vectors algebra

7/29/2019 Vectors Algebra

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Engineering Mathematics 1.2: Vectors Algebra 5

x1 , x2 , x3 axis system

The order and direction of axes are assumed to be 'right handed'. This definition comes from

imagining a right handed screw - if you turn the screw in the direction from axis Ox to Oy the

screw advances along Oz . (and Oy to Oz along Ox, from Oz to Ox along Oy). This is convention

and MUST be adhered to.

The distance from point O to point P, r, can be calculated using the Pythagoras theorem to give

( ) 2/1222 zyx ++=r

2.2 Direction Cosines

We will see later that it is useful to know the angle the line OP makes with each axis.

Fortunately this can be easily calculated. The shaded areas of the figures below each form

right-angled triangles with the axes.

r

Z-axis

Y-axis

X-axis

z

y

x

P(x,y,z)

o

r

-ax s

Y-axis

X-axis

z

y

x

P(x,y,z)

o


So, knowing the length and the appropriate co-ordinate, each angle can be calculated.

The angle POx is given by

r

xl == cos

The angle POy is given by

r

ym == cos

The angle POz is given by

r

zn == cos

r

Z-axis

Y-axis

X-axis

z

y

x

P(x,y,z)

o

r

Z-axis

Y-axis

X-axis

z

y

x

P(x,y,z)

o


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The cosines of these angles, ,, are (often written) nml ,, and these are known as the

direction cosines of the line OP.

It can easily be shown that

12

222

2

2

2

2

2

2222 =

++=++=++

r

zyx

r

z

r

y

r

xnml

2.2.1 Examples

1) If P has co-ordinates (5, 4, -9). What is the length OP and its direction cosines.

122

8116252

=

++=

OP

OP

Direction cosines:

122

9

122

4

122

5

=

=

=

n

m

l

2) The position of the top corner of the lecture room was required. To find this a theodolite

was set up as in the figure below. The distance to the side wall, along the x-axis is, 7m

and the following angles were measured: TOx = 70o and TOy = 50o .

Calculate the co-ordinates of the corner relative to the position of he theodolite.

x

yz

T

7m


Calculate the direction cosines:

6428.050coscos

5.060coscos

====

====

r

ym

r

xl

We need the third direction cosine.

Using 1222 =++ nml

o52.54

58035.0

1cos 22

=

=

===

mlr

zm

The length OT, (r) can be calculated as we know the x co-ordinate, i.e. from l

0.145.0

0.7

5.0

==

==

r

r

xl

From the other direction cosines

125.80.1458035.0

0.90.146428.0

==

==

z

y

Co-ordinates of the corner are: ( ) ( )125.8,0.9,0.7,, =zyx

2.3 A Geometric Representation

We represent vectors geometrically by line segments in space. The length of the line

representing the magnitude and the direction of the line the direction of the vector.

From this definition the starting point is irrelevant.

What is the difference between these two vectors?

a a

O

A A

O


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Their length is the same.

Their direction is the same.

(The arrows indicate the direction along the line)

These two vectors are equivalent (equal).

By the same argument, the two vectors are also equivalent to a vector of the same length and

direction which starts at the origin. We can therefore use a co-ordinate notation of three

numbers - as used previously for OP - to represent any vector (anywhere in three dimensional

space).

We have shown a vector described in two ways: in terms of co-ordinates (x, y, z) and also

represented geometrically by a line from an origin to this point. We say a dual definition can be

used either a geometrical or a co-ordinate(component) one.

