CHAPTER 10
Vectors and the Geometry of Space
1. Vectors in the Plane
We denote the directed line segment from the point P (initial point) to the
!
!
point Q (terminal point) as P Q. The length of P Q is its magnitude, denoted
!
kP Qk.
Definition. A vector is the equivalence class of all directed line segments
(or arrows) having the same magnitude and the same direction (an arrow and
its parallel translates). Any arrow in the class may be used to name the vector.
Example.
a=b=c
since a, b, c name the same vector.
1
2
10. VECTORS AND THE GEOMETRY OF SPACE
Definition. A position vector is the arrow in each class with tail (initial
point) at the origin. If the tip (terminal point) is at the point A(a1, a2), we
write
!
a = OA = ha1, a2i.
In ha1, a2i, a1 and a2 are called the components of a— a1 is the first component,
a2 the second component.
From the Pythagorean Theorem,
q
kak = kha1, a2ik = a21 + a22.
Note. The magnitude of a is often called the norm or 2-norm of a, and is
sometimes denoted kak2.
!
!
For a = OA = ha1, a2i and b = OB = hb1, b2i,
a = b () A = B () a1 = b1 and a2 = b2.
() means “if and only if” (=) is “only if” and (= is “if”).
1. VECTORS IN THE PLANE
3
Basic Operations
Addition:
ha1, a2i + hb1, b2i = ha1 + b1, a2 + b2i.
This is the tip-to-tail or parallelogram law. The sum a+b is called the resultant.
Example.
h4, 3i + h 1, 4i = h3, 7i
4
10. VECTORS AND THE GEOMETRY OF SPACE
Subtraction:
ha1, a2i
hb1, b2i = ha1
b1, a2
For direction of di↵erence vector, make sure b + (a
b2i.
b) = a.
Zero vector:
0 = h0, 0i
k0k = 0 and the zero vector has all directions.
Scalar multiplication:
ca = cha1, a2i = hca1, ca2i
Thus ca is a vector and
q
p
kcak = khca1, ca2ik = (ca1)2 + (ca2)2 = c2a21 + c2a22 =
q
p q
c2(a21 + a22) = c2 a21 + a22 = |c| · kak.
1. VECTORS IN THE PLANE
5
Vectors are parallel if one is a scalar multiple of the other.
ca k a, in the same direction for c
0, the opposite direction for c  0.
Additive inverse:
a=
ha1, a2i = ( 1)ha1, a2i = h a1, a2i
k ak = k( 1)ha1, a2ik = | 1| · kha1, a2ik = kak
Definition. V2 = hx, yi|x, y 2 R , the set of all two-dimensional position vectors.
Definition. A vector space is a set of vectors with the same dimension and
a set of scalars (real or complex numbers) with addition and scalar multiplication such that for any vectors u, v, and w, and any scalars ↵ and ,
(1) v + w = w + v
(2) (u + v) + w = u + (v + w)
(3) ↵( v) = (↵ )v
(4) (↵ + )v = ↵v + v
(5) ↵(v + w) = ↵v + ↵w
(6) 1v = v
(7) 0v = 0
(8) v + 0 = v
(9) For each v there is
v such that v + ( v) = 0
6
10. VECTORS AND THE GEOMETRY OF SPACE
Theorem (1.1). V2 is a vector space.
Proof. (of (2))
(u + v) + w = (hu1, u2i + hv1, v2i) + hw1, w2i =
hu1 + v1, u2 + v2i + hw1, w2i = h(u1 + v1) + w1, (u2 + v2) + w2i =
hu1 + (v1 + w1), u2 + (v2 + w2)i = hu1, u2i + hv1 + w1, v2 + w2i =
hu1, u2i + (hv1, v2i + hw1, w2i) = u + (v + w)
!
Note. For any two distinct points A(x1, y1) and B(x2, y2), AB corresponds
to the position vector hx2 x1, y2 y1i.
Any vector can be written in terms of the standard basis vectors
For any a 2 V2,
i = h1, 0i and j = h0, 1i.
a = ha1, a2i = ha1, 0i + h0, a2i = a1h1, 0i + a2h0, 1i = a1i + a2j.
We call a1 and a2 the horizontal and vertical components of a
Note. kik = kjk = 1.
A unit vector is a vector a where kak = 1.
