MATH 2004 Homework Solution
Han-Bom Moon
Homework 1 Model Solution
Section 12.1 ∼ 12.4.
12.1.11. Find an equation of the sphere with center (−3, 2, 5) and radius 4. What is the
intersection of this sphere with the yz-plane?
Equation of the sphere:
(x − (−3))2 + (y − 2)2 + (z − 5)2 = 42
(x + 3)2 + (y − 2)2 + (z − 5)2 = 16
or
Intersection with yz-plane (⇔ x = 0):
(0 + 3)2 + (y − 2)2 + (z − 5)2 = 16
9 + (y − 2)2 + (z − 5)2 = 16
(y − 2)2 + (z − 5)2 = 7
Intersection = the circle with center (0, 2, 5) and radius
√
7.
12.1.13. Find an equation of the sphere that passes through the point (4, 3, −1) and has
center (3, 8, 1).
Radius = distance from (3, 8, 1) to (4, 3, −1)
p
√
= (4 − 3)2 + (3 − 8)2 + (−1 − 1)2 = 30
Equation of the sphere:
(x − 3)2 + (y − 8)2 + (z − 1)2 = 30
12.1.32. Describe in words the region of R3 represented by x = z.
If we fix y-coordinate as k, i.e., if we look at a plane y = k (which is parallel to
the xz-plane), on this plane x = z defines a line. Because there is no restriction on
y-coordinates, the region is the union of all such lines x = z, y = k. Therefore it is
a plane.
12.1.40. Consider the points P such that the distance from P to A(−1, 5, 3) is twice the
distance from P to B(6, 2, −2). Show that the set of all such points is a sphere,
and find its center and radius.
For P (x, y, z), |AP | = 2|BP |.
p
p
(x + 1)2 + (y − 5)2 + (z − 3)2 = 2 (x − 6)2 + (y − 2)2 + (z + 2)2
1
MATH 2004 Homework Solution
Han-Bom Moon
(x + 1)2 + (y − 5)2 + (z − 3)2 = 4((x − 6)2 + (y − 2)2 + (z + 2)2 )
x2 + 2x + 1 + y 2 − 10y + 25 + z 2 − 6z + 9 = 4(x2 − 12x + 36 + y 2 − 4y + 4 + z 2 + 4z + 4)
x2 + 2x + y 2 − 10y + z 2 − 6z + 35 = 4x2 − 48x + 4y 2 − 16y + 4z 2 + 16z + 176
3x2 − 50x + 3y 2 − 6y + 3z 2 + 22z + 141 = 0
50
22
x + y 2 − 2y + z 2 + z + 47 = 0
3
3
2
25
625
11 2 121
2
x−
−
+ (y − 1) − 1 + z +
−
+ 47 = 0
3
9
3
9
11 2 332
25 2
+ (y − 1)2 + z +
=
x−
3
3
9
√
332
25
11
, radius:
Center:
, 1, −
3
3
3
x2 −
12.2.6. Copy the vectors in the figure and use them to draw the following vectors.
1. a + b
2. a − b
3.
1
2a
4. −3b
2
MATH 2004 Homework Solution
Han-Bom Moon
5. a + 2b
6. 2b − a
12.2.22. Find a + b, 2a + 3b, |a|, and |a − b| for
a = 2i − 4j + 4k,
b = 2j − k.
a + b = (2i − 4j + 4k) + (2j − k) = 2i − 2j + 3k
2a + 3b = 2(2i − 4j + 4k) + 3(2j − k) = 4i − 8j + 8k + 6j − 3k = 4i − 2j + 5k
3
MATH 2004 Homework Solution
Han-Bom Moon
|a| =
p
√
22 + (−4)2 + 42 = 36 = 6
a − b = (2i − 4j + 4k) − (2j − k) = 2i − 6j + 5k
p
√
|a − b| = 22 + (−6)2 + 52 = 65
12.2.24. Find a unit vector that has the same direction as the given vector h−4, 2, 4i.
a = h−4, 2, 4i
p
√
|a| = (−4)2 + 22 + 42 = 36 = 6
Unit vector =
1
a.
|a|
1
1
4 2 4
2 1 2
a = h−4, 2, 4i = h− , , i = h− , , i
|a|
6
6 6 6
3 3 3
12.2.44. Let C be the point on the line segment AB that is twice as far from B as it is
−→
−−→
−−→
from A. If a = OA, b = OB, and c = OC, show that c = 23 a + 31 b.
−−→
A direction vector for the line passing through A and B = AB = b − a
c = a + t(b − a) for some t, because c is on the line passing through A and B.
