Multivariable Calculus
Vectors - Self Assessment A Answers
1. Find a unit vector

u in the direction from A=(2, 1, -3) to (4, -1, -2)
1. __
Solution:
4  2,  1  1,  2  3  2,  2, 1 ,
2 2, 1
,
3
3 3
____
2,  2, 1  3
2. If A=(2, 1, -3) , B= (-1, 2, 5), C= (5, 2, -4) are the three vertices of a triangle In space,
find angle B to the nearest degree. Use the concept of DOT PRODUCT.
2. __14.7 ______
Solution:




1. Let u  BA  3,  1  8 and v  BC  6, 0,  9
 
u  v 18  0  72
90
2. cos B    

uv
74 117
8658
90 
1 
3.  B= cos 
  14.7
 8658 
3. Find the centroid of ∆ABC in problem 2.
3. ________
The centroid is located 2/3 of the distance from a vertex to the midpoint of the opposite
side.
1
 x  x y  y2 z1  z2  
M  1 2 , 1
,
   2, 2, 
2
2  
2
 2
2
If point P is the centroid, then AP  AM
3
2
7
x  2, y  1, z  3  0, 1,
3
2
2
7
5
2
x  2  0, y  1  , z  3   x  2, y  , z  
3
3
3
3
 5 2
Therefore, P   2, ,  
 3 3
The midpoint of BC is
Also, the Centroid is given by
 x  x  x3 y1  y2  y3 z1  z2  z3   5 2 
P 1 2
,
,
   2, ,  
3
3
3

  3 3
-1-
4. Find a point P=(x, y, z) such that
AP 4
 if A=(2, -1, 3) and B=(8, 2, 0).
AB 3
4. _ (1-, 3, -1)________
 4 
4
AP  AB  x  2, y  1, z  3  6. 3,  3  x  10, y  3, z  1
3
3
Therefore, P=(10, 3, -1)
5a) ___ 
5. Find the projection of
a)

u =i - 3j
on

v =2i + j
2 1
i  j ______
5 5
 
u  v  1
2i  j    2 i  1 j
 2 v 
Therefore,
5
5 5
v


1
3
b) u =2i-3j +k on v =i + 2j-3k
a) ____
i  j  k ____
2
2
 
u
v  263
1

Pr ojv u   2 v 
(i  2 j  3k )   (i  2 j  3k )
14
2
v

Pr ojv u
6. Locate the point (-3, 2, -4) in the x-y-z system.
z-axis
P
-3
2
4
y-axis
x-axis
-2-
7. Let




u =2i-3j+ak and v =3i+bj + 5k. Find a and b if u and v are parallel
10
9
and b  
Answer: a 
3
2
Solution: 2i - 3j +ak =r(3i + bj +5k) where r is a scalar.
Therefore, 3r = 2, br = -3 and 5r = a
2
10
9
r  , a
and b  
3
3
2
8. Find the horizontal and vertical components of the vector
3i-4j and the angle ø in degrees that the vector makes with the x-axis.
horizontal _ 3 ______
vertical = __ -4 ________
tan  
b 4
  4

   tan 1 
  53 or 307
a 3
 3 
ø=__ 307  or  53 ____
9. Find the angles  that the vector
 2, 2,1
makes with the x-axis,
the y-axis and the z-axis, respectively.
Solution:
The unit vector in the direction of
 2, 2,1
is
2 2 1

u  , ,
3 3 3
2
2
1
 cos    , cos   , cos  
3
3
3
 2
 2
1
   cos 1   ,   cos 1  , and   cos 1  
 3
 3
 3
Therefore
132  48  71 


