Midterm 1 Solutions
1. In the following, ~u and ~v are vectors in 3-dimensional space.
(a) (8 pts.) What are the maximal and minimal possible lengths of ~u − ~v if |~u| = 2 and |~v | = 3?
Find choices for ~u and ~v that obtain the maximum and minimum values. (You do not need to show
your work on this part.)
The length of ~u − ~v is maximized or minimized if ~u and ~v point in the opposite or same direction,
respectively. We get 5 and 1 for the maximum and minimum, with ~u = 2~i and ~v = ±3~i as examples,
with the + for minimum and − for maximum.
(b) (8 pts.) What are the maximal and minimal possible lengths of ~u × ~v if |~u| = 2 and |~v | = 3?
Find choices for ~u and ~v that obtain the maximum and minimum values. (You do not need to show
your work on this part.)
We have ~u × ~v = |~u||~v | sin θ, so the maximum and minimum are 6 and 0, respectively. We get these
by setting ~u = 2~i and ~v = 3~j for the maximum or ~u = 2~i and ~v = 3~i for the minimum.
2. Multiple choice questions.
(a) Circle the word True or False to indicate your answer. (No explanation needed.)
(i) (2 pts.) If the lines L1 and L2 do not intersect, then there cannot be a plane that contains both
L1 and L2 .
False. L1 and L2 can be parallel.
(ii) (2 pts.) If two ellipses have the same pair of foci and have at least one point of intersection,
then the ellipses are the same.
True. The ellipses are determined by the foci and the sum of distances from a point to them; this
sum is the same if there is a point of intersection.
(iii) (2 pts.) For any pair of vectors ~u, ~v , we always have proj~u (~v ) × ~v = ~0.
False. The first vector in the cross product is proportional to ~u; choosing ~u not parallel or orthogonal
to ~v gives a counterexample.
(b) Classify the solution sets described by the given equations by circling the name of the surface,
or “Nothing” if there are no solutions. (No explanation needed.)
(i) (2 pts.) x2 + 2y 2 + 3z 2 = 4
Ellipsoid.
(ii) (2 pts.) 3y 2 = 4x + z 2
Hyperbolic Paraboloid.
(iii) (2 pts.) x2 − 2 = y 2 − 2y + z 2 + 2z
Cone. (Complete the square to get x2 = (y − 1)2 + (z + 1)2 .)
(iv) (2 pts.) (x + y)2 + (x − y)2 = z 2 + 1.
1-Sheeted Hyperboloid. (Expand the left-hand side and move z 2 to the left to get 2x2 +2y 2 −z 2 = 1.)
3. Let P, Q, R be the three points in 3-dimensional coordinate space given by
P = (2, 5, 7),
Q = (−2, 3, 5),
and
R = (3, 2, 3).
Let T be the triangle whose vertices are P , Q, and R.
(a) (6 pts.) What is the cosine of the angle of the triangle between the sides P Q and P R?
P~Q = h−4, −2, −2i and√P~R = h1, −3, −4i. P~Q · P~R√= −4(1) − 2(−3) − 2(−4) = −4 + 6 + 8 = 10.
√
√
|P~Q| = 16 + 4 + 4 = 24. |P~R| = 1 + 9 + 16 = 26. So we get √10
.
624
(b) (6 pts.) What is the area of the triangle T ?
~i
~k ~j
P~Q × P~R = −4 −2 −2 = 2~i − 18~j + 14~k.
1 −3 −4 1
We have
√
1
1√
Area(T ) = |P~Q × P~R| =
4 + 324 + 196 = 131.
2
2
(c) (6 pts.) Find an equation for the plane containing the triangle T .
2(x − 2) − 18(y − 5) + 14(z − 7) = 0.
4. (a) (9 pts.) Find an equation for the sphere whose diameter is the line segment from P = (3, 3, 1)
to Q = (5, 7, −3).
The
the radius is half the length of the diameter or
√ center is the midpoint or (4,25, −1), and
1
2
4 + 16 + 16 = 3. We get (x − 4) + (y − 5) + (z + 1)2 = 9.
2
(b) (9 pts.) Does the sphere x2 + y 2 + z 2 = 1 intersect the plane
√ x + y = 2? Explain.
|1(0)+1(0)+0(0)−2|
√
= 2. So no point can be on the sphere
The distance from x + y = 2 to the origin is
1+1
of radius 1 centered at the origin.
Alternative solution: Substitute x = 2 − y to get (2 − y)2 + y 2 + z 2 = 2y 2 − 4y + 4 + z 2 = 1. Complete
the square to get 2(y − 1)2 + z 2 = −1, which has no solutions.
5. Let E be the ellipse consisting of the points P such that the sum of the distances from P to the
foci F1 = (0, −2) and F2 = (0, 2) is equal to 8.
(a) (9 pts.) Write an equation for the ellipse E.
We have 2a = 8 or a = 4 and b2 = a2 − c2 = 16 − 4 = 12 in the usual notation, with the foci along
the y-axis. So we get
x2 y 2
+
= 1.
12 16
(b) (9 pts.) Let B be the parabola y = 8x2 . Let L be the line that is parallel to the directrix of
B and passes through the focus of B. The line L intersects B in two points. What is the distance
between these points?
1
1
· y. The focus is (0, 32
), and the intersection points are obtained by solving
We have x2 = 4 · 32
1
1
1
1
2
, 2). So the distance is 18 .
x = 4 · 32 · 32 or x = 16 . This gives intersection points (± 16
6. Consider the lines given by the parametric equations:
L1 : x = 0,
L2 : x = 1,
y = 0,
y = t,
z=s
z=0
(a) (2 pts.) Show that these lines are skew.
They cannot intersect due to the x coordinates. Vectors along them are ~k = h0, 0, 1i and ~j = h0, 1, 0i,
which are not parallel.
(b) (12 pts.) Find an equation describing the surface S consisting of all points whose distance to
L1 is the same as the distance to L2 .
The point (0, 0, 0) is on L1 and (1, 0, 0) is on L2 . So the distance from (x, y, z) to the first line is
p
p
|hx,y,zi×~k|
~i − x~j| = x2 + y 2 and |hx−1,y,zi×~j| = | − z~i + (x − 1)~k| = (x − 1)2 + z 2 . So we get
=
|y
1
1
the equation
p
p
x2 + y 2 = (x − 1)2 + z 2
or x2 + y 2 = (x − 1)2 + z 2 after squaring.
(c) (2 pts.) Classify the surface S by circling its name below.
Hyperbolic Paraboloid. We simplify the equation to
2x − 1 = −y 2 + z 2 .
2