2.4 Vector notation found in text books

A vector can be written down in many ways. Some of the more common (and acceptable) ways

you will come across are:

( )321 ,, aaaOAa a

2.5 Magnitude (or Modulus)

The magnitude, modulus or length of the vectora is written as a or OA and given in

component form by

( ) 2/1232221 aaa ++=a

Geometrically this is the length of the line OA

2.6 The Unit Vector

A vector whose modulus is 1 is called a unit vector, sometimes written as a and

a

aa =


The unit vectors i, j, k

The unit vectors in the co-ordinate directions are denoted i, j,andk.

i = (1,0,0), j = (01,0) and k= (0,0,1)

Any vectorr can be expressed in terms of its componentx, y, zwith respect to the unit vector

(i,j,k) by

zkyjxi ++=r

The notation (x,y,z) is interpreted as (xi, yj, zk).

2.6.1 Example:A vector ( ) kji 434,1,3 +==a

The modulus of ( )( ) 264,1,3 2/1222 === aa

So the unit vector in the direction ofa is

+== kji

26

4

26

1

26

3

a

aa

2.7 Equal Vectors

Two vectors a and b are equal if they have the same magnitude and direction i.e.

a = b

Using the component definition

( )321 ,, aaa=a ( )321 ,, bbb=b

then

332211 bababa ===


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2.8 Parallel Vectors

If is a scalar and a = b

then

if>0, the vectora is in the same direction as b and has magnitude b

if 0

Anti-parallel if < 0.


3 Vector Addition

3.1 The boat problem

A boat steams at 4 knots due East for one hour. The tide is running North-North-East at 3 knots.

Where will the boat be after one hour?

A diagram of the vectors involved looks like that below, where a represents the velocity of the

boat and b represents the velocity of the tide.

The net velocity of the boat is represented by the line OC which is the sum ofa and b.

3.2 Geometric addition laws

This leads to the parallelogram rule for vector addition which may be written:

The sum, or resultant, of two vectors a and b is found by forming a parallelogram with a and b

as two adjacent sides.

The sum a + b is the vector represented by the diagonal of the parallelogram.

In component form

( )332211 ,, bababa +++=+ ba

b

a

a+b

O

C

A

North

NN

East

b

B

B C

AO

b

a

a+b


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Because the vectora is the same as any other vector which is parallel and in the same

direction, a orb could be "moved" in the figure above to form a triangle. This leads to the

triangle law.

If two vectors a and b are represented in magnitude and direction by the two sides of a triangle

taken in order, then their sum is represented in magnitude and direction by the closing third

side.

The triangle rule can be made more general to apply to any geometrical shape - or polygon.

This then becomes the polygon law.

If from a point O, say, as in the figures below, lines are drawn to represent the vectors a, b, c, d,

and e. Then the closing side represents the vectorrand ris the sum of the vectors a, b, c, d,

and e.

edcbar ++++=

b

a

a+b

O A

B

CD

E

a

b

c

d

e

r

O

A

B

C

D

E

a

b

c

d

e

r


3.3 Vector Laws for addition and subtraction

Commutative law

a + b = b + a

Order is NOT important

In component form

(a1 + b1 , a2 + b2 , a3 + b3 ) = (b1 + a1 , b2 + a2 , b3 + a3 )

Associative law

(a + b) + c = a + (b + c)

This follows from the component definition and geometrically - from the triangle or polygon law.

In component form

b

a

a+b

b

a

a+b

a+(b+c)

(a+b)+c

c

b

a

b+c

a+b


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(a1 + b1 ) + c1 = a1 + (b1 + c1)

(a2 + b2 ) + c2 = a2 + (b2 + c2)

(a3 + b3 ) + c3 = a3 + (b3 + c3)

Distributive law

( ) baba +=+

This follows from the component definition or geometrically from similar triangles

In component form

( ) ( ) ( )( ) ( )332211332211 ,,,, babababababa +++=+++

Subtraction

a - b = a + (-b)

In component form:

( )332211 ,, bababa =ba

Geometrically:

AO

B

b

a

a+b

A O

B

b

a

( + ) = +a b a b


Note OBBA = OA

This is an important result as it shows that the vector represented by the line joining two points

is given by taking the vector giving the position of the first point from that for the second

baab =+=+== OAOBBA BOOA

3.4 Vector Addition Examples

(1)

