MATH 2004 Homework Solution Han-Bom Moon
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
= 4
2 or ( x + 3)
2
+ ( y − 2)
2
+ ( z − 5)
2
= 16
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
R
3 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
( x + 1) 2 + ( y − 5) 2 + ( z − 3) 2 = 2 p
( x − 6) 2 + ( y − 2) 2 + ( z + 2) 2
1

MATH 2004 Homework Solution Han-Bom Moon
Center:
( x + 1)
2
+ ( y − 5)
2
+ ( z − 3)
2
= 4(( x − 6)
2
+ ( y − 2)
2
+ ( z + 2)
2
) x
2
+ 2 x + 1 + y
2
− 10 y + 25 + z
2
− 6 z + 9 = 4( x
2
− 12 x + 36 + y
2
− 4 y + 4 + z
2
+ 4 z + 4) x
2
+ 2 x + y
2
− 10 y + z
2
− 6 z + 35 = 4 x
2
− 48 x + 4 y
2
− 16 y + 4 z
2
+ 16 z + 176 x −
25
3
3 x
2
− 50 x + 3 y
2
− 6 y + 3 z
2
+ 22 z + 141 = 0
2 x
2
−
50 x + y
2
3
− 2 y + z
2
+
22 z + 47 = 0
3
−
625
9
+ ( y − 1)
2
− 1 + z +
11
3
2
−
121
9
+ 47 = 0
25
3
, 1 , −
11
3 x −
25
3
, radius:
2
+ ( y − 1)
2
+ z +
11
3
√
332
3
2
=
332
9
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
2 a
4.
− 3 b
2

MATH 2004 Homework Solution Han-Bom Moon
5.
a + 2 b
6.
2 b − a
12.2.22. Find a + b , 2 a + 3 b , | a | , and | a − b | for a = 2 i − 4 j + 4 k , b = 2 j − k .
a + b = (2 i − 4 j + 4 k ) + (2 j − k ) = 2 i − 2 j + 3 k
2 a + 3 b = 2(2 i − 4 j + 4 k ) + 3(2 j − k ) = 4 i − 8 j + 8 k + 6 j − 3 k = 4 i − 2 j + 5 k
3

MATH 2004 Homework Solution
| a | = p
2 2 + ( − 4) 2 + 4 2 =
√
36 = 6 a − b = (2 i − 4 j + 4 k ) − (2 j − k ) = 2 i − 6 j + 5 k
√
| a − b | = p
2 2 + ( − 6) 2 + 5 2 = 65
Han-Bom Moon
12.2.24. Find a unit vector that has the same direction as the given vector h− 4 , 2 , 4 i .
a = h− 4 , 2 , 4 i
| a | = p
( − 4) 2 + 2 2 + 4 2 =
√
36 = 6
Unit vector =
1
| a | a .
1
| a | a =
1
6 h− 4 , 2 , 4 i = h−
4
6
,
2
6
,
4
6 i = h−
2
3
,
1
3
,
2
3 i
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 =
2
3 a +
1 b .
A direction vector for the line passing through A and B =
3
− →
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 ) | ⇒ 2 t | b − a | = | t − 1 || b − a | = (1 − t ) | b − a |
⇒ 2 t = 1 − t ⇒ t =
1
3 c = a +
1
3
( b − a ) =
2
3 a +
1
3 b
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 : cos ∠ A =
− →
AB = h 3 , − 2 , 0 i − h 1 , 0 , − 1 i = h 2 , − 2 , 1 i
−→
AC = h 1 , 3 , 3 i − h 1 , 0 , − 1 i = h 0 , 3 , 4 i
|
− →
AB ·
− →
AB ||
−→
AC
−→
AC |
= p
2 2
2 · 0 + ( − 2) · 3 + 1 · 4
√
+ ( − 2) 2 + 1 2 0 2 + 3 2 + 4 2
2
= −
15
∠ A ; 98
◦
At B :
− →
BA = −
− →
AB = h− 2 , 2 , − 1 i
− →
BC = h 1 , 3 , 3 i − h 3 , − 2 , 0 i = h− 2 , 5 , 3 i
4