Find the norm of the vector
Norm of
 2, 3,1 
 2
2
 2, 3,1
10. ___
 32  12  14
-3-
14 _________
11. Find the Euclidean distance from (-3,2, 1) to (3, -3,5)
77
11. __
distance =
x
__________
 x1    y2  y1   z2  z1   36  25  16  77
2
2
2
2
12. Find two nonzero vectors orthogonal (perpendicular) to
 2, 3, 5
Answer varies: any three numbers a, b, c such that
a, b, c   2, 3, 5  0
For example: a=3, b=2, c=0
__________
Or a=0, b=-5, c=3
__________
13. Find a nonzero vector

w
perpendicular to both
2,  3, 5
and
 1, 2, 3
. Use
the concept of cross product.
13. __
 19,  11, 1 ______
i
j k

w  2  3 5  19i[11 j  k   19,  11, 1
1 2 3
14. Find the angle between the vectors
 1, 2,  3 1, 2,  3


u   1, 2  3 and v  1, 2,  3
 
u  v  1  4  9 12 6
 
2. cos B    
uv
14 14 14 7
1  6 
3.  B= cos    31
7
1. Let
angle: _
-4-
31 ________
15. Find the distance between the lines 3x - 4y=7 and 6x - 8y = 2
Solution. The given lines in the plane are parallel. Therefore, find the distance between
the line 3x- 4x - 7= 0 and any point on the other line.
(-1, -1) is a point on 6x - 8y =2.
dis tan ce  d 
ax1  by1  c
a 2  b2

3(1)  4(1)  7
32   4
2

6
 1.2
5
16. Find the angle between lines L1 and L2 if L1: 2x +3y + 5=0 and L2: x-2y -3=0.
16. ___ 60.3 ________
Solution:

The vector u  B,  A is a direction vector for the line AX+BY = C

The vector v  A, B is a vector normal (perpendicular) to the line AX + BY = C.

The direction of line 2x + 3y + 5 =0 is given by u1  3,  2

The direction of line x - 2y - 3 = 0 is given by u 2   2,  1
  
 4 
62
4
u  v  u v cos   cos  

   cos 1 
  119.7 or 60.3
13 5
65
 65 
17. If L1: x - y -2=0 and L2: 3x - y - 4 = 0, find the equation of the line that
bisects the angle from L1 to L2.
Solution: Let (x, y) be any point on the angle bisector. Any point on the angle bisector is equidistant to the
sides of the angle. Therefore,
by the formula d 
x y2
2

ax1  by1  c
a2  b2
3x  y  4
10
, we obtain that
 5 x  y  2  3x  y  4
 5 ( x  y  2)  (3x  y  4) or

 
 3 5 x 

5 ( x  y  2)  (3x  y  4)
  

5 x   5  1y  2 5  4  0
5  1 y  2 5  4  0 or 3  5 x   5  1 y  2 5  4  0
answer:
3 
-5-
Why?__________
18. Let F1 be a force of 10 pounds acting to the left of a particle.
Let F2 be a force of 8 pounds acting to the left 20 o upward.
Let F3 be a force of 5 pounds acting to the right 50 o downward.
Let F4 be a force of 12 pounds acting upward.
Find the resultant of the system.
Solution:
F1  10i
F2  8 cos160i  8 sin 160 j
F3  5 cos 310i  (5 sin 310) j
F4  12 j
R  F1  F2  F3  F4   10  8 cos160  5 cos 310i  8 sin 160  5 sin 310  12 j
 14.3036i  10.9059 j
 14.30362  10.90592  17.99
10.9059
 180  37  143
 14.3036
The resultant is a force of approximately 17.99 pounds acting to the left and
approximately 37 upward.
R 
and tan  
19. Name three other points of the line through (2,-1, 4,-3) and
(-1, 5, 4, -2)
Solution:
The vector
 1  2, 5  1, 4  4,  2  3   3, 6, 01
gives the direction of the
line.
Any other point can be obtained by
x, y, z, w  2,  1, 4,  3  t  3, 6, 0,1
Answer: (5, -7, 4, -4) for t=-1,


where t is any real number.
(-4, 11, 4, -1) for t=2, (11, -19, 4, -6) for t=-3, etc.
  