Because c is between a and b, 0 < t < 1 (Note that t = 0 ⇒ c = a and t = 1 ⇒
c = b.).
2|AC| = |BC| ⇒ 2|c − a| = |c − b|
⇒ 2|a + t(b − a) − a| = |a + t(b − a) − b|
⇒ 2|t(b − a)| = |(t − 1)(b − a)| ⇒ 2t|b − a| = |t − 1||b − a| = (1 − t)|b − a|
⇒ 2t = 1 − t ⇒ t =
1
3
1
2
1
c = a + (b − a) = a + b
3
3
3
12.3.22 Find, correct to the nearest degree, the three angles of the triangle with three
vertices
A(1, 0, −1), B(3, −2, 0), C(1, 3, 3).
At A:
−−→
AB = h3, −2, 0i − h1, 0, −1i = h2, −2, 1i
−→
AC = h1, 3, 3i − h1, 0, −1i = h0, 3, 4i
−−→ −→
AB · AC
2 · 0 + (−2) · 3 + 1 · 4
2
√
cos ∠A = −−→ −→ = p
=−
2
2
2
2
2
2
15
2 + (−2) + 1 0 + 3 + 4
|AB||AC|
∠A ; 98◦
At B:
−−→
−−→
BA = −AB = h−2, 2, −1i
−−→
BC = h1, 3, 3i − h3, −2, 0i = h−2, 5, 3i
4
MATH 2004 Homework Solution
Han-Bom Moon
−−→ −−→
BA · BC
(−2) · (−2) + 2 · 5 + (−1) · 3
11
p
cos ∠B = −−→ −−→ = p
= √
3 38
(−2)2 + 22 + (−1)2 (−2)2 + 52 + 32
|BA||BC|
∠B ; 54◦
At C:
−→
−→
CA = −AC = h0, −3, −4i
−−→
−−→
CB = −BC = h2, −5, −3i
−→ −−→
CA · CB
0 · (−2) + (−3) · (−5) + (−4) · (−3)
27
p
cos ∠C = −→ −−→ = p
= √
5 38
02 + (−3)2 + (−4)2 22 + (−5)2 + (−3)2
|CA||CB|
∠C ; 29◦
12.3.25 Use vectors to decide whether the triangle with vertices P (1, −3, −2), Q(2, 0, −4),
and R(6, −2, −5) is right-angled.
−−→
QP = h1, −3, −2i − h2, 0, −4i = h−1, −3, 2i
−−→
QR = h6, −2, −5i − h2, 0, −4i = h4, −2, −1i
−−→ −−→
QP · QR = (−1) · 4 + (−3) · (−2) + 2 · (−1) = 0 ⇒ ∠Q = 90◦
It is a right triangle.
12.3.27 Find a unit vector that is orthogonal to both i + j and i + k.
Note: This problem can be solved by using the cross product in next section. Here
I give an alternative solution without using the cross product.
Let v = ai + bj + ck be such a vector.
v · (i + j) = 0 ⇒ a + b = 0
v · (i + k) = 0 ⇒ a + c = 0
⇒ (a, b, c) = (t, −t, −t) for some real number t.
√
p
|v| = 1 ⇒ t2 + (−t)2 + (−t)2 = 3t2 = 1 ⇒ t = ± √13 .
1
1
1
1 1 1
v = h √ , − √ , − √ i or v = h− √ , √ , √ i
3
3
3
3 3 3
12.3.42 Find the scalar and vector projections of b = h5, −1, 4i onto a = h−2, 3, −6i.
proja b =
a·b
(−2) · 5 + 3 · (−1) + (−6) · 4
a=
h−2, 3, −6i
2
|a|
(−2)2 + 32 + (−6)2
=
−37
74 111 222
h−2, 3, −6i = h , −
,
i
49
49
49 49
b·a
−37
37
compa b =
= √ =−
|a|
7
49
5
MATH 2004 Homework Solution
Han-Bom Moon
12.3.45 Show that the vector ortha b = b = proja b is orthogonal to a. (It is called an
orthogonal projection of b.)
a·b
a·b
ortha b · a = b −
a ·a=b·a−
a·a
2
|a|
|a|2
=b·a−
a·b 2
|a| = b · a − a · b = b · a − b · a = 0
|a|2
ortha b is orthogonal (perpendicular) to a.
12.3.55 Find the angle between a diagonal of a cube and one of its edges.
Give a coordinate to the cube and suppose that the vertices are
(0, 0, 0), (a, 0, 0), (0, a, 0), (0, 0, a), (a, a, 0), (a, 0, a), (0, a, a), and (a, a, a)
for some a > 0.
