chapter 10- vector and geometry space

8/10/2019 Chapter 10- Vector and Geometry Space

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The three coordinate axes determine the three

coordinate planes illustrated in the fi"ure &

The xy-plane is the plane that

contains the x- and y-axes'

the y#-plane contains the y-

and #-axes' and x#-plane

contains the x- and #-axes.These three coordinate

planes di%ide space into ei"ht

parts, called octant, in the

fore"round, is determined by

the positi%e axes.

Vectors in the plane

The term vectoris used by scientists to indicate a quantity

(such as displacement or %elocity or force) that has both

ma"nitude and direction. %ector is often represented by

an arrow or a directed line se"ment. The len"th of the

arrow represents the ma"nitude of the %ector and the arrow

points in the direction of the %ector. We denote a %ector by

printin" a letter in boldface ( v) or puttin" an arrow

abo%e the letter (

) .

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For instance , suppose a particle mo%es alon" a line

se"ment from point to point . The correspondin"

displacement vector v, shown in the fi"ure below , has

initial point ( the tail) and terminal point ( the tip) andwe indicate this by writin" % ()= .

*otice that the %ector u +,= has the same len"th and thesame direction as e%en thou"h it is in a different position.

We say that uand vare equivalent( or equal) and we

write u%. Thezero vector, denoted by ! , has len"th 0. t is

only %ector with no specific direction.

Combining Vectors

/uppose a particle mo%es from to , so its displacement

%ector is ( () ). Then the particle chan"es direction and

mo%es from to +, with displacement %ector ()+) as in

the fi"ure&

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n the followin" fi"ure

We start with the same %ectors u and % as in the pre%ious

one and draw another copy of v with the same initial

point as u. completin" the parallelo"ram , we see that u

+ v v + u. This also "i%es another way to construct the

sum& if we place u and % so they at the same point , then

u 0% lies alon" the dia"onal of the parallelo"ram with u

and v as sides . ( This is called the!arallelogram "a#)

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$%ample

raw the sum of the %ectors a and b shown in the

fi"ure &

Solution

First we translate b and place its tail at the tip of a, bein"

careful to draw a copy of b that has the same len"th and

direction. Then we draw the %ector ( a 0b) ( see the

fi"ure)

/tartin" the initial point of a and endin" at the terminalpoint of the copy of b.

lternati%ely , we could place b so it starts where a starts

and construct ( a 0b) by the parallelo"ram law as in the

fi"ure &

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Scalar &ultiplication

t is possible to multiply a %ector by a real number +. ( we

call the real number a scalar to distin"uish it from a %ector). For instance , we want 1% to be the same %ector as ( % 0

%), which has the same direction as % but is twice as lon".

n "eneral , we multiply a %ector by a scalar as follows&

efinition of scalar multiplication

f + is a scalar and vis a %ector, then the scalar multiple

cvis the %ector whose len"th is c times the len"th of vand

whose direction is the same as % if c2! and is opposite to %

if c3 ! . if c! or %! , then cv0

The definition is illustrated in the followin" fi"ure&

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we see that real numbers work like scalin" factors here'that is why we call them scalars. *otice that two non#ero

%ectors are parallel if they are scalar multiplies of one

another.

n particular, the %ector 4% (-5)% has the same len"th as %

but points in the opposite direction. We call it the negative

of %.

!osition vector

/ince the location of the initial point is irrele%ant, we

typically draw %ectors with their initial point located at the

ori"in. /uch a %ector is called a position %ector. For the

position %ector a with initial point at the ori"in and

terminal point at the point (a5,a1) we denote the %ector by

1 2,a 6( a a= = ( see the fi"ure)

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7a"nitude of position %ector a is &2 2

1 2a a a= +

f 0a = then 0a= this is called a #ero %ector. ny %ector

with ma"nitude of 5 i.e. 1u = is called a unit %ector. Wewill occasionally find it con%enient to write %ectors in

terms of some standard %ectors, we define the standard

basis %ectors i and 8 by 1, 0 , , 0,1i and 8= = and thiscan be extended to three dimensional case. *otice that

1i 8= = ( see the fi"ure)

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Difference of t#o vectorsy the difference ' u(v)of two %ectors we mean

u4 v u0 (-v) , so we can construct (u-%) by first drawin"

the ne"ati%e of %,-% , and then addin" it to u by the

parallelo"ram law in the followin" fi"ure&

lternati%ely, since % 0 ( u 4 % ) u, the %ector u 4 %,

when added to % "i%es u so we could construct u-% as infi"ure below by means of trian"le law

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$%ample

f a and b are the %ectors shown in the fi"ure below

,draw a(*b

SolutionWe first draw the %ector ( - 1 b) pointin" in the direction

opposite to b and twice as lon". We place it with its tail

at the tip of a and then use the trian"le law to draw a0(-

1b) as in the fi"ure below&

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Vectors in the space

/e%eral ideas can be extended from 1(91) to : space

(9:). ;oint in 9:is defined by (a,b,c), while x,y,# is the

ri"ht-handed coordinate system. (see the fi"ure )

7a"nitude of2 2 2

1 2 3 1 2 3, ,a a a a is a a a a= = + +

ector addition&

1 2 3 1 2 3

1 1 2 2 3 3, ,

a b a a a b b b

a b a b a b

+ = + + + + +

= + + +

/tandard basis %ectors &

1, 0, 0 , 0,1, 0 , 0, 0,1i 8 and k = = =ny %ector 1 2 3 1 2 3, ,a a a a a i a 8 a k = = + +Findin" the equation of a sphere of radius r centred at