20. If r (t )  2t i  ln t j represents the position of a moving particle in the
plane, find the velocity and speed of the motion when t=2 seconds. Distances are given
in meters.
Solution:

1
1
1
v (t )   2t  2  i    j   i  j is the velocity
2
2
t 
2
1
2

 1 1
v (t )        
 2 2
velocity is
1
2

is the speed
2
2
1 1

v (t )   i  j
2 2
and speed is _
-6-
2
2
meters per second______
21. In problem 20, find the acceleration and rate of change of the speed at t=2
seconds.

3
2
Answer: accelerati on  a (t )  4t i  (t ) j

speed  v (t ) 
 2t
2
 
  t  
2
1 2
4t 4  t 2

d  v (t )  1 4 2 1 / 2
rate of change of speed 
 4t  t  (16t 5  2t 3 )
dt
2
 t 5 (8  t 2 )
8  t2
8 4
12
3
3 2
 4










8
8 44
16 2
4 2
t 4  t 2 
t3 4  t2
22. A projectile is fired from a height of 60 ft with an initial speed of 150 ft/sec at an
angle of 30 degrees. Find the maximum height of the projectile and the time it takes to
reach the ground. Answer to the nearest hundredth.
y(0) = 60 ft
V 0  150(cos 30i  sin 30 j  75 3i  75 j
Working with the vertical component
a(t )   gj  32 ft / sec 2
V (t )  32t  c  V (0)  75  0  c  c  0
Therefore, V (t )  32t  75
The maximum high is obtained when V(t)=0, -32t+75=0, t=75/32=2.34375 seconds
The height is given by y(t) =  16t 2  75t  c , but y(0)=60 implies that c=60
Therefore y(t )  16t 2  75t  60
The maximum high is obtain when t=2.34375 and is given by
2
y (2.34375)  162.34375  752.34375  60  147.890625 ft
Maximum height: 147.89 ft________
It hits the ground when y(t)=0, y(t )  16t  75t  60  0  16t 2  75t  60  0
2
75  75 2  4(16)( 60) 75  9465
t

 it hits the ground when t=5.32 seconds
2(16)
32
time: ___5.32 seconds______
-7-
23. Curves in space are generally represented by parametric equations or, equivalently,

by a vector-valued function r t . When the parameter t represents time, r’(t)=v(t) and

r’’(t)=a(t). The moving trihedral consists of three mutually orthogonal (perpendicular)
 



unit vectors T , N and B , where T is the unit tangent vector, and N is the principal
normal vector and

B
is called the binormal vector and
Find the curvature and the binormal vector
2
r(t)  2ti  t j  (1  t)k

B
 
B T N .
at t=1 for the curve
Solution:



r (t )  2t , t 2 , 1  t , v (t )  2, 2 t ,  1 , a (t )  0, 2 , 0

2 2 1



 at t  1, r (1)  2, 1, 0 , v (1)  2, 2 ,  1 , a (t )  0, 2 , 0 , T  , , 
3 3 3
i j k
 
 
 
v  9  3, v  a  4, v  a  2 2  1  2, 0, 4 , v  a  20  2 5
0 2
2 5
 0.165634665
27
4 2 2 1
8 10 4
0, 2, 0   , , 
 , ,
 
 

va 2 5
v a 4
3 3 3 3
9 9 9
aT 
 , a N

N

v
3
v
3
2 5
2 5
3
3

4
5
2
N 
,
,
3 5 3 5 3 5
 
2 5
27
0
  
 B T  N 
answer : curvature 
i
2
3
4

3 5
j
2
3
5
3 5
k
1
9
18

,0,
3
9 5 9 5
2
3 5
binormal vector = __
-8-

1
2
, 0,
5
5
1
2
, 0,
5
5
____
24. The plane curve x = z -3
resulting surface and sketch.
Solution:
is revolved about the z -axis. Give an equation for the
x  z  3  x 2  ( z  3) 2
The surface that results by revolving about the z-axis is obtain by replacing
x 2 by x 2  y 2
Therefore, the equation of the surface is
x 2  y 2  ( z  3) 2
which is a cone.
Vertex at (0, 0, 3)
3)
3
25. Find the curvature of y= x at (-1, -1)