( ) ( ) ( )( )( ) ( )( )2,1,0

3,0,5302,212,1222

3,3,1

1,2,13,2,10,1,2

=+

=+=

=+

===

cba

ba

ba

cba

Modulus ofc

( ) 61,2,1 2/1222 ==c

The unit vector in the direction ofc

==

6

1,

6

2,

6

1

c

cc

Not that the modulus of 1 == cc

O A

B

ba b-

a

-b


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2) From analysis of the forces in a new bridge it is calculated that the following forces are

imposed at one of the bridge supports (in Newtons).

F1 = (1,1,1)

F2 is 6N and acts in the direction (1,2,-2)

F3 is 10N and acts in the direction (1,2,-2)

In order to design the bridge support the resultant force is required. What is the force the

support must impose on the bridge to reduce the resultant force to zero.

The first force is given in the usual vector form. The second two are in valid forms but need to

be converted so that we can perform the usual vector addition.

First calculate the unit vectors in the direction ofF2 and F3

( )( )

( )( )0,4,3

5

1

0169

0,4,3

2,2,1

3

1

441

2,2,1

2

33

2

22

=++

==

=

++

==

F

FF

F

FF

So the in vector for the two forces are

( )

( )0,8,60,5

4,

5

310

4,4,23

2,

3

2,

3

16

2

2

=

=

=

=

F

F

The resultant force F is found by vector addition

( ) ( ) ( ) ( )3,3,90,8,64,4,21,1,1111 =++=++= FFFF

The force the support must impose on the bridge is equal and in the opposite direction as the

resultant i.e.

( )3,3,9forceReaction =


4 Vector Multiplication

We have seen how vectors can be added, another natural operation in maths is to multiply

quantities together. The idea of multiplication of vectors becomes very useful in engineering. In

particular we will show how to use them to calculate moments and work-done with great ease.

Two types of multiplication exist. They are called the vectorand scalarproducts.

4.1 The Scalar Product

(also known as the dot product or the inner product)

The component of the vectora in the direction of OP is easily calculated as equivalent to the

length ON by

cosa=ON

Considering the constant force F in the figure below, which acts through the point O. If this force

is moved along the line OA, along the vectora, then we can calculate the work done by the

force. (Remember: work done is the component of the force in the direction of movement

multiplied by distance moved by the point of application.)

The component ofF along OA is cosF . This force is moved the distance a , giving,

cosdonework aF=

a

O

A

P

N

O

F

Aa


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This is the scalar product of the two vectors F and a and can be seen as a geometric

definition.

An equivalent component definition can be written.

Scalar product definition:

The scalar product of two vectors ( )321 ,, aaa=a and ( )321 ,, bbb=b is written with a dot ( )

between the to vectors.

It is defined in component form as

332211 bababa ++=ba

And in geometrical form

cosabba =

Where is the angle between the two vectors and 0 .

The two definitions can be proved to be equivalent by the cosine rule for a triangle,

( )( ) cos2222 OBAOOBOAAB +=

Which can be expanded to show that

cos332211 ab=++ bababa

Three important points about a scalar product:

(1) The scalar product of two vectors gives a number (a scalar)

(2) The scalar product is a product of vectors

(it cannot be of two scalars nor a vector and a scalar)

aO A

b

B


(3) Use of the "dot" is essentialto indicate that the calculation is a scalar product.

4.2 Rules for the scalar (or dot) product.

Commutative Law

coscos abba

abba

=

=

Associative Law

You cannot have a scalar product of three vectors as 'dotting' the first two gives a scalar.

Distributive Law (for a scalar multiplier)

Brackets are multiplied out in the usual way.

( ) ( ) ( )bababa ==

Distributive Law (for vector addition)

( ) cabacba +=+

Powers of vectors

2

2

3

2

2

2

1

0cos

a

aa

aa

=

=

++= aaa

No other powers of vectors are possible other than 2.