Theorem (1.2). For any nonzero vector a, a unit vector having the
same direction as a is
1
u=
a.
kak
Proof.
a 6= 0 =) kak > 0 =) u is a positive scalar multiple of a =)
u has the same direction as a.
1
1
1
kuk =
a =
· kak =
· kak = 1.
kak
kak
kak
1. VECTORS IN THE PLANE
Example. Find a unit vector in the same directioin as a = h6, 8i.
p
p
kak = kh6, 8ik = 36 + 64 = 100 = 10.
Therefore
u=
1
3 4
h6, 8i = h , i.
10
5 5
Resolving a vector a into horizontal and vertical components:
a = kak cos ✓i + kak sin ✓j
= kak(cos ✓i + sin ✓j)
= kakhcos ✓, sin ✓i
Note. khcos ✓, sin ✓ik = 1.
7
8
10. VECTORS AND THE GEOMETRY OF SPACE
Example. A plane is flying at 500 km/hr. The wind is blowing at 60 km/hr
toward the SE. In what direction should the plane head to go due east? What
is the plane’s speed relative to the ground?
Solution
Let p = the motion vector for the plane and w = the wind vector.
Find
and kp + wk.
w = kwk cos( 45 )i + kwk sin( 45 )j
p
p
p
p
2
2
= 60 i 60 j = 30 2 i 30 2 j = 42.43i
2
2
Since the j-component of p + w is 0,
p
p = xi + 30 2j
42.43j.
where x is the i-component of p. Since kpk = 500,
q
p
2
500 = x + (30 2)2 =) 5002 = x2 + 1800 =)
x2 = 250, 000
Thus
1800 = 248, 200 =) x ⇡ 498.20.
42.43
= .085 =) = arctan .085 = 4.86 ,
498.20
and since p + w = 540.63i + 0j, kp + wk = 540.63 km/hr.
p = 498.20i + 42.43j =) tan
=
2. VECTORS IN SPACE
9
Maple. See vectorsinplane(10.1).mw or vectorsinplane(10.1).pdf
2. Vectors in Space
We use a right-handed coordinate system.
The coordinate system has 8 octants. The first octant has x > 0, y > 0, z > 0.
Distance Formula
p
d (x1, y1, z1), (x2, y2, z2)} = (x2
x1)2 + (y2
y1)2 + (z2
z1)2 .
Vectors in R3
These are similar to vectors in R2, with one more dimension.
is a vector space with
V3 = hx, y, zi|x, y, z 2 R
a + b = ha1, a2, a3i + hb1, b2, b3i = ha1 + b1, a2 + b2, a3 + b3i,
ca = cha1, a2, a3i = hca1, ca2, ca3i, and
a b = ha1, a2, a3i hb1, b2, b3i = ha1 b1, a2 b2, a3 b3i,
10
10. VECTORS AND THE GEOMETRY OF SPACE
Also,
so
q
kak = kha1, a2, a3ik = a21 + a22 + a23,
kcak = |c|kak, 0 = h0, 0, 0i, and
a = h a1, a2, a3i
Note. The vector with initial point P (a1, a2, a3) and terminal point
Q(b1, b2, b3) corresponds to the position vector
!
P Q = hb1 a1, b2 a2, b3
a3i.
The standard basis consists of
i = h1, 0, 0, i,
j = h0, 1, 0i,
k = h0, 0, 1i with
kik = kjk = kkk = 1.
For any vector a = ha1, a2, a3i,
a = a1i + a2j + a3k.
Also, for any vector a = ha1, a2, a3i 6= 0, a unit vector u in the same direction
as a is
1
u=
a.
kak
The sphere of radius r centered at (a, b, c) consists of all points (x, y, z) such
that
p
d (x, y, z), (a, b, c)} = (x a)2 + (y b)2 + (z c)2
or
(x a)2 + (y b)2 + (z c)2 = r2,
the standard form of the equation of a sphere.
2. VECTORS IN SPACE
11
Problem (Page 711 #28). Identify the geometric shape given by
7
x2 + x + y 2 y + z 2 = .
2
Solution
Rewrite as
(x2 + x
) + (y 2
y
7
) + z2 = .
