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Chapter 5. Three Dimensional Space
5.1 The Coordinate System of the 3D Space
For three dimensional space, we first fix a coordinate
system by choosing a point called the origin, and
three lines, called the coordinate axes, so that each
line is perpendicular to the other two. These lines
are called thex-. y- andz-axes.
Associated with a pointPin three dimensional space
is an ordered triple (a,b,c) wherea,bandcare the
projections ofPon thex-,y- andz-axes respectively.
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This is the Cartesian coordinate system for
three dimensional space. We also call this space the
xyz-space.
By convention, we use the right-handed coordi-
nate system. A right-handed coordinate system
fix the orientation of the axes as follow:
If we rotate the x-axis counterclockwise toward the
y-axis, then a right-handed screw will move in the
positivezdirection.
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5.2 Vectors in xyz-Space
A vector is measurable quantity with a magnitude
and a direction. It is geometrically represented by
an arrow in thexyz-space with an initial point and a
terminal point. The direction of the arrow gives the
direction of the vector; and the length of the arrow
gives the magnitude of the vector.
5.2.1 Terminologies and notations
(1) LetPandQbe points in thexyz-space with co-
ordinates (x1, y1, z1) and (x2, y2, z2) respectively.
Then the vectorP Qis algebraically given by
P Q=
x2 x1y2
y1
z2 z1 .
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The vectorOP = x1y1z1
is called the positionvector ofP.
(2) The zero vector in thexyz-space is O=
00
0
.
(3) The sum ofv1=
x1y1
z1
and v2=
x2y2
z2
is
v1+ v2= x1+x2
y1+y2z1+z2
.
[Note that v1+ O=O + v1=v1.]
(4) The negative ofv1=
x1y1
z1
is v1=
x1y1z1
.
[Note that v1 v1= v1+ v1=O.]
(5) The differencev1 v2is
v1v2
=v1
+(v2
) = x1
y1
z1
+ x2y2z2 =
x1 x2y1 y2z1 z2
.
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(6) Ifcis a real number, the scalarcv1ofv1bycis
cv1=
cx1cy1
cz1
.
Ifc >0, thencv1 is in the same direction as v1.
Ifd
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5.2.2 Example
Let P1,P2,Q1and Q2be the points (3, 2, 1), (0, 0, 0),
(5, 5, 4) and (2, 3, 5) respectively.
P1Q1 = 5 35 24 (1)
= 235
P2Q2 =
2 03 05
0
=
235
.
Hence
P1Q1=
P2Q2.
The magnitude ofP1Q1 is
||P1Q1|| =
(2)2 + (3)2 + (5)2 =
38.
So the magnitude of 5P1Q1 is
5||P1Q1|| = 538.6
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5.2.5 Example
Ifv1=
24
5
and v2=
12
3
, then
v1 v2= (2)(1) + (4)(2) + (5)(3) = 21.
Also
||v1|| =
22 + 42 + 52 =
45,
||v2
||= (
1)2 + 22 + 32 =
14.
Hence
cos = 21
45
14=
7
10.
Thus is approximately 3313.
The vectors w1=
25
1
and w2=
42
2
are per-
pendicular since their dot product
v1 v2= (2)(4) + (5)(2) + (1)(2) = 0.9
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5.2.6 Properties of scalar product
Ifv1, v2 and v3 are vectors in xyz-space and c is a
real number, then
(a)v1 v1= ||v1||2 0.
(b)v1 v2=v2 v1.
(c) (v1+ v2)
v3=v1
v3+ v2
v3.
(d) (cv1) v2=v1 (cv2) =c(v1 v2).
5.2.7 Unit vector
A unit vector inxyz-space is a vector of magnitude
or length 1. The vectors
i=
10
0
, j=
01
0
, k=
00
1
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are unit vectors along the positive x-, y- andz-axes
respectively.
Notice that every vector
xy
z
can be written as
xyz
=xi +yj +zk.For example,
w= 4
5
22= 4i 5j + 22k.
The unit vector with the same direction as wis
1
||w||w= 1
42 + 52 + 222(4i 5j + 22k)
=
4
525i 5
525j + 22
525k.
5.2.8 Projection
The projection of a vector b onto a vector a, de-
noted by projabis illustrated below.
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000000
000000
000000
000000
000000
000000
000000
000000
000000
000000
000000
000000
111111
111111
111111
111111
111111
111111
111111
111111
111111
111111
111111
111111
000000000111111111
b
proj
ab
a
From the definition of the scalar product, we have
a b= ||a||||b|| cos .