MATH 2004 Homework Solution Han-Bom Moon cos ∠ B =
|
− →
BA ·
− →
BA ||
− →
BC
− →
BC |
=
( − 2) · ( − 2) + 2 · 5 + ( − 1) · 3 p
( − 2) 2 + 2 2 + ( − 1) 2 p
( − 2) 2 + 5 2 + 3 2
=
3
√
38
∠ B ; 54
◦
At C :
−→
CA = −
−→
AC = h 0 , − 3 , − 4 i
− →
CB = −
− →
BC = h 2 , − 5 , − 3 i cos ∠ C =
|
−→
CA
−→
·
CA ||
− →
CB
− →
CB |
=
0 · ( − 2) + ( − 3) · ( − 5) + ( − 4) · ( − 3) p
0 2 + ( − 3) 2 + ( − 4) 2 p
2 2 + ( − 5) 2 + ( − 3) 2
∠ C ; 29
◦
=
5
√
38
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 = h 1 , − 3 , − 2 i − h 2 , 0 , − 4 i = h− 1 , − 3 , 2 i
− →
QR = h 6 , − 2 , − 5 i − h 2 , 0 , − 4 i = h 4 , − 2 , − 1 i
− →
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 = a i + b j + c k 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 .
√
| v | = 1 ⇒ p t 2 + ( − t ) 2 + ( − t ) 2 = 3 t 2 = 1 ⇒ t = ± √
3
.
v = h √
3
, − √
3
, − √
3 i or v = h− √
3
, √
3
, √
3 i
12.3.42 Find the scalar and vector projections of b = h 5 , − 1 , 4 i onto a = h− 2 , 3 , − 6 i .
proj a b = a · b a =
| a | 2
( − 2) · 5 + 3 · ( − 1) + ( − 6) · 4 h− 2 , 3 , − 6 i
( − 2) 2 + 3 2 + ( − 6) 2
=
− 37
49 h− 2 , 3 , − 6 i = h
74
49
, −
111
,
49
222 i
49 comp a b = b · a
| a |
=
−
√
37
49
= −
37
7
5

MATH 2004 Homework Solution Han-Bom Moon
12.3.45 Show that the vector orth a b = b = proj a b is orthogonal to a . (It is called an orthogonal projection of b .) orth a b · a = b − a · b a · a = b · a −
| a | 2 a · b a · a
| a | 2
= b · a − a · b
| a |
2
| a | 2
= b · a − a · b = b · a − b · a = 0 orth a 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 = h a, a, a i − h 0 , 0 , 0 i = h a, a, a i .
Take an edge passing (0 , 0 , 0) and ( a, 0 , 0) . Then it gives a vector w = h a, 0 , 0 i . Let
θ be the angle between them.
cos θ = v · w
| v || w |
= √ a · a + a · 0 + a · 0
√ a 2 + a 2 + a 2 a 2 + 0 2 + 0 2
= a
2
√
3 a 2
√ a 2
= a
2
√
3 a 2
=
1
√
3
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 + 3 j − 2 k and b = − i + 5 k and verify that it is orthogonal to both a and b .
a × b = i j k
1 3 − 2
− 1 0 5
= i
3 − 2
0 5
− j
1 − 2
− 1 5
+ k
1 3
− 1 0
= 15 i − 3 j + 3 k
( 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 h 3 , 2 , 1 i and h− 1 , 1 , 0 i .
h 3 , 2 , 1 i × h− 1 , 1 , 0 i = i j k
3 2 1
− 1 1 0
= i
2 1
1 0
− j
3 1
− 1 0
+ k
3 2
− 1 1
= − i − j + 5 k = h− 1 , − 1 , 5 i
All vectors orthogonal to both h 3 , 2 , 1 i and h− 1 , 1 , 0 i are scalar multiple of h− 1 , − 1 , 5 i .
So let v = t h− 1 , − 1 , 5 i = h− t, − t, 5 t i .
| v | = 1 ⇒ p
( − t ) 2 + ( − t ) 2 + (5 t ) 2 =
√
27 t 2 =
√
27 | t | = 1
| t | = √
1
27
⇒ t = ± √
1
27
So h− √
1
27
, − √
1
27
, √
5
27 i , h √
1
27
, √
1
27
, − √
5
27 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 = h 0 , 5 , 2 i − h− 1 , 3 , 1 i = h 1 , 2 , 1 i
−→
P R = h 4 , 3 , − 1 i − h− 1 , 3 , 1 i = h 5 , 0 , − 2 i
− →
P Q ×
−→
P R = i j k
1 2 1
5 0 − 2
= i
2
0
1
− 2
− j
= − 4 i + 7 j − 10 k
− 4 i + 7 j − 10 k is a vector perpendicular to the plane.
1 1
5 − 2
+ k
1 2
5 0
Area of P QR =
1
2
| − 4 i + 7 j − 10 k | =
1
2 p
( − 4) 2 + 7 2 + ( − 10) 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) .
Volume =
− →
P Q = h− 1 , 2 , 5 i − h 3 , 0 , 1 i = h− 4 , 2 , 4 i
−→
P R = h 5 , 1 , − 1 i − h 3 , 0 , 1 i = h 2 , 1 , − 2 i
−→
P S = h 0 , 4 , 2 i − h 3 , 0 , 1 i = h− 3 , 4 , 1 i
− 4 2 4
2 1 − 2
− 3 4 1
= − 4
1
4
−
1
2
− 2
2
− 3
= |− 4 · 9 − 2 · ( − 4) + 4 · 11 | = 16
−
1
2
+ 4
2 1
− 3 4
7