Note that for unit vectors i, j, k

12

== iii 12

== jjj 12

== kkk

Perpendicular Vectors

Ifa and b are perpendicular then2

= and 0

2coscos ==

Hence 0=ba

BUT 0=ba does not imply that a and b are perpendicular, because a could be zero orb could

be zero.

Note: Suppose caba = [ ( ) 0= cba ]

This does not mean that b = c, since a could be zero ora perpendicular to b-c.


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i.e. vectors cannot be cancelled in the same way as scalars.

For unit vectors which are perpendicular

0=== ikkjji

Perpendicularity is important in engineering, for example pressure acts normal to a surface, so

force per unit area is np , wherep is the pressure and n the unit normal vector.

We often have to find a vector that is normal to another vector.

Component of a vector

The component of a vector, F say, in a given direction a is given by cosF .

And we can write

coscos FaFaF ==

to give the component ofF in the direction on a.

4.3 Scalar Product Examples

(1) Find the angle between the two vectors a = (2, 0, 3) and b = (3, 2, 4)

Using cosabba = and 332211 bababa ++=ba

181206 =++=ba

( ) ( )

( ) ( ) 2916494,2,3

139043,0,2

2/1

2/1

=++==

=++==

b

a

=

==

=

02.22

927.02913

18cos

cos291318

(2) Fora = (1, 2, -2), b = (0, 2, 2) c = (3, 2, 1) find

i) ca ii) ( ) cba + iii) ( )cba

i) ( ) 5243 =+=ca

ii) ( ) ( ) ( ) ( ) 110831,2,30,4,1 =++==+ cba

iii)( ) ( )( ) ( ) ( )

0,0,01,2,301,2,3440 ==+= cba

Note that a and b are perpendicular since neithera orb are zero.

Remember that ba is just a number (not a vector).


(3) Find the work done by the forceF = (3, 1, 5) in moving a particle from P to Q where position

vectors of P and Q are (1, 2, 3) and (2, 4, 4) respectively.

( )1,2,1=== OPOQPQr

Work done by the force F = rF

( ) ( ) 105231,2,15,1,3 =++= units of work.

(4) Find the component of the vectorF = (2, -2, 3) in

(i) the idirection

(ii) the direction (3, 1, -2)

(i) The i direction is the vector (1,0,0) so the component in the idirection is

( ) ( ) ( ) 20,0,13,2,20,0,1 ==F

(iii) Is the vector (3, 1, -2) a unit vector?

( ) 14419 2/1 =++ Therefore it is not a unit vector.

Unit vector =( )

14

2,1,3

Component ofFin direction (3, 1, -2) is

( )( ) ( ) 14

2

14/2,1,33,2,214

2,1,3

==

F


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4.4 The Vector (or Cross) Product

The main practical use of the vectorproduct is to calculate the moment of a force in three

dimensions. It is of very limited use in two dimensions.

For two vectors a and b, the vector product is defined as:

( ) nbaba sin=

where is the angle between the vectors a and b, ( 0 ) and n is the unit vector normal to

both a and b.

a , b and n are three vectors which form a right-handed set. (Remember how a right handed

axes system was defined earlier.)

It is very important to realise that the result of a vector product is itself a vector.

Note how from this definition order of multiplication matters:

The vector given by ( )ab is in the opposite direction to ( )ba .

(This follows from the right-hand screw rule), so

( ) ( )abba =

b

a

n

b

a

n


4.5 The Moment of force, F.

If the force F passes through the pointPand r=OP , then the moment of the force about O is

defined as

FrM =

M is a vector in the direction of the normal n .

So, as with all vectors, moments add by the parallelogram rule.

The use of calculating moments by the vector (cross) product comes into itself when used in

situations (particularly in three dimensions) which are very difficult to visualise.