Therefore the length of the projection ofbonto ais
||projab|| = ||b|| cos =(a b)
||a|| .
So
projab= (||projab||) (unit vector along a)
=(a b)
||a||
a
||a||
=(a b)||a||2 a.
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5.3 Vector Product
Ifv1=
x1y1
z1
and v2=
x2y2
z2
,
then their vector product or cross product is
the vector
v1 v2=
i j k
x1 y1 z1x2 y2 z2
= (y1z2 y2z1)i (x1z2 x2z1)j + (x1y2 x2y1)k.
For example, if v1 =
212
and v2 = 31
3
,then their vector product is the vector
v1
v2= i j k
2 1 2313
= i + 12j 5k.
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5.3.1 Properties of vector product
Let v1, v2and v3be vectors inxyz-space, and let c
be a real number. Then
(a) v1 v2= v2 v1.
(b) v1 (v2+ v3) =v1 v2+ v1 v3.
(c) (v1+ v2)
v3=v1
v3+ v2
v3.
(d) c (v1 v2) = (cv1) v2=v1 (cv2) .
(e) v1 v1=O.
(f) O v1=v1 O=O.
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5.3.2 Direction of v1
v2
Let v1 and v2 be two (non-parallel) vectors which
determine a plane . i.e. is the plane that contains
both v1and v2.
Using the definition of vector product, we can check
that
(v1 v2) v1= 0 and (v1 v2) v2= 0
i.e. v1 v2 is perpendicular to v1and v2.
Hencev1 v2 is perpendicular to the plane .
..................................................................................................................................................................
................
v1
........................................................................................................................................................ ................
v2...................................................................................................................................................................................
................
v1 v2
.........................................................
...
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5.3.3 Magnitude of v1
v2
Let be the angle between v1and v2.
We have
||v1 v2|| = ||v1||||v2|| sin .5.4 Lines in 3D Space
5.4.1 Vector equation of a line
LetLbe a line passing through a point P0 with po-
sition vector r0 = x0i+y0j+z0k and parallel to a
vectorv=ai + bj + ck. Then any pointP onLhas
position vector
OP =r=r0+tv
= (x0i +y0j +z0k) +t(ai +bj +ck) (3)
for somet R.17
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(3) is called a vector equationof the lineL.
............................................................................................................................................................................................
................
P0
...........................
............................
...........................
...........................
...........................
............................
...........................
...........................
...........................
............................
...........................
...........................
...........................
...........................
............................
.................... ................
P................................................................................................................................................................................... ........
........v
rr0
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
0
L
5.4.2 Parametric equation of a line
Writing
r=xi +yj +zk
the vector equation (3) becomes
xi +yj +zk= (x0i +y0j +z0k) +t(ai +bj +ck).
Equating the three components, we get
x=x0+at, y=y0+bt, z =z0+ct.
These are called theparametric equationsof the
lineLdue to the parametertin the equations.
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5.4.3 Example
The pointsAandB have position vectors
3i + 2j 3k and i j + 4k
respectively. Write down the parametric equations of
the line passing throughAandB.
Solution: The position vectors ofAandB are
OA = 3i + 2j 3k, OB =i j + 4k
respectively. So the line is parallel to the vector
AB = (i j + 4k) (3i + 2j 3k) = 4i 3j + 7k.
If we use the position vector ofA as r0, the vector
equation is given by
r= (3i + 2j 3k) +t(4i 3j + 7k). (4)19
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Alternatively, if we use the position vector ofB as
r0, the vector equation is given by
r= (i j + 4k) +s(4i 3j + 7k). (5)
To get the parametric equations of the line, let
r=xi +yj +zk,
and substitute in the LHS of equation (4) or (5).
xi + yj + zk= (3 + 4t)i + (2 3t)j + (3 + 7t)k.
Hence
x= 3 + 4t, y= 2 3t, z= 3 + 7t.5.4.4 Example.
Given the following lines whose vector equations are
L1: r=i +t1(i + 2j + 3k),
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L2: r= (2i +j) +t2 3i + 92j + 92kandL3: r= (2i +j) +t3(3i +j).
(a) Find the position vector of the point of intersec-
tion ofL1andL2.
(b) Show thatL1andL3are skew, i.e. do not inter-
sect each other.
Solution:
(a) Eliminating r from the vector equations ofL1
andL2, we get
i +t1(i + 2j + 3k) = (2i +j) +t2
3i +
9
2j +
9
2k
.
