MATHEMATICS 226, FALL 2014, PROBLEM SET 1
Solutions1
1. Specify the boundary and the interior of the sets S in 3-space whose points
(x, y, z) satisfy the given conditions. Is S open, closed, or neither?
(a) x2 + y 2 + z 2 ≥ 16: The boundary is the sphere x2 + y 2 + z 2 = 16, and the
interior is the set x2 + y 2 + z 2 > 16, consisting of all points on the outside of
the sphere. The boundary of S is contained in S, so the set is closed.
(b) z ≥ 0, x2 + (y − 2)2 + z 2 < 1: The interior is the open half-ball x2 + (y −
2)2 + z 2 < 1, z > 0 (the half of the ball of radius 2, centered at (0, 2, 0), that
lies above the xy-plane). The boundary consists of the half of the sphere
x2 + (y − 2)2 + z 2 = 1 that lies above the xy-plane, and the closed disk
x2 + (y − 2)2 ≤ 1, x = 0 in the xy-plane. Since part of the boundary (the
open disk) is contained in S, but the half-sphere is not, the set is neither
open nor closed.
2. Find all values of t for which the vector v = 4ti − tj + 6k is perpendicular
to the vector w = 2i + (4 − 2t)j + (1 − 2t)k.
We need 0 = v · w = 4t · 2 + (−t)(4 − 2t) + 6(1 − 2t) = 8t − 4t + 2t2 + 6 − 12t =
2t2 − 8t + 6, so that t2 − 4t + 3 = 0, t = 1 or t = 3.
3. Find vectors a, b, c in R2 such that the set of points whose position vector
r satisfies the inequalities
r · a ≤ 1, r · b ≤ 1, r · c ≤ 1
(1)
is the triangle with vertices (1, 2), (2, −2), (−3, 0).
The triangle is the common part of three closed half-planes:
• above the line through (−3, 0) and (2, −2),
• below the line through (1, 2) and (2, −2),
• below the line through (−3, 0) and (1, 2),
1
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1
all with the boundary line included. We need to find the inequalities defining
the three half-planes. Here we present the proof for the first one, the other
two are similar. We first find the equation of the line through (−3, 0) and
(2, −2). The line must be parallel to h5, −2i, so its equation is
x+3
y
−
, −2(x + 3) = 5y.
5
−2
The half-plane is then given by 5y ≥ −2(x + 3). Similarly, the other two
half-planes are y ≤ −4x + 6 and 2y ≤ x + 3.
We now need to rewrite these inequalities in a form that matches (1):
−2x − 5y ≤ 6, 4x + y ≤ 6, −x + 2y ≤ 3,
−2x − 5y
4x + y
−x + 2y
≤ 1,
≤ 1,
≤ 1.
6
6
3
So we can let
a=
−1 −5
,
3 6
, b=
2 1
,
3 6
, c=
−1 2
,
3 3
.
These are unique up to relabelling.
4. Find two unit vectors each of which makes equal angles with the vectors
u = 4i − j − k, v = i + j, and w = 2i + j + k.
Let r = hx, y, zi, then x makes equal angles with u, v, w if
u·r
v·r
w·r
=
=
|u||r|
|v||r|
|w||r|
Plugging in and multiplying everything by |r|, we get
x+y
2x + y + z
4x − y − z
√
√
= √ =
.
18
2
6
√
√
Multiplying the first equation by 18 = 3 2, we get 4x − y − z = 3x + 3y,
so that z = 4x − y − 3x − 3y = x − 4y. Then from the second equation we
have
√
√
√
√
3x + 3y = 2x + y + z = 2x + y + x − 4y = 3x − 3y, x + y = 3x − 3y,
2
√
√
√
√
so that x( 3 − 1) = y( 3 + 1). Let x =√
( 3 + 1)t √
and y = ( 3 − 1)t,
√ for
some t to be determined later; then z = ( 3 + 1 − 4 3 + 4)t = (5 − 3 3)t.
We want r to be a unit vector, so that
√
√
√
√
1 = |r|2 = x2 +y 2 +z 2 = (3+2 3+1+3−2 3+1+25−30 3+27)t2 = (60−30 3)t2 .
Therefore t = ± √
1
√ ,
60−30 3
*
r=±
so that
+
√
√
3+1
3−1
5−3 3
p
√ ,p
√ ,p
√
60 − 30 3
60 − 30 3
60 − 30 3
√
3