4.6 Area calculation

The area of the parallelogram ABCD is given by

ABADABADABh === sinArea

n

r

OP

F

A B

D

C

h


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The area of the triangle PQR is given by

PRPQ

PRPQ

PR

=

=

=

2

1

sin21

h2

1Area

4.7 Laws for the Vector (or Cross) products

Anti-Commutative

( ) ( )abba =

This follows from the definition that includes the 'right-handed axes', n changes direction when

the multiplication is reversed.

Non-associative

(The triple vector product)

( ) ( ) cbacba

The vectorbcis perpendicular to both b and c and the plane containing b and c.

Similarly, ( ) cba is in the plane ofa and b, so ( )cba and ( ) cba are different vectors.

Brackets must always be used for more than two vectors in a vector product.

Distributive law with multiplication by a scalar

( ) ( ) baba =

P

R

h


Distributive law with addition

( ) ( ) ( )cabacba +=+

4.8 Parallel vectors.

From the definition of the vector product, i fa and b are parallel, then

0=

and

0=ba

So we can say that if 0=ba then

a = 0

or

b = 0

or

a and b are parallel.

If caba = it cannot be deduced that b = c. You must first show that a 0 and that a is not

parallel to b - c.

4.9 Cartesian (component) form.

For unit vectors i,j,k

( )0sin,0as0 ===== kkjjii

Because of perpendiculatiry

jikikjkji === ,,

jkiijkkij === ,,

Component form of the vector product:

( )

( ) kbjbibbbb

kajaiaaaa

321321

321321

,,

,,

++==

++==

b

a

( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( )ibajbaiba

kbajbakba

kkbajkbaikba

kjbajjbaijba

kibajibaiiba

kbjbibkajaia

bbbaaa

+++

++=

+++

+++

++=

++++=

=

231332

123121

332313

322212

312111

321321

321321 ,,,,ba

So, how do you remember this!!?


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There are several ways, the determinant form is shown here

21

21

31

31

32

32

321

321

bb

aak

bb

aaj

bb

aai

bbb

aaa

kji

+=

=ba

( ) ( ) ( )kbabajbabaibaba 122113312332 +=ba

4.10 Examples

(1) Given the three vectors a = (3, 1, 1), b = (2, -1, 1), c = (0, 1, 2)

( )5,1,2

12

13

12

13

11

11

112

113

=

+

=

=

kji

kji

ba

This vector is perpendicular to the plane containing a and b.

( )

( )2,4,3

10

12

20

52

21

51

210

512

=

+=

=

kji

kji

cba

This lies in the plane containing a and b.

( ) ( )( ) ( )( )2,4,3

1,1,33,3,6

13

=

=

= abacbbca

This is the same as the solution for ( ) cba above.


It can be shown that in general,

( ) ( ) ( )acbbcacba =

It can also be shown that

( ) ( ) ( )cbabcacba =

(2) Find the area of the triangle having vertices at

P= (3, 2, 2), Q= (1, -1, 2), R = (2, 1, 1)

Area PQR = PRPQ 2

1

( ) ( )1,1,10,3,2 == PRPQ

( )1,2,3

11

32

11

02

11

03

111

032

=

+

=

=

kji

kji

PRPQ

Area = ( )2

14149

2

1

2

1=++=PRPQ

P

R


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(3) A force F of 6 units acts through the point P=(2,3,1) in the direction of the vector (4,1,2). Find

the moment of the force about the point Q=(1,2,1).

The unit vector in the direction of the force is

=

++

++

21

2,

21

1,

21

4

4116

24 kji

So the force F, has components

=

21

2,

21

1,

21

46F

The position vector of P relative to Q is

( )0,1,1=QP

i.e. r

So the moment of the force about Q is

( )

=

=

+=

==

21

18,

21

12,

21

12

3,2,2

14

11

24

01

21

01

214

011

214

01121

6

M

M

kji

kji

kji

FQP

Q

P


4.11 Triple products of Vectors

Definitions of the two triple products are

( ) cba is the Triple Vector product.

