Solutions to Midterm 1
10:30 Section
You must show all your work. The number of points earned on each problem will be
determined by how well you have justified your work.
1. The line segment AB with endpoints A(3,1,6) and B(1,5,9) is shown below.
a. Plot the projection of A onto the xy-coordinate plane and name the point C. What are
the coordinates of C (3,1,0)
b. Plot the projection of B onto the xy-coordinate plane and name the point D. What are
the coordinates of D? (1,5,0)
c. What is the distance between the points A and B?
22  42  32  29
d. What is the distance between the points C and D?
22  42  20  2 5
2. Let a  3i  4 j  7k and b  3j  7k .
a. Find a  2b .
(3i  4 j  7k )  2(3j  7k ) 
3i  4 j  7k  6 j  14k 
3i  2 j  21k
b. Find 3 b a .
3 3 j  7k (3i  4 j  7k ) 
3 9  49(3i  4 j  7k ) 
3 58(3i  4 j  7k )
3. What are the direction angles of the vector v  8, 2, 7 . Express the answers in
radians.
8, 2, 7  64  4  49  117  3 13
8
   2.40 radians
3 13
2
cos  
   1.385 radians
3 13
7
cos  
   2.275 radians
3 13
cos  
4. Find a vector orthogonal to the plane containing the points (0,3,0), (0,0,7), and
(-3,0,5).
To find such a vector we first find two vectors on the plane and then find their cross
product.
u  0  0,3  0, 0  7  0,3, 7
Let
v  0  3,3  0, 0  5  3,3, 5
Then u  v  6i  21j  9k .
5. Find the volume of the parallepiped determined by the vectors
a  3, 2,0 , b  4,0,1 , and c  1,1,-1 .
We can find the triple scalar product by finding the determinant of the matrix
= 3(1)  2(3)  0  9 cubic units
.
6. a) Find parametric equations for the line that passes through the point (2,0,6) and is
parallel to the line with the parametric equations x  2  3t , y  t , z  1  3t .
b) At what point does the line intersect the xy-coordinate plane?
a)
x  2  3t , y  t , z  3  3t
b) The line intersects the xy-coordinate plane when z = 0. That is, when
z  6  3t  0 or t  2  x  8, y  2
The point of intersection is (8,2,0).
7. Find an equation of the plane containing the point (-4,2,0) that is parallel to the plane
5 y  2z  9 .
A normal vector to the plane is <0,5,-2>. We then have the equation 0(x+4) +5(y-2) -2(z0) = 0, or 5y -2z -10= 0.
8. What is the distance between the point (3,1,0) and the line represented by the
parametric equations x  2  t , y  1  5t , z  2t ?
Let (2,1,0) be a point chosen from the given line. Let w  3  2,1 1,0  0  1,0,0 ,
the vector with initial point (2,1,0) and terminal point (3,1,0). The vector v  1,5, 2
is parallel to the given line. We use the formula distance =
i
j
vw
v
.
k
v  w  1 5 2  0i  2 j  5k
1
vw
v

0
0
4  25
29
.

1  25  4
30
9. a) Draw the traces of the surface with equation z  3  3x 2  3 y 2 on the planes
z  3, z  6, z  15, x  0, and x  1 .
b) Sketch the surface.