(a,b,c) & the sphere consists of all points (x,y,#) whose

distance from (a,b,c) is r &

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2 2 2

2 2 2 2

( ) ( ) ( )

( ) ( ) ( )

x a y b # c r

x a y b # c r

+ + =

+ + =

,he Dot !roduct

/o far we ha%e added two %ectors and multiplied a %ector

by a scalar. The question arises ' s it possible to multiply

two %ectors so that their product is a useful quantity< 6ne

such product is the dot product, whose definition follows

(5)efinition

f { } 3,,,, 21321 bbbbandaaaa == , then the dot product of a and b

is the number a .b"i%en by&

a . b a5b50 a1b10 a:b:

Thus to find the dot product of a and b we multiply

correspondin" components and add. The result is not a

%ector. t is a real number, that is , a scalar, for this reason ,the dot product is sometimes called the scalar product ( or

inner product). lthou"h definition (5) is "i%en for three-

dimensional %ectors , the dot product of two-dimensional

%ectors is defined in a similar fashion&22112121 ,., bababbaa +=

=xample

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7)1)(3()2(2)0(12.32

6)2

1(4)2(7)6)(1(

2

1,2,6.4,7,1

2)1)(4()3(21,3.4,2

=++=+

=++=

=+=

k8k8i

The dot product obeys many of the laws that hold for

ordinary products of real numbers. These are stated in

the followin" theorem.

(1) ;roperties of the ot ;roduct

f a, b, and c are %ectors in :and c is a scalar, then&

0.05

).().(4

..).(3

..2

.1 2

==

+=+==

a

cbabca

cabacba

abba

aaa

These properties are easily pro%ed usin" definition 5.

For instance , here are proofs of properties 5 and :&

caba

cacacabababa

cabacabacaba

cbacbacba

cbcbcbaaacba

aaaaaa

..

)()(

)()()(

,,.,,).(3

.1

332211332211

333322221111

333222111

332211321

22

3

2

2

2

1

+=

+++++=

+++++=

+++++=

+++=+

=++=

The dot product a .b can be "i%en a "eometric

interpretation in terms of the an"le > between a and b,

which is defined to be the an"le between the representation

of a and b that start at the ori"in, where ! ? > ? @ in other

words > is the an"le between the line se"ment 6( and6) in the fi"ure below &

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'-) ,heorem

f > is the an"le between the %ectors a and b, thencos. baba =

;roof

f we apply the law of cosines to trian"le 6 in the

fi"ure , we "et(A) cos2222

6)6(6)6(() +=

*ote that the law of cosine still applies in the limitin" cases

when > ! or @ or ( a! or b!) . but B6BBaB, B6BBbB

, and BB Ba-bB , so equation (A) becomes

(C) Ba-bB1 BaB1 0 BbB1-1BaBBbBcos>

$sin" properties of dot product we can rewrite the left side

of this equation as follows&

22

2

.2

....

)).((

bbaa

bbabbaaa

bababa

+=

+=

=

Therefore, equation (C) "i%es

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or

cos.

cos2.2

cos2.2 2222

baba

baba

bababbaa

=

=

+=+

=xample

f the %ectors a and b ha%e the len"th A and D, and the

an"le between them is @E:, find a . b.

/olution

$sin" theorem (:) , we ha%e

122

1.6.4)

3cos(. ===

baba

The formula in theorem : also enables us to find the

an"le between two %ectors.

+orollary & f > is an an"le between the non#ero

%ectors a and b , then

ba

ba+os

.=

$%ample

Find the an"le between %ectors 1,2,2 =a and

2,3,5=b .

/olution

3)1(22 222 =++=a and

382)3(5 222 =++=b

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and since a . b 1(C)01(-:)0(-5)11

We ha%e from the definition of the dot product&

. 2cos

(3)( 38)

a b

a b= = , so the an"le between a and b is

( )1 02

cos 1.46, 84(3)( 38)

or

=xample

(i)etermine the an"le between a and b if it is known

that 3, 4, 1 , 0,5, 2a and b= =(ii) +heck the ortho"onality of

1 12

2 4u i 8 and % i 8= = +r r

/olution(i)

. 3, 4, 1 0, 5, 2 3(0) (5)( 4) ( 1)(2) 22

9 16 1 26

0 25 4 29

a b

a

b

= = + + =

= + + =

= + + =

r

r

the an"le is then &

. 22cos 0.8011927

26 29

a b

a b

= = = r r that means >1.C

radians 5A:.1A!

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(ii)

1 1 1 5. 2, 1, 0 , , 0 12 4 4 4

u %= = = r r

( not ortho"onal)

55,

4u %= =r r

notice that1

4% u=r r

they

are parallel.

lso & 05 5

. 5. cos 1 1804 4

u % u % = = = = =r r r r

=xample

force is "i%en by the %ector F:i0A80Ck and mo%es a

particle from the point ;(1,5,!) to the point (A,D,1).

Find the work done.

/olution

The displacement %ector is

(2,5,2), ;F= = , so that work done is. 3, 4, 5 . 2,5, 2

6 20 10 36

W F ,= =

= + + =

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Components and !roections

For any two non#ero position %ectors a and b, let thean"le between the %ectors be >. f we drop a

perpendicular line se"ment from the terminal point of a

to the line containin" the %ector b, then from elementary

tri"onometry, the base of the trian"le ( in case where

(!3>3@E1) has the len"th "i%en by cosa ( see the

fi"ure)

6n the other hand, notice that if )2

(

, the len"th of

the base is "i%en by cosa ( see the fi"ure)

n either case, we refer to cosa as the component of a

alon" b, denoted +ompba . we can write this expression

as &

b

baa+omp

orbab

bab

b

baaa+omp

b

b

.