( ) cba is the Triple Scalar product.

The triple scalar product has an interesting geometrical meaning:

We know that

( )

( )ba

nbaba

andbydefinedramparallelogtheofarea

sin

=

=

Thus

( ) ( )

( ) cosramparallelogtheofarea

ramparallelogtheofarea

cn

cncba

=

=

But h=cosc =height of the parallelepiped normal to the plane containing a and b . ( is the

angle between c and n ).

So

( ) cba = area of the parallelepiped defined by a, b and c.

It follows then that:

(a) If any two vectors are parallel ( ) 0= cba (zero volume)

(b) If the three vectors a co-planar then ( ) 0= cba (zero volume)

A

BC

O

c

b

a

h


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(c) If ( ) 0= cba then either

i) a = 0, or

ii) b = 0or

iii) c =0 or

iv) two of the vectors are parallel or

v) the three vectors are co-planar

(d) ( ) ( ) ( )acbcbacba == = the same volume.


5 Vector Equations of Lines and Planes

5.1 The Vector Equation of a Line

Consider the line which passes through the points A and B with position vectors a and b

respectively and a point P, which has position vectorr, and lies on this line as shown

rba === OPOBOA

From this diagram we can see that

AP

APOAOP

+=

+=

ar

Ift is some multiple of AB such that

ABt+= ar

as ba =+AB , then

( )abar += t

So the equation of the line AB is:

( ) bar tt += 1

for


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( )

11

1

111

ab

axt

tbatx

=

+=

Similarly fory andz

33

3

22

2

ab

azt

ab

ayt

=

=

Hence the Cartesian equation of a line passing through the points (a1,a2,a3) and (b1,b2,b3).is

tab

az

ab

ay

ab

ax=

=

=

33

3

22

2

11

1

Note: For any equation in Cartesian form we can readily express the equation in vector form.

e.g.

tzyx === 12

19

3

In vector form

( )( ) ( )1,1,32,9,0 +=

+=

t

t

r

abar

5.2 Equation of line examples:( ) ( )ababar +=+= ttt1

(1) Find the equation of the line r1 through the points (0,1,-2) and (3,4,3).

( )( ) ( )( )ttt

ttttt

tt

52,31,3

3,4,322,1,0

1

++=

++=

+= bar

i.e. tztytx 52313 +=+==

In Cartesian form

tzyx

tzyx

=+

=

=

=++

=

=

5

2

2

1

3

0

23

2

13

1

03

0

(2) Find the equation of the line r2 through the points (1,1,0) and (-3,-2,-7)

( )( ) ( )

( )sss

sssss

ss

7,31,41

7,2,30,1,1

1

=

+=

+= bar


In Cartesian form

szyx

=

=

=

73

1

4

1

(3) Do r1 and r2intersect?

For the lines to intersect then t and s should be such that:

ss

stst

752

3131413

=+

=+=

are all satisfied.

Solving the first two simultaneously s = 1 and t = -1.

Substituting these values into the third gives

-7 = -7

i.e. the three equations are satisfied, so the two lines intersect.

Substituting the values fors ort into the respective Cartesian equations gives the point of

intersection as (-3,-2,-7)

(4) Find the angle between the two lines which are in the same direction as the two vectors c =

(3,3,5) and d = (-4,-3,-7)

5635912

cos

==

=

dc

dcdc

( )

( ) 7449916

432599

2/1

2/1

=++=

=++=

d

c

==

=1.173

9927.07443

56

cos


18/19


5.3 The Vector Equation of a Plane

We can use the fact that a line joining any two points in the plane is perpendicular to the normal

to the plane.

i.e. n and AP are perpendicular.

The vectorn is normal to the plane, a is the position vector of A, r is the position vector of P (r =

(x, y, z)).