The diagonal is the vector v = ha, a, ai − h0, 0, 0i = ha, a, ai.
Take an edge passing (0, 0, 0) and (a, 0, 0). Then it gives a vector w = ha, 0, 0i. Let
θ be the angle between them.
cos θ =
a·a+a·0+a·0
1
a2
a2
v·w
√
=√
=√
=√ √ =√
2
2
2
2
2
2
2
2
2
|v||w|
3a
3
a +a +a a +0 +0
3a a
θ ; 54.7◦
12.3.64 Show that if u + v and u − v are orthogonal, then the vectors u and v must
have the same length.
u + v and u − v are orthogonal ⇒ (u + v) · (u − v) = 0.
0 = (u + v) · (u − v) = u · (u − v) + v · (u − v) = u · u − u · v + v · u − v · v
= |u|2 − u · v + u · v − |v|2 = |u|2 − |v|2
⇒ |u|2 = |v|2 ⇒ |u| = |v|
12.4.3 Find the cross product a × b of a = i + 3j − 2k and b = −i + 5k and verify that
it is orthogonal to both a and b.
i j k 1 −2 1 3 3 −2 a × b = 1 3 −2 = i + k
− j
−1 5 −1 0 0 5 −1 0 5 = 15i − 3j + 3k
(a × b) · a = 15 · 1 + (−3) · 3 + 3 · (−2) = 0 ⇒ a × b and a are orthogonal.
(a × b) · b = 15 · (−1) + (−3) · 0 + 3 · 5 = 0 ⇒ a × b and b are orthogonal.
6
MATH 2004 Homework Solution
Han-Bom Moon
12.4.19 Find two unit vectors orthogonal to both h3, 2, 1i and h−1, 1, 0i.
= i
i j k h3, 2, 1i × h−1, 1, 0i = 3 2 1 −1 1 0 3 1 3 2 2 1 − j
+ k
= −i − j + 5k = h−1, −1, 5i
−1 0 −1 1 1 0 All vectors orthogonal to both h3, 2, 1i and h−1, 1, 0i are scalar multiple of h−1, −1, 5i.
So let v = th−1, −1, 5i = h−t, −t, 5ti.
√
p
√
|v| = 1 ⇒ (−t)2 + (−t)2 + (5t)2 = 27t2 = 27|t| = 1
1
1
|t| = √ ⇒ t = ± √
27
27
So h− √127 , − √127 , √527 i, h √127 , √127 , − √527 i are what we desire.
12.4.32 Find a nonzero vector orthogonal to the plane through the points P (−1, 3, 1),
Q(0, 5, 2), and R(4, 3, −1), and find the are of triangle P QR.
−−→
P Q = h0, 5, 2i − h−1, 3, 1i = h1, 2, 1i
−→
P R = h4, 3, −1i − h−1, 3, 1i = h5, 0, −2i
i j k 2 1 1 1 1 2
−−→ −→ PQ × PR = 1 2 1 = i
− j
+ k
0 −2 5 −2 5 0
5 0 −2 = −4i + 7j − 10k
−4i + 7j − 10k is a vector perpendicular to the plane.
1
1p
(−4)2 + 72 + (−10)2 =
Area of P QR = | − 4i + 7j − 10k| =
2
2
√
165
2
12.4.36 Find the volume of the parallelepiped with adjacent edges P Q, P R, and P S,
where P (3, 0, 1), Q(−1, 2, 5), R(5, 1, −1), and S(0, 4, 2).
−−→
P Q = h−1, 2, 5i − h3, 0, 1i = h−4, 2, 4i
−→
P R = h5, 1, −1i − h3, 0, 1i = h2, 1, −2i
−→
P S = h0, 4, 2i − h3, 0, 1i = h−3, 4, 1i
−4 2 4 2 −2
1 −2 Volume = 2 1 −2 = −4 − 2
−3 1
4 1 −3 4 1 = |−4 · 9 − 2 · (−4) + 4 · 11| = 16
7
2 1 + 4
−3 4 MATH 2004 Homework Solution
12.4.45
Han-Bom Moon
(a) Let P be a point not on the line L that passes through the points Q and R.
Show that the distance d from the point P to the line L is
d=
|a × b|
|a|
−−→
−−→
where a = QR and b = QP .
−−→
−−→
Let θ be the angle between QP and QR. Then
−−→ −−→
−−→ −−→
−−→
−−→ |QR × QP |
|QR × QR|
|a × b|
d = |QP | sin θ = |QP | −−→ −−→ =
=
−−→
|a|
|QR||QP |
|QR|
−−→ −−→
−−→ −−→
because |QR × QP | = |QR||QP | sin θ.