.1

cos1

coscos

=

==

==

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*otice that +ompba is a scalar and that we di%ide the

dot product by b and not by a . one way to keep this

strai"ht is to reco"ni#e that the components in fi"uresabo%e depend on how lon" a is but not on how lon" b is.

We can %iew the last formula as the dot product of the

%ector a and a unit %ector in the direction of b, "i%en by

b

b.

6nce a"ain, consider the case where the %ector a

represents a force. 9ather than the component of a alon"b, we are often interested in findin" a force %ector

parallel to b ha%in" the same component alon" b as a.

we call this %ector the pro8ection of a onto b, denoted by

;ro8ba , as indicated in the fi"ures below. /ince the

pro8ection has ma"nitude acompb and points in the

direction of b, for !3>3@E1 and opposite b , for

@E13>3@, we ha%e that

bb

baapro8

b

b

b

ba

b

bacompapro8

b

bb

2

.

.)(

=

==

Where bb

represents a unit %ector in the direction of b.

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=xample

For 5,1,3,2 == banda , find the component of a alon"

b and the pro8ection of a onto b.

/olution

We ha%e 2613

251

152

5,1

5,1.3,2.=

++

=

==

b

baacompb

/imilarly we ha%e

2

5,

2

15,1

2

15,1

26

13

26

5,1

26

13.

===

=

=

b

b

b

baapro8b

,he cross product

n this section , we define a second type of product of

%ectors, the cross product or %ector product. While the dot

product of two %ectors is a scalar, the cross product of two

%ectors is another %ector. The cross product has many

applications, from physics and en"ineerin" mechanics to

space tra%el.

/ome important definitions

efinition

The determinant of a 1 x 1 matrix of real numbers is

defined by

1 2

1 2 2 11 2

2 2x matrix

a a

a b a bb b = 14 2 43

=xample ' e%aluate the determinant1 2

3 4

From the abo%e definition we ha%e&

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2)3)(2()4)(1(43

21==

efinition

The determinant of a : x : matrix of real numbers isdefined as a combination of three 1 x 1 determinants, as

follows&

1 2 3

2 3 1 3 1 2

1 2 3 1 2 3

2 3 1 3 1 2

1 2 3

3 3x matrix

a a ab b b b b b

b b b a a ac c c c c c

c c c

= + +

1 4 2 4 3

this equation is referred to as an expansion of the

determinant alon" the first row. *otice that the multipliers

of each of the 1 x 1 determinants are the entries of the first

row of the : x : matrix. =ach 1 x 1 determinant is the

determinant we "et by eliminatin" the row and column in

which the correspondin" multiplier lies.

=xample

=%aluate the determinant

1 2 4

3 3 1

3 2 5

/olution

=xpandin" alon" the first row, we ha%e

[ ] [ ]

[ ]

1 2 4

3 1 3 1 3 33 3 1 (1) (2) (4)2 5 3 5 3 2

3 2 5

(1) (3)(5) (1)( 2) (2) ( 3)(5) (1)(3)

(4) ( 3)( 2) (3)(3) 41

= +

+ =

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we use the determinant notation as a con%enient de%ice for

definin" the cross product, as follows&

efinitionFor two %ectors 1 2 3 1 2 3, , , ,a a a a and b b b b= = in :,we define the cross product ( or %ector product) of a and b

to be &

2 3 1 3 1 2

1 2 3

2 3 2 3 1 2

1 2 3

i 8 ka a a a a a

axb a a a i 8 k b b b b b b

b b b

= = +

notice that a x b is also a %ector in :. To compute a x b ,

you must write the component of a in the second row and

the component of b in the third row' the order is important.

=xample

+ompute 1, 2, 3 4, 5, 6x

2 3 1 3 1 21, 2,3 4,5, 6 1 2 3

5 6 4 6 4 54 5 6

3 6 3 3, 6, 3

i 8 kx i 8 k

i 8 k

= = +

= + =

note that the cross product is defined only for %ectors in

:, there is no correspondin" operation for %ector in 1.

TheoremFor any %ector 3 , 0, 0 0a - axa and ax = =We pro%e the first one &

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2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

1 2 3

2 3 3 2 2 2 3 2 1 2 2 1( ) ( ) ( ) 0

i 8 ka a a a a a

axa a a a i 8 k a a a a a a

a a a

a a a a i a a a a 8 a a a a k

= = +

= + =

Theorem

For any %ectors a and b in :, a x b is ortho"onal to both a

and b.

;roof

9ecall that two %ectors are ortho"onal if and only if their

dot product is #ero.

[ ] [ ] [ ]

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1

1 2 3 1 3 2 1 2 3 2 3 1 1 3 2 2 3 1

.( ) , , .

0

a a a a a aa axb a a a i 8 k

b b b b b b

a a a a a aa a a

b b b b b ba a b a b a a b a b a a b a b

a a b a a b a a b a a b a a b a a b

= +

= +

= +

= + +

=/o that a and (a x b ) are ortho"onal.