MATH 2004 Homework Solution Han-Bom Moon
12.4.45
(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 d = |
− →
QP | sin θ = |
− →
QP |
|
− →
QR ×
|
− →
QR ||
− →
QP |
− →
QP |
=
|
− →
QR ×
−
− →
QR |
|
−
QR |
=
| a × b |
| a | 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 = h 1 , 1 , 1 i − h 0 , 6 , 8 i = h 1 , − 5 , − 7 i
− →
QR = h− 1 , 4 , 7 i − h 0 , 6 , 8 i = h− 1 , − 2 , − 1 i
− →
QP ×
− →
QR = i j k
1 − 5 − 7 = − 9 i + 8 j − 7 k
− 1 − 2 − 1 d =
| − 9 i + 8 j − 7 k |
|h− 1 , − 2 , − 1 i|
= p
( − 9) 2 + 8 2 + ( − 7) 2 p
( − 1) 2 + ( − 2) 2 + ( − 1) 2
=
√
194
√
6
Discovery Project 1. Let v
1
, v
2
, v
3
, and v
4 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 v
1
+ v
2
+ v
3
+ v
4
= 0 .
Note: I recommend to refer the picture on the textbook 840p.
Let a =
− →
P Q , b =
−→
P R , and c =
−→
P S .
v
2
1) is perpendicular to the face P RS ,
2) points outward, 3) has length area ( P RS ) . Thus v
2
=
1
2 b × c .
Similarly,
Now
− →
QR = b − a and
1 v
3
= c × a , v
2
−→
QS = c − a . Thus
4
=
1
2 a × b .
v
1
=
1
2
−→
QS ×
− →
QR =
1
2
( c − a ) × ( b − a ) =
1
2
( c × b − c × a − a × b + a × a )
=
1
2
( − b × c − c × a + b × a ) .
Note that c × b = − b × c and a × a = 0 . So v
1
+ v
2
+ v
3
+ v
4
=
1
2
( − b × c − c × a − a × b ) +
1
2 b × c +
1
2 c × a +
1
2 a × b = 0 .
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 , area of tetrahedron =
1
3 height · area of base
=
1
3 height of parallelepiped ·
=
1
2 area of base of parallelepiped
1
6 volume of parallelepiped =
1
6
| a · ( b × c ) | .
(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 = h 1 , 2 , 3 i − h 1 , 1 , 1 i = h 0 , 1 , 2 i c b =
=
−→
P R = h 1 , 1 , 2 i − h 1 , 1 , 1 i = h 0 , 0 , 1 i
−→
P S = h 3 , − 1 , 2 i − h 1 , 1 , 1 i = h 2 , − 2 , 1 i b × c = i j k
0 0 1
2 − 2 1
= h 2 , 2 , 0 i volume =
1
6
(0 · 2 + 1 · 2 + 2 · 0) =
1
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
= A
2
+ B
2
+ C
2
.
(This is a three-dimensional version of the Pythagorean Theorem.)
From v
1
+ v
2
+ v
3
= − v
4
, v
4
· v
4
= ( − v
4
) · ( − v
4
) = ( v
1
+ v
2
+ v
3
) · ( v
1
+ v
2
+ v
3
)
= v
1
· v
1
+ v
2
· v
2
+ v
3
· v
3
+ 2 v
1
· v
2
+ 2 v
1
· v
3
+ 2 v
2
· v
3
But because three faces meeting at S are perpendicular to each other, v
1
· v
2 v
1
· v
3
= v
2
· v
3
= 0 . Therefore
=
D
2
= | v
4
|
2
= v
4
· v
4
= v
1
· v
1
+ v
2
· v
2
+ v
3
· v
3
+ 2 v
1
· v
2
+ 2 v
1
· v
3
+ 2 v
2
· v
3
= v
1
· v
1
+ v
2
· v
2
+ v
3
· v
3
= | v
1
| 2
+ | v
2
| 2
+ | v
3
| 2
= A
2
+ B
2
+ C
2
.
9