(b) Use the formula in part (a) to find the distance from the point P (1, 1, 1) to
the line through Q(0, 6, 8) and R(−1, 4, 7).
−−→
QP = h1, 1, 1i − h0, 6, 8i = h1, −5, −7i
−−→
QR = h−1, 4, 7i − h0, 6, 8i = h−1, −2, −1i
i
j
k −−→ −−→ QP × QR = 1 −5 −7 = −9i + 8j − 7k
−1 −2 −1 p
√
(−9)2 + 82 + (−7)2
| − 9i + 8j − 7k|
194
d=
=p
= √
2
2
2
|h−1, −2, −1i|
6
(−1) + (−2) + (−1)
Discovery Project 1. Let v1 , v2 , v3 , and v4 be vectors with lengths equal to the areas
of the faces opposite the vertices P, Q, R, and S, respectively, and directions
perpendicular to the respective faces and pointing outward. Show that
v1 + v2 + v3 + v4 = 0.
Note: I recommend to refer the picture on the textbook 840p.
−−→
−→
−→
Let a = P Q, b = P R, and c = P S. v2 1) is perpendicular to the face P RS,
2) points outward, 3) has length area(P RS). Thus
1
v2 = b × c.
2
Similarly,
1
1
v3 = c × a, v4 = a × b.
2
2
−−→
−→
Now QR = b − a and QS = c − a. Thus
1 −→ −−→ 1
1
v1 = QS × QR = (c − a) × (b − a) = (c × b − c × a − a × b + a × a)
2
2
2
1
= (−b × c − c × a + b × a).
2
Note that c × b = −b × c and a × a = 0. So
1
1
1
1
v1 + v2 + v3 + v4 = (−b × c − c × a − a × b) + b × c + c × a + a × b = 0.
2
2
2
2
8
MATH 2004 Homework Solution
Han-Bom Moon
2. The volume V of a tetrahedron is one-third the distance from a vertex to the
opposite face, times the are of that face.
(a) Find a formula for the volume of a tetrahedron in terms of the coordinates of its vertices P, Q, R, and S.
The height (or the distance from a vertex to the opposite face) of a tetrahedron is equal to that of the parallelepiped generated by P, Q, R, and
S. Also the area of the base of the tetrahedron is the half of the are of the
−−→
−→
−→
base of the parallelepiped. Therefore if a = P Q, b = P R, and c = P S,
1
area of tetrahedron = height · area of base
3
1
1
= height of parallelepiped · area of base of parallelepiped
3
2
1
1
= volume of parallelepiped = |a · (b × c)|.
6
6
(b) Find the volume of the tetrahedron whose vertices are P (1, 1, 1), Q(1, 2, 3),
R(1, 1, 2), and S(3, −1, 2).
−−→
a = P Q = h1, 2, 3i − h1, 1, 1i = h0, 1, 2i
−→
b = P R = h1, 1, 2i − h1, 1, 1i = h0, 0, 1i
−→
c = P S = h3, −1, 2i − h1, 1, 1i = h2, −2, 1i
i j k b × c = 0 0 1 = h2, 2, 0i
2 −2 1 1
1
volume = (0 · 2 + 1 · 2 + 2 · 0) =
6
3
3. Suppose the tetrahedron in the figure has a trirectangular vertex S. (This
means that the three angles at S are all right angles.) Let A, B, and C be the
areas of the three faces that meet at S, and let D be the area of the opposite
face P QR. Using the result of Problem 1, or otherwise, show that
D 2 = A2 + B 2 + C 2 .
(This is a three-dimensional version of the Pythagorean Theorem.)
From v1 + v2 + v3 = −v4 ,
v4 · v4 = (−v4 ) · (−v4 ) = (v1 + v2 + v3 ) · (v1 + v2 + v3 )
= v1 · v1 + v2 · v2 + v3 · v3 + 2v1 · v2 + 2v1 · v3 + 2v2 · v3
But because three faces meeting at S are perpendicular to each other, v1 ·v2 =
v1 · v3 = v2 · v3 = 0. Therefore
D2 = |v4 |2 = v4 · v4
= v1 · v1 + v2 · v2 + v3 · v3 + 2v1 · v2 + 2v1 · v3 + 2v2 · v3
= v1 · v1 + v2 · v2 + v3 · v3 = |v1 |2 + |v2 |2 + |v3 |2 = A2 + B 2 + C 2 .
9