*otice that since a x b is ortho"onal to both a and b, it is

also ortho"onal to e%ery %ector lyin" in the plane

containin" a and b. ut, "i%en a plane, out of which side ofthe plane does a x b point< We can "et an idea by

computin" some simple cross products.

*otice that

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0 0 1 0 1 01 0 0

1 0 0 0 0 10 1 0

i 8 k

ix8 i 8 k k = = + =

likewise 8 x k i

/ight(hand rule

There are illustrations of the ri"ht-hand rule& f you ali"n

the fin"ers of your ri"ht hand alon" the %ector a and bend

your fin"ers around in the direction of rotation from a

toward b ( throu"h an an"le of less than 5G!!), your thumb

will point in the direction ofa x b ( see the fi"ure )

*ow , followin" the ri"ht-hand rule , b x a will point in the

direction opposite a x b ( see the fi"ure )

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in particular, notice that

0 1 01 0 0

i 8 k

8xi k= =

efinition & Two non#ero %ectors 3,a b - parallel if andonly if a x b!

9ecall that a and b are parallel if and only if the an"le >

between them is either ! or @ . in either case , sin> ! and

so from one of the pre%ious theorems we ha%e that &

sin (0) 0axb a b a b= = =The result then follows from the fact that only %ector with

#ero ma"nitude is the #ero %ector.

The same theorem pro%ides us with the followin"

interestin" "eometric interpretation of the cross product.

For any two non#ero %ectors a and b, as lon" as a and b

are not parallel, they form two ad8acent sides of the

parallelo"ram , as seen in the fi"ure below

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notice that the area of the parallelo"ram is "i%en by the

product of the base and the altitude, we ha%e

rea ( base ) ( altitude )

sina b axb=

That is, the ma"nitude of the cross product of two %ectors"i%es the area of the parallelo"ram with two ad8acent sides

formed by the %ectors.

=xample

Find the area of the parallelo"ram with two ad8acent sides

formed by %ectors 1, 2, 3 , 4, 5, 6a and b= =/olution

First notice that&

2 3 1 3 1 21 2 3 3, 6, 3

5 6 4 6 4 54 5 6

i 8 k

axb i 8 k = = + =

from the formula for the area of the parallelo"ram we ha%ethat& 3, 6, 3 54 7.348(rea axb= = =

From the same theorem we can also find the distance from

a point to a line in 9:, as follows. Het d represent the

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For any three noncoplanar %ectors a,b, and c ( three

%ectors that do not lie in a sin"le plane), consider the

parallelepiped formed usin" %ectors as three ad8acent

ed"es ( see fi"ure )

9ecall that the %olume of such solid is "i%en by&

olume ( rea of base)(altitude)

Further , since two ad8acent sides of the base are formed

by the %ectors a and b, we know that the area of the base is

"i%en by axb . 9eferrin" to the fi"ure abo%e , notice that

the altitude is "i%en by&

.( )axb

c axbcomp c

axb= , the %olume of the parallelepiped is

then&.( )

.( )c axb

-olume axb c axbaxb

= =

the scalar c.(a x b) is called the scalar triple product of the%ectors a ,b , and c. it is possible to e%aluate the scalar

triple product by computin" a sin"le determinant. *ote that

for 1 2 3 1 2 3 1 2 3, , , , , , , ,a a a a b b b b and c c c c= = = ,we ha%e &

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1 2 3

1 2 3

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

1 2 3

1 2 3

1 2 3

.( ) .

, ,

i 8 k

c axb c a a a

b b b

a a a a a ac c c i 8 k

b b b b b b

a a a a a ac c c

b b b b b b

c c c

a a a

b b b

=

= +

= +

=

=xample

Find the %olume of the parallelepiped with three ad8acent

ed"es formed by three %ectors1, 2,3 , 4,5, 6 , 7,8, 0a b and c= = =

/olution

First , note that.( )-olume c axb= , we ha%e that

7 8 02 3 1 3 1 2

.( ) 1 2 3 7 8 05 6 4 6 4 5

4 5 6

( 3)(7) ( 6)(8) 21 48 27

c axb = = +

= = + =

so, the %olume of the parallelepiped is 1I

!roperties of cross product

f candba, are %ectors and e is a scalar, then&

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1. ( )

2. ( ) ( ) ( )

3. ( ) ( )

4. ( ) ( )

5. .( ) ( ). ( )

6. ( ) (

a xb bxa anticommuta%ity

ea xb e axb ax eb

a x b c axb axc distributi%e law

a b xc axc bxc distributi%e law

a bxc axb c scalar triple product

a x bxc a

=

= =

+ = ++ = +

=

=

r r r r r r

r r r r r r r

r r r r r r r

r r r r r r

r r r. ) ( . ) ( )c b a b c %ector triple product

r r r r r r

;roof (5)

We ha%e 1 2 3 1 2 3, , , , ,a a a a and b b b b= this can bewritten in form

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

2 3 1 3 1 2

( )

i 8 ka a a a a a

axb a a a i 8 k b b b b b b

b b b

b b b b b bi 8 k bxa

a a a a a a

= = +

= + =

since swappin" two rows in a 1x1 matrix ( or in a :x:

matrix, for this matter) chan"es the si"n of its determinant.