2
Complete squares:
⇣
1⌘ ⇣ 2
1⌘
7 1 1
2
x +x+
+ y
y+
+ z 2 = + + =)
4
4
2 4 4
⇣
⌘
⇣
⌘
1 2
1 2
x+
+ y
+ z2 = 4
2
2
⇣
⌘
1 1
Thus we have the sphere of radius 2 centered at
2, 2, 0 .
Problem (Page 711 #20). Find a vector of magnitude 3 in the same direction as v = 3i + 3j k.
Solution
Thus
p
p
2
2
2
kvk = 3 + 3 + ( 1) = 19
1
u=p v
19
is a unit vector in the same direction of v, so
3
9
9
3u = p v = p i + p j
19
19
19
has magnitude 3 in the same direction as v.
3
p k
19
Problem (Page 711 #40). An equation for the x-axis is
y 2 + z 2 = 0 or y = z = 0.
12
10. VECTORS AND THE GEOMETRY OF SPACE
Problem (Page 712 #62). Three ropes are attached to a 300-pound crate.
Rope A exerts a force of h10, 130, 200i pounds on the crate, and rope B exerts
a force of h 20, 180, 160i pounds on the crate. What force must rope C exert
on the crate to move it up and to the right with a constant force of h0, 30, 20i
pounds.
Solution
Force of rope A is a = h10, 130, 200i, rope B is b = h 20, 180, 160i, rope C
is c = hc1, c2, c3i, and the weight is w = h0, 0, 300i.
We want the net force
a + b + c + w = h0, 30, 20i =)
h 10 + c1, 50 + c2, 60 + c3i = h0, 30, 20i =)
c1 = 10, c2 = 20, c3 = 40 =)
c = h10, 20, 40i =) kck ⇡ 45.8.
Thus rope C exerts a force of about 45.8 pounds in the direction h1, 2
4i.
Maple. See vectorsinspace(10.2).mw or vectorsinspace(10.2).pdf
3. The Dot Product
This is also often referred to as the scalar product.
In V2:
a · b = ha1, a2i · hb1, b2i = a1b1 + a2b2
In V3:
a · b = ha1, a2, a3i · hb1, b2, b3i = a1b1 + a2b2 + a3b3
Example. If a = h 3, 2, 5i and b = h3, 5, 6i,
a · b = h 3, 2, 5i · h3, 5, 6i = ( 3)(3) + 2(5) + 5( 6) =
9 + 10
30 =
29
3. THE DOT PRODUCT
13
Note.
(1) The dot product of two vectors is a number.
(2) A vector ha, bi in V2 can be thought of a vector ha, b, 0i in V3, so results we
prove for V3 also hold for V2.
Theorem (3.1). For vectors a, b, c and any scalar d,
(1) a · b = b · a
(2) a · (b + c) = a · b + a · c
(3) (da) · b = d(a · b) = a · (db)
(4) 0 · a = 0
(5) a · a = kak2
Proof.
(2) a · (b + c) = ha1, a2, a3i · hb1, b2, b3i + hc1, c2, c3i
= ha1, a2, a3i · hb1 + c1, b2 + c2, b3 + c3i
= a1(b1 + c1) + a2(b2 + c2) + a3(b3 + c3)
= a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3
= a1b1 + a2b2 + a3b3 + a1c1 + a2c2 + a3c3
= ha1, a2, a3i · hb1, b2, b3i + ha1, a2, a3i · hc1, c2, c3i
= a · b + a · c.
(5) a · a = ha1, a2, a3i · ha1, a2, a3i = a21 + a22 + a23 = kha1, a2, a3ik2 = kak2.
Note. a · b = 0 does not imply a = 0 or b = 0, so this di↵ers from our
usual multiplication.
14
10. VECTORS AND THE GEOMETRY OF SPACE
For a 6= 0 and b 6= 0 in V3, the angle ✓ (0  ✓  ⇡) between a and b is the
smaller of the two angles in the plane formed by giving both vectors the same
initial point.
If a and b have the same direction, ✓ = 0.
If a and b have the opposite direction, ✓ = ⇡.
a and b are orthogonal (or perpendicular) if ✓ =
⇡
.
2
We say 0 is orthogonal to every vector.
Law of Cosines
ka
bk2 = kak2 + kbk2
2kakkbk cos ✓
Theorem (3.2). Let ✓ be the angle between nonzero vectors a and b.
