Math B6C – Chapter 10 Quiz
* Solutions *
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1. Find a unit vector in the direction of ( 3, 1 ).
The y–component equals what?
Just divide this vector by it’s length, and you have a unit vector in the same direction:
(3,1)
(3,1)  3 1 


,

10  10 10 
32 12
The y-component of this vector is thus
1
1
 10 2 .
10
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2. A = (–1, 2) and B = (2, 3). The vector projection of A onto B has x–component equal to what?
The vector projection of A onto B is given by:

A






B B
B B
  A B
 2
 
 3
B   1 2 
   
2
  2   3  22  32
B 

The x-component of this projection vector is thus

 2  6
 2
 
 3
 13
8


4 2 13 
  
13  3 12 
 
13 
8
13
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3. A = (1, 2, 1) and B = (1, 1, –1). Find a unit vector in the direction of B x A.
Its z–component equals what?
 1  1 
     1
 1    2  
 1 1   1
   
 3
2
1 1 1 1 
 
,
,
 1 (2), (1 (1)),2 1  2
1 1 1 1 2 
 1
 
Dividing this vector by its length yields a unit vector in this direction:
(3, 2,1)
(3, 2,1)  3 2 1 

 
,
,

14
32  (2)2 12
 14 14 14 
The z-component of this vector is thus
1
1
 14 2 .
14
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4. Find the cosine of the (acute) angle between vectors A and B in problem #3.
Let  denote this angle.
Then
cos 
A B  A B cos , so
1 
 
 2
1 
 
 1
 
 1
 1
 
AB
1  2 1 2
2.




3
AB
1 4 1 111
6 3 3 2
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5. If A = (2, –3, 5), then
2
2
A  22  (3)2  52  4  9  25  38 .
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

6. Find a unit tangent vector to the curve y  x2  sin x at the point  , 1  2  .
2

The absolute value of the x–component of this tangent vector equals what?
dy
 2 x  cos x ,
dx
which at the given point becomes
dy


 2  cos   .
dx
2
2
1
Thus a tangent vector to the curve is given by   , since it has the same slope as the tangent to the curve
 
at the given point.
Thus a unit tangent is
component of this vector is
1
1  2




1  2 


1
 
 
1 

1   2  , and the absolute value of the x

2
1  

.
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

7. Find a unit normal vector to the curve y  x2  sin x at the point  , 1  2  .
2

The absolute value of the x–component of this normal vector equals what?
1
 
We found in problem #6 that   is tangent to the curve at the given point. Since   is
 1
 
perpendicular to the tangent vector (since their dot products is zero), it must be a normal vector to the
 

 


 
1
 1  2 


curve at the given point. Normalizing it we get a unit normal to the curve,
,
2
2

1
 1 

 1  2 


hence the absolute value of the x-component of this normal vector is

1  2
.
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8. Find the area of the parallelogram determined by the vectors A = (–3, 1, 0) and B = (5, –2, 1).
This is just the length of the cross product of these vectors:
 3  5 
3 5 3 5 
     1 2

1


2

,

,

    

0
1
0
1
1

2
 0  1  

   
1 0, (3  0),6  5
1
 
 3
1
 
Thus the area of the parallelogram is 1 9 1  11
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 t  1


9. Find the distance from the point A = (3, 1, –2) to the line parametrized by r (t )   2t  .
3  t 


Consider the diagram:
Let A=(3,1,–2), and P=(1,0,3) (corresponding to taking t=0 in our parametrization). Let vector W be in
the direction of the line, which looking at the parametrization we see can be chosen to be the vector
(1,2,–1). The vector V is given by A–P=(2,1,–5). We seek the distance d, which by basic trigonometry is
given by
Thus,
d  v sin .
d
But a basic property of the cross product states that
v w
w
, and so we first must compute the cross product.
 2   1
     1
v  w   1    2  
 5  1  5
   