y 
1   y 
2 3/ 2

6x
 
1  3x 2 2 


3/ 2

Ans: __________________
6
1  3 
2 3/ 2

6
3
3 10


50
10 10 5 10
26. Find the curvature, tangential and normal components of the acceleration of the
vector function f (t )  t , t 2 , t 3



r (t )  t , t 2 , t 3  v (t )  1, 2t , 3t 2  a (t )  0, 2, 6t



 at t  2, r (2)  2, 4, 8 , v (2)  1, 4, 12 , a (2)  0, 2, 12 , v  161
i j k




v (2)  a (2)  0  8  144  152, v (2)  a (2)  1 4 12  24,  12, 2
0 2 12


v (2)  a (2)
576  144  4
776
 0.013636
3
161 161
v3
161




v
(
2
)

a
(2)
v
(
2
)

a
(
2
)
152
776
aT 

 11.979 , a N 

 2.1954
v
v
161
161
 




Curvature: 0.013636____Tangential component: _11.979___Normal component: __2.1954____
-9-
27. Find the arc-length parametrization of the helix
x(t)=3cost, y(t)=3sint and z(t)=4t

v (t )   3 sin t , 3 cost , 4  v  9 sin 2 t  9 cos2 t  16  25  5
ds
s
 5  ds  5dt  s  5t  t 
dt
5
4s
s
s
anwer : x( z )  3 cos , y ( z )  3 sin  , z 
5
5
5
28. Convert
a) (2, 1, -2) to cylindrical coordinates
a) __


5 , 26.57,  2 ________
r 2  x2  y2  r 2  5  r  5
y
1
  Arc tan  Arc tan  26.57
x
2


3 , 135, 125.26 ________
y
1
 2  x 2  y 2  z 2    3,   Arc tan  Arc tan
 135
x
1
z
 1 
  Arc cos   Arc cos   125.26
 3

b) (-1, 1, -1) to spherical coordinates
c)
  3 
2, ,
 from spherical
 3 2 
b) __
to cylindrical coordinates
 3 
r   sin   2 sin    2
 2 
  

3
 3 
z   cos   2 cos   0
 2 
c) __(-2,
-10-

, 0)_______
3
d)
  3 
2, ,
 from spherical
 3 2 
to rectangular
 3    
x   sin  cos   2 sin   cos   1
 2  3
 3    
y   sin  sin   2 sin   sin     3
 2  3
 3
z   cos   2 cos
 2

0



d) ___  1, 3 , 0 _______
e)
 

2, ,  3 from cylindrical
 6

to rectangular coordinates
 
x  r cos   2 cos   3
6
 
y  r sin   2 sin    1
6
z  z  3
e) _
f)
 

2, ,  3 from cylindrical
 6


to spherical coordinates
 2  r 2  z 2  4  9    13

     30
cos  
6
z



3 , 1,  3 _________
 3 
,    arccos
  146.31
13
 13 
3
29. Convert the given equation to cylindrical and spherical
a) x  2 y  z  5
Solution : r cos   2r sin   z  5
b) x 2  y 2  2 x
Solution : r 2  2r cos 0  r  2 cos 
-11-
30. Convert the given equations to rectangular coordinates
a)   
a) ____y=-x________
4
 