Then
a · b = kakkbk cos ✓.
Corollary. For nonzero vectors a and b,
a·b
cos ✓ =
.
kakkbk
Example. Find the angle between a = h2, 3, 6i and b = h 2, 1, 4i.
2( 2) + 3(1) + 6(4)
23
p
cos ✓ = p
= p =)
4 + 9 + 36 4 + 1 + 16 7 21
⇣ 23 ⌘
✓ = arccos p
⇡ .7713R ⇡ 44.19
7 21
3. THE DOT PRODUCT
15
Corollary. a and b are orthogonal (a ? b) () a · b = 0.
Theorem (3.3 — Cauchy-Schwarz Inequality). For any vectors a and b,
|a · b|  kakkbk
or
Proof.
q
q
2
2
2
|a1b1 + a2b2 + a3b3|  a1 + a2 + a3 b21 + b22 + b23.
|a · b| = kakkbk cos ✓ = kakkbk|cos ✓|  kakkbk.
Theorem (3.4 — Triangle Inequality). For any vectors a and b,
ka + bk  kak + kbk.
Proof.
ka + bk2 = (a + b) · (a + b) = a · a + a · b + b · a + b · b
= kak2 + 2a · b + kbk2
 kak2 + 2kakkbk + kbk2
= (kak + kbk)2
Taking square roots then gives the result.
16
10. VECTORS AND THE GEOMETRY OF SPACE
Component of a along b
⇣
⇡⌘
compb a
0<✓<
2
Note. cos(⇡
✓) =
compb a = kak cos ✓ =
-compb a
cos ✓
⇣⇡
2
<✓<⇡
⌘
kakkbk
1
1
cos ✓ =
kakkbk cos ✓ =
a · b =)
kbk
kbk
kbk
compb a =
a·b
kbk
is a number, a directed length, that depends on the length of a since
unit vector.
b
is a
kbk
3. THE DOT PRODUCT
17
Projection of a onto b
This is a force vector parallel to b having the same component along b as a.
⇣
⇡⌘
projb a
0<✓<
2
projb a = compb a
projb a =
is a vector.
projb a
b
=
kbk
!
⇣⇡
!
2
<✓<⇡
a·b b
=)
kbk kbk
a·b b
kbk kbk
⌘
18
10. VECTORS AND THE GEOMETRY OF SPACE
Problem (Page 721 #63b). Find the component of the weight vector along
the road vector for a 2500-pound car on a 15 bank.
Let b be the direction of the banked road with kbk = 1.
The weight
b = hcos 15 , sin 15 i.
w = h0, 2500i.
The component of w in the direction of b is
w·b
compb w =
= h0, 2500i · hcos 15 , sin 15 i = 2500 sin 15 =
kbk
Thus the force is to the inside of the track, allowing the car to turn.
647.
Problem (Page 720 #56). The orthogonal projection of a vector a along
a vector b is defined as
orthb a = a
projb a.
Sketch a picture showing a, b, projb a, and orthb a, and explain what is
orthogonal about orthb a.
Maple. See dotprod(10.3).mw or dotprod(10.3).pdf
4. THE CROSS PRODUCT
19
4. The Cross Product
This is a product that exists only in 3 dimensions. We fist need an introduction
to determinants for computational purposes.

a a
Definition. For a 2 ⇥ 2 (“2 by 2”) matrix M = 1 2 , the determinant
b1 b2
of M is
a a
|M | = 1 2 = a1b2 a2b1.
b1 b2
Definition. The determinant of a 3 ⇥ 3 matrix M is
a1 a2 a3
b b
b b
bb
|M | = b1 b2 b3 = a1 2 3
a2 1 3 + a3 1 2
c2 c3
c1 c3
c1c2
c1 c2 c3
= a1b2c3 a1b3c2 + a2b3c1 a2b1c3 + a3b1c2
Example.
2 3 4
5 1
1 =6
6 2 3
4 + 18 + 45
40
24 = 1
Area of the Parallelogram formed by a and b
Area = (base) · (height) = kakkbk sin ✓
a3b2c1.
20
10. VECTORS AND THE GEOMETRY OF SPACE
Definition. The cross product a ⇥ b is the vector orthogonal to a and
b by the right-hand rule, whose magnitude is the area of the parallelogram
determined by the vectors a and b, i.e.,
ka ⇥ bk = kakkbk sin ✓.