;roof(:)

For 1 2 3, ,c c c c= , we ha%e

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1 1 2 2 3 3, ,b c b c b c b c+ = + + + , and so

1 2 3

1 1 2 2 3 3

( )

i 8 k

ax b c a a ab c b c b c

+ =+ + +

lookin" only at the i component of this, we ha%e

2 3

2 3 3 3 2 2

2 2 3 3

2 3 3 2 2 3 3 2

2 3 2 3

2 3 2 3

( ) ( )

( ) ( )

a aa b c a b c

b c b c

a b a b a c a c

a a a a

b b c c

= + + +

= +

= +

which is also the i component of a x b 0 a x c. similarly it

is possible to show that the 8 and k components also match.

"ines and planes in space*ormally , we specify a line in the xy-plane by selectin"

any two points on the line or a sin"le point on the line and

its direction, as indicated by the slope of the line. n three

dimensions, specifyin" two points on a line will still

determine the line. n alternati%e is to specify a sin"le

point on the line and its direction. n three dimensions,

direction should make you think about %ectors ri"ht away.HetJs look for the line that passes throu"h the point

;5(x5,y5,#5) and that is parallel to the position %ector

1 2 3, ,a a a a= ( see fi"ure).

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For any other point ;(x,y,#) on the line, obser%e that the

%ector 1; ;will be parallel to a. Further , two %ectors are

parallel if and only if one is a scalar multiple of the other,

so that& 1; ; ta= for some scalar t. The line then

coincides of all points ;(x,y,#) for which ( 1; ; ta= ) holds.

/ince1 1 1 1

, ,; ; x x y y # #= , we ha%e that&

1 1 1 1 2 3, , , ,x x y y # # ta t a a a = = , finally , since two%ectors are equal if and only if all of their components are

equal, we ha%e&

1 1 2 2 1 3, ,x x ta y y ta and # # ta = = =the last equation is called parametric equation for the line ,

where t is the parameter. s in two dimensional case , a

line in space can be represented by many different sets ofparametric equations. ;ro%ided none of a5,a1, and a:are

#ero., we can sol%e for the parameter in each of the three

equations, to obtain&

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1 1 1

1 2 3

x x y y # #

a a a

= =

we refer to the last equality as symmetric equations of the

line.

=xample

Find an equation of the line throu"h the point ( 5,C,1) and

parallel to the %ector 4,3,7 . lso, determine where the

line intersects the y#-plane.

/olution

From the pre%ious equations we know that the parametric

equations for the line are

1 4 , 5 3 , 2 7x t y t and # t = = =The symmetric equations of the line are &

1 5 2

4 3 7

x y # = =

The followin" fi"ure representin" the case&

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the line intersects the y#-plane where x!. settin" x!

yields that17 1

,4 4

y and #= =

alternati%ely, we could sol%e x-5At for t ( a"ain where

x!) and substitute this into the parametric equations for y

and # , so the line intersects the y#-plane at the point

( ! , 5IEA, 5EA ).

=xample

Find an equation of the line passin" throu"h the points

;(5,1,-5) and (C,-:,A).

/olution

First, a %ector that is parallel to the line is5,5,4))1(4,23,15( ==;F

pickin" either point will "i%e us equations for the line. Kere

we use ;, so that parametric equations for the line are&t#andtytx 51,52,41 =+== ,similarly , symmetric equations

of the line are

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1 2 1

4 5 5

x y # += =

the followin" fi"ure shows the "raph of the line

/ince we ha%e specified a line by choosin" a point on the

line and a %ector with the same direction , we can state the

followin" definition&

efinition

Het H5 and H1 be two lines in 9:, with parallel %ectors a

and b, respecti%ely, and let > be the an"le between a and b.

5-The lines H5and H1are parallel whene%er a and b are

parallel.

1-f H5and H1intersect, then

(i)The an"le between H5and H1is > and

(ii)The lines H5and H1are ortho"onal whene%er a and b

are ortho"onal.

n two dimensions, two lines are either parallel or they are

intersect. This is not true in three dimensions.

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$%ample& /how that the lines &t#andtytxl 25,21,2:1 === and

s#andsysxl 31,2,1:2 ===are not parallel, yet do notintersect.

/olution

We used different letters ( t and s )for the two lines as the

parameters . sol%in" the first parametric equation of each

line for the parameter in terms of x , we "etxt =2 , and 1= xs ,

respecti%ely , this means that the parameter representssomethin" different in each line' so we must use different

letters. *otice from the "raph below ,

that the lines are most certainly not parallel, but it is

unclear whether or not they intersect. Lou can read fromthe parametric equations that a %ector parallel to H5is

2,2,11 =a while the %ector parallel to H1is 3,1,12 =a .

/ince a5is not a scalar multiple of a1, the %ectors are not

parallel and so, the lines H5and H1 are not parallel'. The

lines intersect if there is a choice of the parameters s and t

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that produces the same point, that is, that produces the

same %alue for all x, y, and #. settin" the x-%alues equal, we

"etst += 12

, so thatts = 1. settin" that y-%alues equal andsettin" ts = 1 , we "et

tt

st

+===+

1)1(2

221

/ol%in" this for t yields 0=t , which further implies that 1=s

. settin" the #-components equal "i%es1325 +=+ st , but this equation is not satisfied when 0=t and

1=s, so, H5and H1are not parallel, yet do not intersect.

efinition

*onparallel, nonintersectin" lines are called skew lines.

*ote that itJs easy to %isuali#e skew lines. raw two planes

that are parallel and draw a line in each plane ( so that it

lies completely in the plane). s lon" as two lines are not

parallel, these are skew lines ( see the fi"ure below).