Hence it follows that
t1= 1 + 3t2, 2t1= 1 +9
2t2, 3t1=9
2t2
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from which we obtain
t1= 1, t2= 2/3.
Putting t1 =1 into the vector equation ofL1, we
obtain
r=i + (1)(i + 2j + 3k) = 2j 3k.
So the position vector of the point of intersectionP
of the two lines:
OP = 2j 3k.
(b) Eliminating r from the vector equations ofL1
andL3, we get
i +t1(i + 2j + 3k) = (2i +j) +t3(3i +j).
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Hence it follows that
t1= 1 + 3t3, 2t1= 1 +t3, 3t1= 0
Solving the first two equations above gives t1 = 2/5
but the last equation says t1 = 0, thus there is a
contradiction. So there is no solution to the equations
and we conclude that L1andL3do not intersect.
5.4.5 Example
Find the shortest distance from the point A with
position vector 4i+ 5j to the line L whose vector
equation is
r= (3i +j) +t(i + 2j).
Solution: Lpasses through the pointP(3, 1, 0)23
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and is parallel to the vector a=i + 2j. Letbbe the
vector
P A=
OA OP = (4i + 5j) (3i +j) = 7i + 4j.
(4,5,0)
P(3,1,0)
L
b = 7i + 4j
Q a = i + 2j
From Section ??, the length of the projection of b
ontoais
|P Q| = a b||a|| =(i + 2j) (7i + 4j)
12 + 22= 15
5.
Now the shortest distance from A to L is given by
|AQ|.
Applying Pythagoras theorem on the right triangle
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AP Q,
|AP|2 = |P Q|2 + |AQ|2,
we get
|AQ| = ||b||2 1552
=
72 + 42 15
2
5
= 2
5.
5.5 Planes in 3D Space
Suppose we wish to find the vector equation of a
plane passing through a given point R with posi-
tion vectorr0 relative to the originO and such that
has n as a normal vector to it. Let P be a gen-
eral point in the plane with position vector r. Then
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and
n=ai +bj +ck,
so that
r r0= (x x0)i + (y y0)j + (z z0)kand
(r r0) n=a(x x0) +b(y y0) +c(z z0).
Therefore, the vector equation of the plane can be
written in the form
ax+by+cz=d, whered=ax0+by0+cz0.
The Cartesian equation of a plane passing through a
point (x0, y0, z0) and with normal vectorai + bj + ck
is
ax+by+cz=ax0+by0+cz0.
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5.5.2 Example
Find the equation of the plane passing through the
point (0, 2, 1) normal to the vector 3i + 2j k.
Solution: The required equation is
3x+ 2y z= 3(0) + 2(2) (1), or
3x+ 2y z= 5.
5.5.3 Example
Find the vector equation of the plane passing through
the pointsA(0, 0, 1),B(2, 0, 0) andC(0, 3, 0).
Solution: The following vector is perpendicular to
the plane:
AB AC= i j k
2 010 31 = 3i + 2j + 6k.
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The plane passes through (0, 0, 1). So an equation of
the plane is
3x+ 2y+ 6z= 3(0) + 2(0) + 6(1),
or
3x+ 2y+ 6z= 6.
The plane also passes through (2, 0, 0), so we will get
the same equation
3x+ 2y+ 6z= 3(2) + 2(0) + 6(0) = 6.
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5.5.4 Distance from a point to a plane
The shortest distance from a point S(x0, y0, z0) to
a plane : ax+by+cz=d,is given by
|ax0+by0+cz0
d
|a2 +b2 +c2 (6)
Indeed for any point P(x1, y1, z1) on the plane ,
the length of the projection ofP S onto a normal
vectornof the plane gives the distance fromSto .
SinceP(x1, y1, z1) lies on , soax1+ by1+ cz1=d.
Now
P S=
OSOP = (x0x1)i+(y0y1)j+(z0z1)k.
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A normal vector to the plane is n=ai +bj +ck.
Hence the distance from S(x0, y0, z0) to is
Projn
P S
=|P S
n
|||n|| (Section ??)=
|[(x0 x1)i + (y0 y1)j + (z0 z1)k] (ai +bj +ck)|a2 +b2 +c2
=|a(x0 x1) +b(y0 y1) +c(z0 z1)|
a2 +b2 +c2
=|ax0+by0+cz0 d|a2 +b2 +c2
sinceax1+by1+cz1=d.
5.5.5 Example
Find the distance of the point (2, 3, 4) to the plane
x+ 2y+ 3z= 13.
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Solution: Using (6), we have (x0, y0, z0) =
(2, 3, 4) anda= 1, b= 2, c= 3.