Solution:
b)  



y
 
 tan   tan      1  y   x
4
x
 4
b) ___ x 2  y 2  z 2 _________
4
r
 
tan   tan     1 
z
4
c) r=4cos
r2
z
2
 1  r 2  z 2  x 2  y 2  z 2 which is a cone
c) __ x 2  y 2  4 x _____
Solution: r 2  4r cos   x 2  y 2  4 x
2
2
d) z  2r  4
Solution: z 2  2 x 2  y 2  4  2 x 2  2 y 2  z 2  4
d) ____ 2 x 2  2 y 2  z 2  4 ________

e)

r 2  2z 2  4
e) _ x 2  y 2  2 z 2  4 ___________
x 2  y 2  2z 2  4
 2  5  4  0
 2  5  4  0    1(   4)  0    1 or   4
f)
x 2  y 2  z 2  1 or x 2  y 2  z 2  4
f ) x 2  y 2  z 2  1 or x 2  y 2  z 2  4
2
2
31. The hyperbola y  z  1 , x=0 is rotated around the z-axis. Write a cylindircal
equation for the surface thereby generated.
2
y 2 by x 2  y 2 to obtain x
r 2  z 2  1 in cylindrical coordinates
Solution: replace
equivalent
-12-
 y 2  z 2  1 which is
32. A sphere of radius 2 is centered at the origin. A hole of radius 1 is drilled through
the sphere, with the axis of the hole lying on the z-axis. Describe the solid region that
remain in
a) Cylindrical coordinates
a) ___ 1  r 2  4  z 2 _______________
b). Spherical coordinates
b) ____ csc     2 ______________
33. Find the point of intersection of the lines
L1:
x 5 y2 z 1
x 9 y5 z 7


and L2 :


3
1
4
1
2
4
Solution:
 x  5  3t

L1 :  y  2  t
 z  1  4t

 x 9s
 5  3t  9  s
 3t  s  4


and L2 :  y  5  2 s  2  t  5  2 s  
t  2 s  7
 z  7  4 s
1  4t  7  4 s

 t  3 and s  5
 x  14

y5
 z  13

Therefore, the po int of int er sec tion is (14,5, 13)
34. Find the point of intersection of the lines:
L1:
x  7 y  6 z  11
x1 y1 z 8


and L2 :


3
1
4
1
2
3
Solution:
 x  7  3t

L1 :  y  6  t
 z  11  4t

 x  1 s
  7  3t  1  s
3t  s  8


and L2 :  y  1  2 s   6  t  1  2 s  
t  2 s  5
 z  8  3s
11  4t  8  3s

 t  3 and s  1
 x2

 y  3
 z  1 or 5

Therefore, the lines are skewed , they do not int er sec t.
-13-
35. Find the parametric form of the line of intersection of the planes
2x - y + 3z - 2=0 and x + 3y + 2z - 4 = 0
Solution:

The vector u  2,  1, 3 Is normal to the plane 2x - y +3z -2 =0

The vector v  1, 3, 2 Is normal to the plane x +3y +2z -4 =0
i
j k
 
The direction of the line is given by the vector u  v  2  1 3  11i  j  7k
1 3 2
Find any point on the intersection of the lines by using any number for one of the
variables and finding the values of the other two variables.
2 x  y  1
1
5
Let z=1  
 x   and y 
7
7
 x  3y  2
Therefore, the line goes thought the point (-1/7, 5/7, 1) with direction number given by -11i - j + 7k
Answer:
1

 x  7  11t

5
Line of int er sec tion is given by  y   t
7

z

1
 7t


36. Find the equation in parametric form of the line through (-2, 3, 1) parallel to the line
x 1 y  2 z


3
1
4
Solution: the given line goes through the point (1, -2, 0) with direction vector
3,  1, 4 .
The parallel line goes thought the point (-2, 3, 1) and has the same direction vector.
Therefore, the parametric equation is given by:
 x   2  3t

t
y  3
 z  1  4t

-14-
37. Classify each of the following ellipsoids as PROLATE or OBLATE
a) 2 x  2 y  3z  6
a) __oblate___
x2 y2 z 2

 1
b)
5
4
5
b) ___Oblate___
2
2
2
x 2  y  3 z  2
c)