Theorem (4.4). For nonzero vectors a and b in V3, if ✓ is the angle
between a and b (0 < ✓ < ⇡), then
ka ⇥ bk = kakkbk sin ✓.
Theorem (4.2). For any vectors a and b in V3,
a ⇥ b ? a and a ⇥ b ? b.
Definition (Algebraic Definition of Cross Product).
i j k
a a
a ⇥ b = a1 a2 a3 = 2 3 i
b2 b3
b1 b2 b3
a1 a3
a a
j + 1 2 k.
b1 b3
b1 b2
4. THE CROSS PRODUCT
21
Example.
i j k
h5, 1, 2i ⇥ h 1, 3, 4i = 5 1 2 =
1 3 4
20)j + (15 + 1)k = h10, 18, 16i
p
p
p
kh10, 18, 16ik = 100 + 324 + 256 = 680 = 2 70 ⇡ 26.08.
(4 + 6)i + (2
Theorem (4.1). For any vector a in V3,
a ⇥ a = 0 and a ⇥ 0 = 0.
Example.
i j k
i ⇥ j = 1 0 0 = k,
0 1 0
but
i j k
j⇥i= 0 1 0 =
1 0 0
so the cross product is not commutative.
i⇥j=k
j⇥k=i
k⇥i=j
k,
j⇥i= k
k⇥j= i
i⇥k= j
22
10. VECTORS AND THE GEOMETRY OF SPACE
Theorem (4.3). For any a, b, c 2 V3, and any scalar d,
(1) a ⇥ b =
(b ⇥ a) (anti-commutativity)
(2) (da) ⇥ b = d(a ⇥ b) = a ⇥ (db)
(3) a ⇥ (b + c) = a ⇥ b + a ⇥ c
(4) (a + b) ⇥ c = a ⇥ c + b ⇥ c
(5) a · (b ⇥ c) = (a ⇥ b) · c (scalar triple product)
(6) a ⇥ (b ⇥ c) = (a · c)b
(a · b)c (vector triple product)
Corollary (to Theorem 4.4). Two nonzero vectors a and b in V3 are
parallel if and only if a ⇥ b = 0.
Proof.
a k b () ✓(\ between a and b) = 0 or ✓ = ⇡ () sin ✓ = 0 ()
ka ⇥ bk = kakkbk sin ✓ = 0 () a ⇥ b = 0.
Volume of a Parallelepiped having noncoplanar vectors a, b, c as adjacent edges.
Volume = (area of base) · altitude = ka ⇥ bk · |compa⇥b c| =
|c · (a ⇥ b)|
ka ⇥ bk ·
= |c · (a ⇥ b)|
ka ⇥ bk
the absolute value of a scalar triple product.
4. THE CROSS PRODUCT
23
Corollary (to Theorem 4.3). The scalar triple product of a, b, c is
a · (b ⇥ c) = b · (c ⇥ a) = c · (a ⇥ b).
Note.
i j k
a · (b ⇥ c) = a · b1 b2 b3
c1 c2 c3
b2 b3
i
c2 c3
= ha1, a2, a3i ·
= a1
b2 b3
c2 c3
a2
!
b1 b3
b b
j+ 1 2 k
c1 c3
c1 c2
b1 b3
b b
+ a3 1 2
c1 c3
c1 c2
a1 a2 a3
= b1 b2 b3
c1 c2 c3
Example. Find the volume of the parallelepiped with 3 adjacent edges
formed by h 2, 1, 2i, h1, 1, 2i, and h2, 1, 3i.
2 1 2
1 2 | = |6 + 4
V =| 1
2 1 3
4 + 3 + 2 + 4| = 15
24
10. VECTORS AND THE GEOMETRY OF SPACE
Example. Find the distance from a point Q to the line through P and R.
!
!
!
!
kP Q ⇥ P Rk = kP QkkP Rk sin ✓
=
|{z}
!
d=kP Qk sin ✓
!
kP Rk · d =)
!
!
kP Q ⇥ P Rk
d=
!
kP Rk
Problem (Page 732 #20). Find the distance from the point Q = (1, 3, 1)
to the line through P = (1, 3, 2) and R = (1, 0, 2).
!
!