So the distance is
|1(2) + 2(
3) + 3(4)
13
|12 + 22 + 32 = 5
14
5.6 Vector Functions of One Variable
Let f(t),g(t) and h(t) be real-valued functions of a
real variablet. A vector function
r(t) =
f(t)g(t)
h(t)
=f(t)i +g(t)j +h(t)k
is a function such that the images (output) are vec-
tors (instead of scalars). The three functions f(t),
g(t) andh(t) are called the component functions
ofr(t).
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5.6.1 Example
Consider the vector function
r(t) =ti + (t2 + 1)j + (2 7t)k.
Then
r(2) = 2i + 5j 12k.
5.6.2 Derivatives of vector functions
The derivativeof a vector function
r(t) =f(t)i +g(t)j +h(t)k,
wheref,g andhare differentiable functions, is
(r)(t) =f(t)i +g(t)j +h(t)k. (7)
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5.6.3 Example
Consider the vector function
r(t) =ti + (t2 + 1)j + (2 7t)k.
Then by (7), since
d
dt(t) = 1,
d
dt(t2 + 1) = 2t,
d
dt(2 7t) = 7,
we have
r(t) =i + (2t)j 7k.
5.6.4 Definite integral of a vector function
The definite integral of a continuous vector function
r(t) =f(t)i +g(t)j +h(t)k
on the interval [a, b] is
ba
r(t) dt= b
af(t) dt i+
ba
g(t) dtj+ b
ah(t) dt k.
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For example,
2
0
(2ti + 3t2j)dt=
t2t=2
t=0i +
t3t=2
t=0j= 4i + 8j.
5.7 Space curves
A curve in xyz-space can be represented by some
continuous function
r(t) =f(t)i +g(t)j +h(t)k
such that a point P lies on the curve if its position
vectorOP is the image of the vector function, i.e.,
OP =r(t0) for somet0 R.
We call
r(t) =f(t)i +g(t)j +h(t)k
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the vector equationof the curve and
x=f(t), y=g(t), z=h(t)
the parametric equationof the curve.
5.7.1 Example
The vector equation
r(t) = (1 +t)i + (2 +t)j + (3 +t)k
represents the straight line in the xyz-space that
passes through the point (1, 2, 3) and is parallel to
the vector i +j + k.
5.7.2 Smooth curves
A vector function r(t) represents a smooth curve
on an intervalIifr(t) is continuous and r(t) is never
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5.7.4 Example
The following vector function represents a piecewise
smooth curve:
r(t) =
ti +t2
j +t3
k if 0 t 1(2t 1)i +t2j + (t2 +t 1)k if 1< t 2.
5.7.5 Tangent vector and tangent line to a
curve
The tangent line to a curve r(t) at a point P
whose position vector isr(t0) is defined to be the line
throughPparallel to the tangent vector r(t0) (here
it is assumed that r(t0)= 0). The unit tangent
vector to the curve att=t0is
r(t0)
||r(t
0
)||.
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Thus
r(
2) = (1)i + (0)j + (1)k= i + k
is the tangent vector to the circular helix at (0, 1,2
),
the point on the helix corresponding to t = 2 . The
unit tangent vector to the curve at (0, 1,2
) is
12
(i + k).
The tangent lineto the helix at (0, 1,2
) is parallel
to
r(
2
) = (
1)i + (0)j + (1)k.
Therefore the parametric equations of the tangent
line at the point (0, 1,2
), are
x= t, y= 1, z=
2+t.
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5.7.7 Arc length of a space curve
Suppose that a curve has the vector equation
r(t) =f(t)i +g(t)j +h(t)k,
or alternatively, parametric equations
x=f(t), y=g(t), z=h(t),
wheref(t),g(t) andh(t) are continuous functions.
If this curve is traversed exactly once as t
increases fromatob, then its arc length is
L= b
a
(f(t))2 + (g(t))2 + (h(t))2 dt
=
ba
dx
dt
2+
dy
dt
2+
dz
dt
2dt.
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A more compact formula of both arc length formulas
is
L=
ba
||r(t)|| dt.
5.7.8 Example
Recall the circular helix of Example ??:
r(t) = (cos t)i + (sin t)j +tk,
r(t) = ( sin t)i + (cos t)j + k.
Hence we can find the arc length fromt= 0 tot= 2
as follows:
||r(t)|| =
( sin t)2 + (cos t)2 + 12 =
2,
L= 2
0
||r(t)|| dt= 2
0
2dt= 2
2.