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!lanes in /-

6bser%e the followin" fi"ure &

*ote that the y#-plane is a set of points in space such that

e%ery %ector connectin" two points in the set is ortho"onal

to i. Kowe%er, e%ery plane parallel to the y#-plane satisfies

this criterion . n order to select the one that corresponds

to the y#-plane, you need to specify a point throu"h which it

passes ( any one will do).

n "eneral, a plane in space is determined by specifyin" a

%ector 1 2 3, ,a a a a= that is normal to the plane ( i.e.,ortho"onal to e%ery %ector lyin" in the plane) and a point

;5(x5,y5,#5) lyin" in the plane ( see the fi"ure)

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n order to find an equation of the plane, let ;(x,y,#)

represent any point in the plane. Then, since ; and ;5are

both points in the plane , the %ector

1 1 1 1, ,;; x x y y # #= lies in the plane and so, must beortho"onal to a. y theorem that we used earlier we ha%e

1 1 2 3 1 1 1

1 1 2 1 3 1

0 . , , . , ,

0 ( ) ( ) ( )

a ;; a a a x x y y # # or

a x x a y y a # #

= =

= + +

the last equation is equation for the plane passin" throu"h

the point ( x5,y5,#5) with normal %ector 1 2 3, ,a a a .

$%ample

Find an equation of the plane containin" the point (5,1,:)

with normal %ector (A,C,D) and sketch the plane.

/olution

From the equation of the plane we ha%e

!A(x-5)0C(y-1)0D(#-:) , which is the equation of the

plane. To draw the plane, we locate three points lyin" in

40


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the plane, n this case , the simplest way to do this is to

look at the intersections of the plane with each of the

coordinate axes. . when y#!, we "et that &

!A(x-5)0C(!-1)0D(!-:)Ax-A-5!-5G, so that Ax:1 andxG . the intersection of the plane with the x-axis is the

point ( G,!,!) , similarly we can find the intersections of the

plane with y- and #-axes& (!,:1EC , !) and ( !,!,5DE:)

respecti%ely. $sin" these three points , we can draw the

plane seen in fi"ure below&

we start by drawin" the trian"le with %ertices at the three

points' the plane we want is the one containin" this

trian"le. *otice that since the plane intersects all three of

the coordinate axes, the portion of the plane in the firstoctant is the indicated trian"le.

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1inding the equation of a plane given three points

*ot that we can expand the equation of the plane

1 1 2 1 3 10 ( ) ( ) ( )a x x a y y a # # = + + to the followin" form&1 2 3 1 1 2 2 3 3

tan

0 ( )

cons t

a x a y a # a x a x a x= + + + 1 4 44 2 4 4 43

we refer to the last equation as a linear equation in the

three %ariables x, y,and #. n particular, this says that e%ery

linear equation of the form

!ax0by0c#0d is the equation of a plane with normal

%ector , ,a b c , where a,b,c,and d are constants.We noticed earlier that three points determine a plane, but

how can we find an equation of a plane "i%en only three

points< We need first to find a normal %ector then proceed

easily as in the example

$%ample

Find the plane containin" three points ;(5,1,1), (1,-5,A)

and 9(:,C,-1).

/olution

First, we will need to find a %ector normal to the plane.

*otice that two %ectors lyin" in the plane are

1, 3, 2 , 1, 6, 6;F and F9= = , consequently, a

%ector ortho"onal to both of ,;F and F9 is the cross

product

42


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1 3 2 6,8,9

1 6 6

i 8 k

;x9= =

uuur uuur

since ,; and 9 are not parallel, ;FxF9 must be

ortho"onal to the plane, as well an equation for the plane

is then 0 6( 1) 8( 2) 9( 2)x y #= + + , in the followin"fi"ure we can see the trian"le with %ertices at the three

points. The plane in question is the one containin" the

indicated trian"le.

,he equation of a plane given a point and a parallel

plane

n three dimensions, two planes are either parallel or they

intersect in a strai"ht line.. suppose that two planes ha%in"

43


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normal %ectors a and b , respecti%ely, intersect. Then the

an"le between the planes is the same as the an"le between

a and b ( see the fi"ure)

With this in mind, we say that the two planes are parallelwhene%er their normal %ectors are parallel and the planes

are ortho"onal whene%er their normal %ectors are

ortho"onal.

$%ample

Find an equation for the plane throu"h the point (5,A,-C)

and parallel to the plane defined by 1x-Cy0I#51

/olution

First, notice that a normal %ector to the "i%en plane is

2, 5,7 . /ince the two planes are to be parallel, this%ector is also normal to new plane, then we can write down

the equation of the plane as follows&

!1(x-5)-C(y-A)0I(#0C)

1inding intersection of t#o planes

The intersection of two nonparallel planes should be a line

$%ample

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Find the intersection of the planes 32 =++ #yx and534 =+ #yx .

/olution & sol%in" both equations for x , we "et#yx = 23 and #yx 345 +=

settin" these expressions for x equal "i%es us#y#y 34523 +=

sol%in" for # "i%es us262 += y# or 13 += y#

returnin" to either plane equations , we can sol%e for x

( also in terms of y) , we ha%e18)13(223223 +=+== yyy#yx

Takin" y as the parameter ( i.e., lettin" ty= ) , we obtain

parametric equations for the line of intersection&13,,18 +==+= t#andtytx

/ee the fi"ure below

Distance bet#een parallel planes

The distance from the plane ax0by0c#! to a point ;!