P Q = h0, 0, 3i and P R = h0, 3, 0i =)
kh0, 0, 3i ⇥ h0, 3, 0ik 1 i j k
1
1
d=
= k 0 0 3 k = k9ik = (9) = 3.
kh0, 3, 0ik
3 0 3 0
3
3
4. THE CROSS PRODUCT
25
Example (Torque applied by a wrench).
F = force applied at end of handle.
⌧ = torque applied to the bolt.
r = position vector for handle.
Then
⌧ = r ⇥ F =) k⌧ k = kr ⇥ Fk = krkkFk sin ✓
Problem (Page 732 #28). If you apply a force of magnitude 40 pounds at
⇡
the end of an eighteen-inch-long wrench at an angle of to the wrench, find
3
the magnitude of the torqe applied to the wrench.
⇡
kFk = 40, krk = 18, ✓ = =)
3
p
p
⇡
3
k⌧ k = 18 · 40 sin = 720
= 360 3 ⇡ 623.54.
3
2
Maple. See crossprod(10.4).mw or crossprod(10.4).pdf.
26
10. VECTORS AND THE GEOMETRY OF SPACE
5. Lines and Planes in Space
Lines in 3-space:
(1) Vector equation specified by two points or a point and a direction.
!
P = OP1 + ta = hx1, y1, z1i + tha1, a2, a3i = hx1 + ta1, y1 + ta2, z1 + ta3i
or
!
!
P1P k a =) P1P = ta =) hx x1, y y1, z z1i = tha1, a2, a3i
Note. If given two points, use one point with the direction vector determined by the two points.
(2) Parametric equations
x = x1 + ta1,
y = y1 + ta2,
z = z1 + ta3
or
x x1 = ta1,
y y1 = ta2,
z z1 = ta3
(3) Symmetric equations (solve for t in the three equations above, and equate
the results).
x x1 y y1 z z1
=
=
a1
a2
a3
5. LINES AND PLANES IN SPACE
27
Problem (Page 740 #8). Find the vector, parametric and symmetric equations of the line through ( 3, 1, 0) and perpendicular to both h0, 3, 1i and
h4, 2, 1i.
i j k
h0, 3, 1i ⇥ h4, 2, 1i = 0 3 1 = h1, 4, 12i
4 2
1
is perpendicular to bothh0, 3, 1i and h4, 2, 1i. Then the vector equation is
hx + 3, y
1, zi = th1, 4, 12i,
the parametric equations are
x + 3 = t,
y
1 = 4t,
z = 12t,
and the symmetric equations are
x+3 y 1
z
=
= .
1
4
12
For t = 3, x = 0, y = 13, and z = 36, so (0, 13, 36) is on this line and each
quotient of the symmetric equations is 3.
Definition. Let l1 and l2 be two lines in R3, with parallel (or direction)
vectors a and b, respectively, and let ✓ be the angle between a and b.
(1) l1 and l2 are parallel whenever a and b are parallel.
(2) If l1 and l2 intersect, then
(a) the angle between l1 and l2 is ✓.
(b) l1 and l2 are orthogonal whenever a and b are orthogonal.
(3) Nonparallel, nonintersecting lines are called skew lines.
28
10. VECTORS AND THE GEOMETRY OF SPACE
Problem (Page 740 #14). Determine whether the following lines are parallel, skew, or intersect:
8
8
>
>
<x = 1 2t
<x = 3 + 2s
y = 2t
y= 2
>
>
:z = 5 t
:z = 3 + 2s
We have
hx
1, y, z
5i = th 2, 2, 1i and hx
3, y + 2, z
3i = sh2, 0, 2i.
Since h 2, 2, 1i 6= ch2, 0, 2i, lines are not k.
Do lines intersect? Equate x’s:
1
2t = 3 + 2s =) 2t =
2s
2.
Equate y’s and substitute:
2t =
2 =)
2=
2s
2 =) s = 0 and t =
Substitute into z’s:
5 + 1 6= 3 + 0,
so the lines do not intersect, and thus we have skew lines.
1.
5. LINES AND PLANES IN SPACE
29
Planes in R3
A plane in space is determined by a vector a = ha1, a2, a3i that is normal to
the plane and a point P1(x1, y1, z1) lying in the plane.
A vector is normal to a plane if it is orthogonal to every vector in the plane.