1
10
5
5
c) __ Prolate ____
d) 3x  y  3( x  3)  1
d) ___Prolate___
2
2
2
2
2
38. Find the surface generated when the following curves in the specified plane is revolved
around the specified axis.
a) 2x + y =6 in the xy-plane is revolved about the x-axis.
2
2
Note: y  6  2 x  y 2  6  2 x 
a __ y 2  z 2  6  2 x  ___________
b) 2x + y=6 in the xy-plane is revolved about the y-axis.
Note : x 
c)
6 y
(6  y ) 2
 x2 
2
4
(6  y ) 2
4
z  2x 2 in the xz-plane is revolved about the x-axis
z  2x 2  z 2  4x 4
d)
b) x 2  z 2 
c) y 2  z 2  4x 4
z  2x 2 in the xz-plane is revolved about the z-axis


d) __ z  2 x 2  y 2 __________
e) z  y  3 in the yz-plane is revolved about the y-axis e) ___ x 2  z 2  y 2  3 _________
2
2
-15-
f) z  y  3 in the yz-plane is revolved about the z-axis f) ___ z 2  x 2  y 2  3 ___
2
2
g) x=3 in the xy-plane is revolved about the y-axis
g) __ x 2  z 2  9 _________
h) x=3 in the xy-plane is revolved about the x-axis
h) __the plane x=3_________
i) x=3 in the xy-plane is revolved about the z-axis
x 2  y 2  9 ________
i) ___the outside of the cylinder
j) x  ( y  5)  9 in the xy-plane is revolved about the x-axis
2
2

x 2  ( y  5) 2  9  y  5  9  x 2  y 2  5  9  x 2

2
k) x  ( y  5)  9 in the xy-plane is revolved about the y-axis
2

j) y 2  z 2  5  9  x 2

2
2
k) __ x 2  z 2  ( y  5) 2  9 ___________
39. Match the equation on the left with the appropriate space description on the right
___e__ 1) xz + 3 = 0
a) empty set
___c____ 2) x + 1 = 0, x = y
b) Point.
___b___ 3) y= 5, z+3=0, x + y + z = 2
c) Line
___c___ 4) x=3, 2x - y + 3z = 5
d) Plane
___f___ 5) y   4  x
2
 z2
e) Cylinder
___a___ 6) x2 + 4y2 + 9z2 + 36 =0
f) Sphere
2
___e___ 7) y = x (parabolic cylinder)
g) Prolate spheroid
__ h__ 8) x2 + 4y2 + z2 =16
h) Oblate spheroid
__ k___ 9) x2 - z2= y
i) Elliptical cone
__ l ___ 10) x2 + y2 + 4x = z2
j) Elliptic paraboloid.
__k___ 11) 0 = y +1 - x2 + 9z2
k) Hyperbolic paraboloid
__i__ 12) x2 = y2 - 9(z -1)2
l) Hyperboloid of one sheet.
m) Hyperboloid of two sheets
-16-
40. Find the equation of the plain that contains the parallel lines
x 

L1  y 

z 
x 

2t 5 and L2  y 

t
4
z 
3t
2
3t
5
2t 2
t
1
Solution:

The vector v  3,  2, 1 gives the direction of both lines and lies on the plane.
The points (-2, 5, -4) and (5, 2, -1) belong to the plane because they belong to the lines

and the plain contains the lines. Therefore the vector w  5  2, 2  5,  1  4  7,  3, 3
lies on the plane.
i
j k
 
The vector v  w  3  2 1  3i  2 j  5k   3,  2, 5 is normal to the plane.
7 3 3
Let A=-3, B=-2 and C=5.
The equation of the plane is given by -3x -2y +5z +D=0. Using the point (5, 2, -1), or the
other point, we obtain -15-4-5+D=0 or D=24
The equation of the plane is given by -3x -2y +5z +24=0 or 3x + 2y +5z -24=0
-17-