( x!,y!,#!) not on the plane is measures alon" a line

se"ment connectin" the point to the plane that is

ortho"onal to the plane ( see the fi"ure)

45


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To compute this distance, pick any point ;5( x5y5#5) lyin"

in the plane and let , ,a a b c= denote a %ector normal tothe plane. From the fi"ure notice that the distance from ;!

to the plane is simply 1 0acomp ;; , where&

1 0 0 1 0 1 0 1, ,;; x x y y # #=

from the rule of distance we ha%e that the distance is

1 0 1 0

0 1 0 1 0 1

2 2 2

0 0 0 1 1 1

2 2 2

0 0 0

2 2 2

.

, ,, , .

, ,

( ) ( ) ( )

( )

a

a

comp ; ; ; ; a

a b cx x y y # #

a b c

a x x b y y c # #

a b c

ax by c# ax by c#

a b c

ax by c# d

a b c

=

=

+ + =

+ ++ + + +

= + ++ +

=+ +

uuuur uuuur

46


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since (x5, y5,#5) lies in the plane and ax0by0c#-d, for

e%ery point (x,y,#) in the plane

$%ampleFind the distance between the parallel planes

1

2

: 2 3 6

: 4 6 2 8

; x y #

; x y #

+ = + =

/olution

First, obser%e that the planes are parallel, since their

normal %ectors 2, 3,1 , 4, 6, 2and are parallel.Further, since planes are parallel, the distance from the

plane ;5to e%ery point in the plane ;1is the same , so ,

pick any point in ;1, say )4,0,0( . The distance d from the

point )4,0,0( to the plane ;5is then "i%en by&

2 2 2

(2)(0) (3)(0) (1)(4) 6 2

142 3 1d

+ = =

+ +

Surface in spacen this section we take a first look at functions of two

%ariables and their "raphs, which are surfaces in three-

dimensional space.

2uadric surface

The "raph of a second-de"ree equation in three %ariables

x,y,and # is called quadric surface. The "raph of theequation &2 2 2 0ax by c# dxy ey# fx# "x hy 8# k + + + + + + + + + =

47


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n three-dimensional space ( where a,b,c,d,e,f,",h,8,and k

are all constants and at least one of a,b,c,d,e or f is

non#ero) is referred to as a quadric surface.

The most familiar quadric surface is the sphere&2222 )()()( rc#byax =++ , of radius r centred at the point),,( cba , to draw the sphere centred at )0,0,0( , first draw a

circle of radius r, centred at the ori"in in the planey# .

Then, to "i%e the surface its three-dimensional look, draw

circles of radius r centred at the ori"in, in both the planex#

and planexy , as in the fi"ure

*ote that due to the perspecti%e, these circles will look like

ellipses and will be only partially %isible ( the hidden parts

of the circles will be represented by dashed lines)

"enerali#ation of the sphere is the ellipsoid&2 2 2

2 2 2

( ) ( ) ( )1

x a y b # c

d e f

+ + =

notice that when def, the surface is a sphere.

48


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$%ample

Mraph the ellipsoid2 2 2

11 4 9

x y #

+ + =

/olution

first draw the traces in the three coordinate planes. n the

y#-plane, x ! , which "i%es us the ellipse&2 2

14 9

y #+ =

shown in the fi"ure below&

*ext, add to the last fi"ure the traces in the xy- and x#-

planes, these are&2 2 2 2

1, 11 4 1 9

x y x #and+ = + = , which are both ellipses

(see the fi"ure)

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To sketch the problem in the hand for instance we can sol%e

for # and plot the two functions&2 2

2 23 1 , 3 14 4

y y# x and # x= = , to obtain surface

S3etching a paraboloid

raw a "raph of the quadric surface

2 2x y #+ =To "et an idea of what the "raph looks like, first draw its

traces in the three coordinate planes. n the planey# , we

ha%e0=xand so

#y =2

( a parabola). n the planex#

, we ha%e0=y and so, #x =2 ( a parabola) . n the planexy , we ha%e0=# and so, 022 =+ yx ( a point- the ori"in). We sketch the

traces in the fi"ure below

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Finally since the trace in the xy-plane is 8ust a point, weconsider the traces in the planes )0( >= kk# . *otice that these

are the circles kyx =+ 22 , where for lar"er %alues of # ( i.e.,

lar"er %alues of k) , we "et circles of lar"er radius.. we

sketch the surface as in the fi"ure below

51


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/uch surface is called paraboloids and since the traces in

planes parallel to the xy-plane are circles, this is called a

circular paraboloid.Mraphin" utilities with three-dimensional capabilities

"enerally produce a "raph like the fi"ure below&

For 22 yx# += . *otice that the parabolic traces are %isible,

but not the circular across sections we draw in the second

fi"ure. The four peaks %isible in the last fi"ure are due tothe rectan"ular domain used to plot ( in this case

55 x and 55 y ).

This can be impro%ed by restrictin" #-%alues.

With 150 # , we can clearly see the circular cross section

in the plane 15=# in the followin" fi"ure

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We can plot the parametric surface to the same problem ,

we ha%e tx cos= , and ty sin= and 2s#= , with 50 s and20 t . The fi"ure below shows the circular cross

sections in the plane k#= for 0>k .

S3etching an elliptic cone

raw a "raph of quadric surface2

2 2

4

yx #+ =

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while this equation may look a lot like that of an ellipsoid,

there is a si"nificant difference ( look where the #1is), we

start by lookin" at the traces in the coordinate planes. For

theplaney#

, we ha%e0=xand so &2

2

4

x#= or 22 4#y = , so that #y 2= . That is, the trace is a

pair of lines & #y 2= and #y 2= ( see the fi"ure)

Hikewise, the trace in the planex# is a pair of lines& #x 2= .