(1) Linear equation:
!
a · P1P = ha1, a2, a3i · hx
x1, y
y1, z z1i =
a1(x x1) + a2(y y1) + a3(z
|
{z
linear equation f or the plane
Simplifying, every equation of the form
z1) = 0
}
ax + by + cz + d = 0
is the equation of a plane with normal vector ha, b, ci 6= 0.
(2) Vector equation:
!
a · P1P = 0
Definition. Two planes are parallel when their normal vectors are parallel,
and orthogonal when their normal vectors are orthogonal.
30
10. VECTORS AND THE GEOMETRY OF SPACE
Example. Three noncollinear points determine a plane. Find the plane
containing P ( 1, 3, 0), Q(3, 2, 4), and R(1, 1, 5).
We first find two vectors in the plane by using the points given.
!
!
P Q = h4, 1, 4i and QR = h 2, 3, 1i.
We then find a normal vector n by taking their cross product.
i
!
!
P Q ⇥ QR = 4
2
j k
1 4 = h11, 12, 14i.
3 1
Therefore, the equation is
11(x + 1)
12(y
3)
14(z
0) = 0
or
11x
Intercepts are
x=
47
,
11
12y
14z + 47 = 0.
y=
47
,
12
z=
47
.
14
5. LINES AND PLANES IN SPACE
31
Problem (Page 740 #24). Find the plane containing the point (3, 0, 1)
and ? to the planes x + 2y z = 2 and 2x z = 1.
Normal vectors to the two given planes are n1 = h1, 2, 1i and n2 = h2, 0, 1i.
Then a normal vector to the sought plane is
i j k
n = n1 ⇥ n2 = 1 2 1 = h 2, 1, 4i.
2 0 1
Then the equation is
2(x
3)
y
4(z + 1) = 0
or
2x + y + 4z = 2.
32
10. VECTORS AND THE GEOMETRY OF SPACE
Distance from a point P0 to a plane ax + by + cz + d = 0
We measure the distance orthogonal to the plane.
(1) Choose any point P1 in the plane.
(2) a = ha, b, ci is a normal vector to the plane.
(3) The distance d is then
!
d = |compa P1P0|.
Problem (Page 741 #40). Find the distance from (1, 3, 0) to 3x+y
5z =
2.
P0 = (1, 3, 0), a normal vector is h3, 1, 5i, and we choose P1 = (0, 2, 0) since
!
it satisfies the equation. Then P1P0 = h1, 1, 0i, and
d = |comph3,1,
5i h1, 1, 0i| =
h1, 1, 0i · h3, 1, 5i
4
4
= p
=p
kh3, 1, 5ik
35
35
Note. You can see the derivation of the distance formula on page 739 of
the text.
5. LINES AND PLANES IN SPACE
33
Problem (Page 740 #36 — Finding the intersection of 2 planes). Find the
intersection of 3x + y z = 2 and 2x 3y + z = 1.
(1) Possibilities are that the intersection is a line, a plane, or does not exist (the
planes are parallel.
Since normal vectors to the planes, h3, 1, 1i and h2, 3, 1i, are not parallel,
the intersection here is a line.
(2) Solve each equation for the same variable – z seems easiest here.
z = 3x + y
2 and z =
2x + 3y
1
(3) Set these equal to each other.
3x + y
2=
2x + 3y
1
(4) Solve for a second variable – we choose y.
5
1
1 =) y = x
2
2
(5) Substitute this into either first equation.
⇣5
1⌘
11
5
z = 3x + x
2 =) z = x
2
2
2
2
or
⇣5
1⌘
11
5
z = 2x + 3 x
1 =) z = x
2
2
2
2
(6) Set the third variable equal to a parameter t.
2y = 5x
x=t
(7) Then parametric equations for the line are
8
>
<x = t
y = 52 t 12
>
:z = 11 t 5
2
2
Note. Compare this problem to Example 5.8 on page 738.
34
10. VECTORS AND THE GEOMETRY OF SPACE
Maple. See Lines and Planes(10.5).mw or Lines and Planes(10.5).pdf
Also see Matching Lines.
6. Sufaces in Space
Maple. See surfaces(10.6).mw or surfaces(10.6).pdf
Also see Drawing Surfaces, Drawing 3D Surfaces, and Matching Quadric Surfaces.