And so,
v  w  v w sin .
 9
2
2 1 2 1 
 
,
,
  1  (10), (2  (5)),4 1  3
1  5 1 1 2 
 3
 
v  w  92  (3)2  32  99  3 11.
d
NOTE: This can also be written as
v w
w

Since
w  12  22  (1)2  6,
3 11
66

.
2
6
66
66
33


.
2
4
2
we have
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10. Find the distance from the point (3, 1, –2) to the plane 5x – 2y + z = 2.
Consider the diagram:
Let q = (3, 1, –2), and let let P be the point (1,1,–1) (any other point on the plane would do just as well).
Let
 5
(5, 2,1)
1  
n̂  2

 2 .
2
2
30
5  (2) 1
 1

We seek the distance d from q to the plane.

But d is just the length of the projection of vector
v q
 2
 
p   0  onto the normal vector n.
 1
 
 2
 5
  1   10  0 1
d  v nˆ   0
 9
 2 
30
30
 1 30  1 
 
 
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11. Parametrize the line through point (1, 5, –3) parallel to vector (–1, 3, 2).
 1  1 
   
r (t )  t  3    5  .
 2   3
   
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12. Find the cosine of the angle between the planes x + 3y – z = 5 and 2x – y + 5z = 3.
We seek the cosine of the angle between the normal vectors to these curves. As in problem #4, we
compute:
cos 
 1
 
 3
 1
 
 2
 
 1
 5
 
AB
2 35
6



AB
1  9  1 4  1  25 11 30
330
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13. Find the distance from the line of intersection of the planes in problem #12 to the point (1,1,1).
First we’ll find a parametric representation of the line of intersection of the planes. The direction of this
line of intersection is the direction of the cross products of the normal vectors to the planes:
 1   2
14 

1
2
1
2
     3 1
 
  15 1, (5  (2)), 1  6  7 
A B   3  1  
,
,
1 5 3 1 
 1  5   1 5
 7 
   
 
 2 
 
We could use this vector for our line, or the simpler vector  1  (which we get by factoring –7 out of
1
 
that vector). All we need now is a point on the line of intersection of the planes. Here’s one way to find
a point on this line of intersection… Let z=0, and solve the system of equations for x and y:


2x – y = 3.
x + 3y = 5,
The solution is x=2, y=1. So the point (2,1,0) satisfies both plane equations, and is thus on the line of
intersection of the two planes. Thus the line of intersection is parametrized by
 2  2
   
r (t )  t  1   1 .
 1  0
   
Finally, to find the distance from this line to point (1,1,1), we proceed as in problem #9,
with A=(1,1,1), P=(2,1,0), W=(–2,1,1), and V=A–P=(–1,0,1).
 1  2
    0
v w   0  1   
 1   1   1
   
And so,
 1
1 1 2 1 2 
 
,
,
  0 1, (1 (2)), 1 0   1
1
1 1 0 1 
 1
 
v  w  (1)2  (1)2  (1)2  3.
d
v w
w
Since

w  (2)2 12 12  6,
3
1

.
6
2
we have
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14. Convert (1, 2, 3) from rectangular coordinates to cylindrical coordinates:
r  12  22  5 &   arctan 2  arctan 2, so in cylindrical coordinates this point is given by
1

5, arctan 2, 3
.
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15. Convert (1, 2, 3) from rectangular coordinates to spherical coordinates.
(approximately)
We have
  12  22  32  14  3.741657386773941385583749, and
r  12  22  5 &   arctan 2  arctan 2  1.107148717794090503017065.
1
Also,
  arctan r  arctan 5  0.6405223126794245741435580.
Thus
 ,,   (3.74166,
z
3
0.64052, 1.10715) .
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16. Convert the equation
   cot  csc
from spherical coordinates to cylindrical coordinates.
   cot  csc   cos 1   cos2
sin  sin 
sin 
Multiplying both sides of the equation by
sin 2 
(to clear the denominator), we get:
 sin 2    cos.
Since
 sin  r &  cos  z, if we multiply both sides of the equation by  , we get
 2 sin 2    cos
2
  sin     cos
r 2  z .
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