The vector

ar == OAOPAP

Now r - ais perpendicular to n if

( ) 0= nar

or

nanr =

or

=nr

The above two are the general form of the vector equation of a plane.

If we can take n = (, , ) the equation in Cartesian form is

pzyx =++

since r = (x, y, z)

n

O

AP


5.4 Examples involving both lines and planes

(1) Find the equation of the plane through the points

a = (3,2,0) b = (1,3,-1) c = (0,-2,3)

a - b lies in the plane

a-b

= (2, -1, 1)a - c lies in the plane

a - c = (3, 4, -3)

A normal, n, to the plane is

( ) ( ) ( ) ( )3,4,31,1,2 == caban

( ) ( ) ( )( )( )11,9,1

38,36,43

43

12

33

12

34

11

343112

=

+=

+

=

=

n

kji

kji

Equation of the plane is nanr =

( ) ( ) ( )

15

0183

11,9,10,2,311,9,1

=++=

=r

In component form this is

( ) ( )15119

1511,9,1,,

=++

=

zyx

zyx

Check

1533180:

1511271:150183:

=+

=+=++

c

ba

All are consistent.


19/19


(2) (a) Find the point where the plane ( ) 41,2,1 =r (or 42 =++ zyx ) meets the line

r = (2, 1, -1) + t(1, 1, 2) or ( tzyx

=+

=

=

2

1

1

1

1

2)

(b) Find the angle that the line makes with the plane.

[Remember the equations of a line

( )abar += t

tab

az

ab

ay

ab

ax=

=

=

33

3

22

2

11

1

]

The point of intersection must satisfy both the equation of the plane AND the equations of the

line.

i.e (using the vector form)

( )[ ] ( )( ) ( )

5/1

453

4221122

41,2,1

=

=+

=++++

=+

t

t

t

t aba

Substituting into the equation of the line gives the point of intersection

( )

( ) ( )

=

+=

+=

5

3,

5

6,

5

11

2,1,15

11,1,2

abar t

(b) Find the angle that the line makes with the plane.

The normal to the plane is (1, 2, 1) = n.

A vector in the direction of the line is (1, 1, 2) = b - a

( )

( ) 5221

cos

=++=

=

nab

nabnab

( )

( ) 6141

6411

2/1

2/1

=++=

=++=

n

ab

=

==

56.33

833.06

5cos


(3) Find the perpendicular distance from the point P (2,-3,4) to the planex+2y+2z=13

The equation of the plane in vector form is

( ) 132,2,1 =r

A vector normal to the plane is

( )2,2,1=n

Hence the equation of a line perpendicular to the plane and passing through P is

( )( ) ( )2,2,14,3,2 t

t

+=

+= abar

This line meets the plane when

( ) ( ) ( ) ( ) ( ) 132,2,12,2,12,2,14,3,22,2,1 =+= tr

i.e.

( ) ( )

1

1394

13441862

==+

=++++

t

t

t

Thus the line meets the plane at N with co-ordinates

( ) ( ) ( )6,1,32,2,114,3,2 =+

Hence the perpendicular distance PN

( ) ( ) ( )[ ] units3463123 2/1222 =+++

(3) Find the equation of the line of intersection of the two planes given by:

x + y + z = 5 and 4x + y + 2z = 15

In vector from these planes can be written:

( )( ) 152,1,4

51,1,1

=

=

r

r

The line of intersection must be perpendicular to both the vectors (1,1,1) and (4,1,2).

Hence a vector in the direction of the line must be in the direction

(1,1,1)(4,1,2) = (1,2,-3)

To complete the equation of the line we need to find any one point on the line. Choosing x=0

then, from the Cartesian equations of the two planes,

y + z = 5

y + 2z = 15

Hence forx=0, y=-5 andz = 10

And the point (0, -5, 10) is a point on the line.

The equation of the line can then be written r = (0, -5, 10) + t (1, 2, -3)

vectors algebra

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