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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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2 MA1505 Chapter 5. Three Dimensional Space

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


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


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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.

nus - ma1505 (2012) - chapter 5

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