Multivariable Calculus
Exam 1
You must show all your work. The number of points earned on each problem
will depend upon how well you have justified your solution.
1. Plot the point P(1,5,6).
What is the distance between P and the xy-coordinate plane?
6
What is the distance between P and the yz-coordinate plane?
1
What is the distance between P and the xz-coordinate plane?
5
What is the distance between P and the origin?
2. Let a  5i  2 j  3k .
a. What is |a| ?
52  22  32  38
12  52  62  62
b. Find a vector u in the same direction as a with length 1.
a 5i  2 j  3k

a
38
c. Find a vector b in the opposite direction from a with length 3.
a
3
3 
 5i  2 j  3k 
a
38
3. a. Find the scalar projection of a  3,0, 2 onto b  5,3, 1 .
b. Find the vector projection of b onto a.
Be careful noticing the order of a and b
scalar projection =
vector projection =
3, 0, 2
5,3, 1
5,3, 1
3, 0, 2
5,3, 1
3, 0, 2
2

15  2
17

25  9  1
35
3, 0, 2 
15  2
94
2
3, 0, 2 
17
3, 0, 2
13
4. Use the triple scalar product to verify the vectors
u  i  2 j  3k
v  4i  5 j  6k
w  7i  8 j  9k
are coplanar.
We find the volume of the parallelepiped induced by these three vectors.
1 2 3
5 6
4 6
4 5
4 5 6 1
2
3

8 9
7 9
7 8
7 8 9
1(45  48)  2(36  42)  3(32  35)  3  12  9  0
Since the volume of a parallelepiped is determined by three vectors, these vectors have
produced a parallelepiped of volume 0. Therefore, the vectors must be coplanar.
5. Find the angle between the two planes
3x  4 y  5 z  7
x  y  6z  8
Normal vectors to the two planes are 3, 4,5 and 1,1, 6 .
We can use the formula a b  a b cos to find the angle  between the two planes.
a b  3  4  30  31
a  9  16  25  50  5 2
b  1  1  36  38 
31
31


5 2 38 5 76
  135.3o or 2.36 rad.
cos( ) 
6. A line named l1 contains the two points (3,0,-1) and (2,5,0). Find parametric
equations and symmetric equations for a second line named l 2 that is parallel to l1 and
contains the point (1,2,3).
The direction of l1 is 1, 5, 1 . Any line parallel to l1 has the same direction. Since l 2
also contains the point (1,2,3) parametric and symmetric equations are
x  1  t , y  2  5t , z  3  t and
x-1 y  2 z  3


1
5
1
7. Reduce the equation to one of the standard forms and classify the surface:
x 2  y 2  6 x  8 y  z  20  0
x 2  y 2  z  6 x  8 y  z  20  0
x 2  6 x  9  y 2  8 y  16  z  20  9  16
 x  3   y  4 
2
2
 z 5
This is the standard form of elliptic paraboloid.
8. Find the domain of the vector function r (t )  (
1
)i  ln(5  t ) j  sin(3t  5)k .
t 4
2
We consider each component of the vector function and then find the intersections of the
domains of each of the components.
The domain of the first component is the set of all real numbers except 2 and -2.
The domain of the second component is the set of all real numbers t such that t > -5.
The domain of the third component is the set of all real numbers.
Therefore, the domain of the vector function is
t | t  real numbers and t  5 and t  2 .
9. Find the parametric equations of a line tangent to the space curve of
r (t )  et  2 , t 2  2, 6t  1 at the point (1,2,11).
r (t )  et  2 , t 2  2, 6t  1 
r(t )  et  2 , 2t , 6
At some value of t, et  2  1, t 2  2  2, 6t  1  11. This value is t  2.
At t  2, r(2)  1, 4, 6 .
The parametric equations of the tangent line are
x  1 t
y  2  4t
z  11  6t