The trace in the planexy is simply the ori"in(why)< Finally,

the traces in the planes )0( = kk# , parallel to the planexy ,

are the ellipses &2

2 2

4

yx k+ = . ddin" these to the drawin" "i%es us the

double-cone seen in the fi"ure below&

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/ince the traces in planes parallel to the xy-plane are

ellipses, we refer to this as an elliptic cone. 6ne way toplot this with a +/ is to "raph the two functions&

2 22 2

4 4

y y# x and # x= + = + .

in the followin" fi"ure

55


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We restrict the #-ran"e to 1010 # to show the elliptical

cross sections. *otice that this plot shows a "ap between

the two hal%es of the cone. lternati%ely, the parametric

plot shown in the followin" fi"ure

With 2 2cos , 2 sinx s t y s t= = and s# = , with 55 s

and 20 t , shows the full cone with its elliptical and

linear traces.

S3etching a hyperboloid of one sheet

raw a "raph of the quadric surface

2 22 14 2

x #y+ =

/olution

The traces in the coordinate plane are as follows&

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22

22

2 2

( 0) : 1 ( )2

( 0) : 1 ( )4

( 0) : 1 ( )4 2

#y# plane x y hyperbola

xxy plane # y ellipse

x #and x# plane y hyperbola

= =

= + =

= =

( see the fi"ure )

Further notice that the trace of the surface in each planek#= ( parallel to the planexy ) is also an ellipse&

2 22 1

4 2

x ky+ = +

Finally, obser%e that the lar"er k is, the lar"er the axes ofthe ellipses are. ddin" this information to the last fi"ure ,

we draw the surface seen in the followin" fi"ure

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We call this surface a hyperboloid of one sheet.

$sin" the +/ , we can "raph the two functions &2 2

2 22 1 2 14 4

x x# y and # y

= + = +

as in the fi"ure below&

(where we ha%e restricted the #-ran"e to1010 # , to show the elliptical cross sections.)

lternati%ely , the parametric plot seen in the fi"ure

58


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with 2 cos cosh , sin cosh , 2 sinhx s t y s t and # t= = = ,with 0 2 5 5s and t , shows the fullhyperboloid with its elliptical and hyperbolic trace

S3etching a hyperboloid of t#o sheets

raw a "raph of the quadric surface&2 2

2 14 2

x #y =

/olution

*otice that this is the same equation as in the last example,

except for the si"n of the termy . s we ha%e done before,

we first look at the traces in the three coordinate planes.The trace in the )0( = xplaney# is defined by&

22

12

#y = , since it is clearly impossible for two ne"ati%e

numbers to add up to somethin" positi%e, this is a

59


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contradiction and there is no trace in the planey# . That is ,

the surface does not intersect the planey# . The traces in the

other two coordinate planes are as follows&2

2

2 2

( 0) : 1 ( )4

( 0) : 1 ( )4 2

xxy plane # y hyperbola

x #x# plane y hyperbola

= =

= =

we show these traces in the followin" fi"ure '

Finally, notice that for kx= , we ha%e that2 2

2 12 4

# ky + =

so that the traces in the plane kx= are ellipses for 42 >k . t

is important to notice here that if 42


61/66

sheets. ;uttin" this all to"ether , we ha%e the surface seen

in the fi"ure below&

we call this surface a hyperboloid of two sheets.

We can plot this on +/ by "raphin" the two functions

2 22 2

2 1 2 14 4

x x# y and # y

= = , as in

the fi"ure '

61


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where we ha%e restricted the #-ran"e to 1010 # , to show

the elliptical cross sections. This plot shows lar"e "aps

between the two hal%es of the hyperboloid.

The parametric plot with

2 cosh , sinh cos 2 sinh sin ,x s y s t and # s t= = =for 4 4 0 2s and t , produces the left half of

the hyperboloid with its elliptical and hyperbolic traces.The ri"ht half of the hyperboloid has parametric equations

2 cosh , sinh cos , 2 sinh sinx s y t and # s t= = = ,

with 4 4 0 2s and t . s shown in thefi"ure

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S3etching a hyperbolic paraboloid

/ketch the "raph of the quadric surface defined by theequation

2 22# y x=

/olution

We consider first the traces in planes parallel to each of the

coordinate planes&

;arallel to kxyk#planexy == 222:)( ( hyperbola , for )0( k .

;arallel to 22

2:)( kx#kyplanex# +== (parabola openin"down).

and parallel to 222:)( ky#kxplaney# == ( parabola openin"

up).

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first we start by drawin" the traces in the x#-planes as seen

in the fi"ure

/ince the trace in the xy-plane is the de"enerate hyperbola

)2( 22 xy =

( two lines & 2x y=m

), we instead draw thetrace in se%eral of the planes k#= . For )0( >k ), these are

hyperbolas openin" toward the positi%e and ne"ati%e y-

direction and for )0(


65/66

We refer to this fi"ure as a hyperbolic paraboloid

wireframe "raph of 222 xy# = is shown in the fi"ure below

with )5555( yandx where we limited the #-ran"e to128 # .

*ote that only the parabolic cross sections are drawn, but

the "raph shows all the features of the fi"ure. ;lottin" this

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surface parametrically is fairly tedious because it requires

four different sets of equations) and does not impro%e the

"raph noticeably .

chapter 10- vector and geometry space

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