). The
vector from midpoint to midpoint is now M a + 7az). The unit
vector is therefore
MAB − MBC (−2ax − ay + 7az)
MM = = = − 0. 27ax − 0. 14ay +0 . 95az
|M AB − M BC | 7. 35
a
where factors of 1/2 have cancelled.
pg. 1
, 2
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the unit
vector is therefore parallel to AC. First we find AC = 2ax +ay −7az, which we recognize as −7.35aMM.
The vectors are thus parallel (but oppositely-directed).
1.3. The vector from the origin to the point A is given as (6, 4), and the unit vector directed from
the origin toward point B is (2,−2,1)/3. If points A and are ten units apart, find the
coordinates of point B.
With
1),
we use the fact that |B − A| = 10, or
Expanding, obtain
or
B2 − 8B − 44 = 0. Thus 2 75 (taking positive
option) and so
B
1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector
in rectangular components that lies in the xy plane, is tangent to the circle at ( 0), and is in the
general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is t = aφ. Its x and y
components are tx = aφ · ax = −sinφ, and ty = aφ · ay =√cosφ. At the point (√3,1), φ = 30◦, and so t.
◦ ◦
= − sin30 a + cos 30 a =0 . 5(− a + 3a )
1.5. A vector field is specified as xG = 24xyax + y12(x2 + 2)ayx+ 18z2azy. Given two points, P(1,2,−1) and Q(−2,1,3),
find:
a) G at P: G(1,2
b) a unit vector in the direction of G at Q: G(−2,1,3) = (−48,72,162), so
(− 48, 72, 162)
G = =( − 0. 26, 0. 39, 0. 88)
|(− 48, 72, 162)|
a
c) a unit vector directed from Q toward P:
P− Q (3, − 1, 4)
QP = = =(0 . 59, 0. 20, 0. 78) a − |
|P
√ −
Q 26
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy,12(x2 + 2),18z2)|, or
pg. 2
,3
10 = |(4xy,2x2 + 4,3z2)|, so the equation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. If a is a unit vector in a given direction, B is a scalar constant, and r = xax + yay + zaz, describe the surface
r · a = B. What is the relation between the the unit vector a and the scalar B to this surface? (HINT:
Consider first a simple example with a = ax and B = 1, and then consider any a and B.):
We could consider a general unit vector, a = A1ax + A2ay + A3az, where
Then r·a = A1x+A2y +A3z = f(x,y,z) = B. This is the equation of a planar surface, where f = B. The
relation of a to the surface becomes clear in the special case in which a = ax. We obtain r · a = f(x)
= x = B, where it is evident that a is a unit normal vector to the surface (as a look ahead (Chapter
4), note that taking the gradient of f gives a).
1.7. Given the vector field E = 4zy2 cos2xax + 2zy sin2xay + y2 sin2xaz for the region |x|, |y|, and |z| less than
2, find:
a) the surfaces on which Ey = 0. With Ey = 2zy sin2x = 0, the surfaces are 1) the plane , with |x z|
< 2; 3) the plane , with |y| < 2,
|z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which : This occurs when 2zy sin2x = y2 sin2x, or on the plane 2z
= y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos2x = zy sin2x = y2 sin2x = 0. This
condition is met on the plane , with |x| < 2, |z| < 2.
1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between two
vectors by finding the angle between A = 3ax − 2ay + 4az and B = 2ax + ay − 2az. Does this ambiguity exist
when the dot product is used?
We use the relation A × B = |A||B|sinθn. With the given vectors we find
A n
±n
where n is identified as shown; we see that n can be positive or negative, as sinθ can be positive
or negative. This apparent sign ambiguity is not the real problem, however, as we really want the
magnitude of the angle anyway. Choosing the positive sign, we are left with
Two values of θ (75.7◦ and 104.3◦) satisfy this equation, and
hence the real ambiguity.
pg. 3
, 4
In using the dot product, we find√ A · B = 6 − 2 − 8 = −4 = |A , or
75.7◦. Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one θ value results when using
the dot product, so no ambiguity.
1.9. A field is given as
G
Find:
a) a unit vector in the direction of G at P(3,4,−2): Have Gp = 25/(9+16)×(3,4,0) = 3ax+4ay, and |Gp| =
5. Thus aG = (0.6,0.8,0).
b) the angle between G and ax at P: The angle is found through aG · ax = cosθ. So cosθ = (0.6,0.8,0) ·
(1,0,0) = 0.6. Thus .
c) the value of the following double integral on the plane y = 7:
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A ) and B = b(ax − ay
+ az). Now use A · B = |A||B|cosθ, or b2(1 − 1 + 1) = (√3b)(√3b)cosθ ⇒ cosθ =
. This result (in magnitude) is the same for any two diagonal vectors.
1.11. Given the points M(0.1,−0.2,−0.1), N(−0.2,0.1,0.3), and P(0.4,0,0.1), find:
a) the vector RMN: RMN = (−0.2,0.1,0.3) − (0.1,−0.2,−0.1) = (−0.3,0.3,0.4).
b) the dot product RMN · RMP: RMP = (0.4,0,0.1) − (0.1,−0.2,−0.1) = (0.3,0.2,0.2). RMN · RMP = (−0.3,0.3,0.4)
· (0.3,0.2,0.2) = −0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RMN on RMP:
R
d) the angle between RMN and RMP:
pg. 4
