chapter5 vectors

8/8/2019 Chapter5 Vectors

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5.1 Introduction to Vector

A) Concept of vectors

Vectors Representation

Vectors may be represented by usingdirected line segments orarrows.

The magnitude of a vector is represented by its length and its direction is given by the

direction of the arrow.

The tail of the arrow is called the initial point of the vector and the head of the arrow the

terminal point.

A

This vector from O to A is written

OA or OA ora or a or a

.

a

O

Equal Vectors

a Two vectors are equal if they have the same magnitude and

b the same direction .

Negative Vectors

aIfa is any nonzero vector, then a , the negative ofa is

defined to be the vector having the same magnitude as a- a but oppositely directed.


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Position vectors

A vector that starts at the origin is called a position vector.So, the position vector of B is the vector that starts from the

origin and ends at B; b = OB

. Similarly, position vector of

P is p = OP

Zero vector

The zero vector , denoted by 0, has magnitude zero. Contrary to all the other vectors, it has no

specific direction.

Scalar Multiplication Of Vectors

Ifa is a vector and is a nonzero real number ( scalar ),a then theproducta is a vector whose length is times

2a 2

1a the length ofa and the direction is the same as a if 0

and opposite to that ofa if 0.

Parallel vectors

If the vectors u1 and u2 are parallel, then they are scalar multiple of each other.

Thus, u1 = u2 ( )For example ;

a = i +j is parallel to b = 3i + 3j because a =1

3b

y

xp

b

P B

O


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a

Perpendicular vector

If vector a and b are perpendicular , hence the angle between a and b is 90

.

Unit Vectors

Any vectors of magnitude 1 unit is a unit vector.

a The unit vector of a vector a is a vector whose magnitude

is 1 unit in the direction ofa.

a = 1 The unit vector ofa ,

a =a

a.

B) Addition and subtraction of vectors using the triangle law

Addition Of Vectors

Ifa and b are any two vectors, then the suma+b is the vector determined as follows :

Position the vector b so that its initial point coincides with the terminal point ofa. The vector

a+b is represented by the arrow from the initial point ofa to the terminal point ofb.

b b

a + b a

a a + b a b + a

b

From the right figure, it is evident that a + b = b + a, and the sum coincides with the diagonal of

the parallelogram when the two vectors are positioned so they have the same initial point.

b


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Subtraction Of Vectors

Ifa and b are any two vectors, then the difference ofb from a is defined by

ab = a + (b ).

To obtain the difference a

b without constructing

b , position a and b so their initial pointcoincide. The vector from the terminal point of b to the terminal point of a is then the vector

ab.

a

ab a ab

b b b

Example 1

1- The points A,B and C have position vectors a,b and c. Point P is the mid point of AB andpoint Q is the midpoint of BC. Find

a. the position vectors of P and Qb. PQ

2- The point O is the centre of the regular hexagon ABCDEF. Given that OA = a andAB = b, find

a) OB

b) BD

c) CF


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5.2 Vectors In Two and Three Dimensions

Vectors In Two Dimensions

For the moment, we shall restrict the discussion to the vectors in 2-dimensional space (theplane). Consider the Cartesian1xy plane consisting of an origin O and a pair of perpendicular

vectors.

y

The two are the unit vector i in the positive direction of the

x-axis and the unit vectorj in the positive direction of the

y-axis.

j

x

O i

y If A is the point with coordinate (3 , 4) ,A(3 , 4)

3i + 4j OB = 3 , so OB

= 3i

BA = 4 , so BA

= 4j

4j

Therefore ,

OA = OB

+ BA

= 3i + 4j

xO 3i B ie. the position vector of A(3 , 4) is r = 3i + 4j

Similarly, the position vector of any point P(a , b) is ai + bj.

One way of writing this position vector isa

=b

OP


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Addition and Scalar Multiplication of Vectors in Two Dimensions

Vector Arithmetic

i) Magnitude : The magnitude or length of v = ai + bj is v = 22 ba .

1- Find the magnitude of v = 3i + 4j .

ii) Additionand : If v 1 = a 1i + b 1j and v 2= a 2i + b 2j , then

Subtractionv 1+ v 2=(a 1 + a 2)i + (b 1 + b 2)jv 1v 2=(a 1 a 2)i + (b 1b 2)j

2- If u = 2i5j and v = 4i + 3j , find u + v and uv3- Find AB where A is the point (3 ,1) and B is the point (2 , 3).

Note that for any two points A(x 1 ,y 1) and B(x 2 ,y 2) ,

AB =

OB

OA

= (x 2x 1)i + (y 2y 1)j

iii) Scalar multiplication : Ifkis a scalar and v = ai + bj is a vector , then

kv = (ka)i + (kb)j

4- If u = 2i5j and v = 4i + 3j , find 3u +2v.


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Vectors In 3-Dimensions

The Cartesian coordinate for space are often called rectangular coordinate .

z This consist of a fixed point O, the origin, and threemutually perpendicular axes, Ox , Oy and Oz. The axes

are placed in such a way that they form a right-handedset as shown in figure 2.1.

y Each pair of coordinate axes determines a plane calledO a coordinate plane. These are referred to as thexy-

plane, thexz-plane and theyz-plane.x

Figure 2.1

zP(1 , 2 , 3) Any point P in space can be specified by an ordered

triple of numbers (a , b , c) where a , b and c are thesteps in the direction ofx,y andz axes

3 respectively, to P.

1 yIn figure 2.2, we have constructed the point P(1 , 2 , 3).

2

x

Figure 2.2

We now take unit vectors i,j and k in the direction ofx,

z y andz axes respectively.

P(1 , 2 , 3)If P(a , b , c) is any point in the space, then the position

3k vector of P is

y

OP = ai + bj + ck

i O

In figure 2.3, the position vector of the point (1 , 2 , 3)2j is i + 2j + 3k. Conversely, the point whose position

x vector is 2i4j + k has coordinates (2 ,4 , 1).

Figure 2.3


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Addition and Scalar Multiplication of Vectors in Three Dimensions

Vector Arithmetic

Vectors in space apply the same rules of addition , subtraction, scalar multiplication and also the

magnitude just as they are in the plane.

For any vectors v 1= a 1i + b 1j + c 1k and v 2=a 2i + b 2j + c 2k , and for any scalar k,

i) v 1 =2

1

2

1

2

1cba

ii) v 1+ v 2 = (a 1 + a 2)i + (b 1 + b 2)j + (c 1 + c 2)k

v 1v 2 = (a 1 a 2)i + (b 1 b 2)j + (c 1 c 2)k

iii) kv 1=ka 1i + kb 1j + kc 1k

Example 2

1- Find 2ab where a = i +j + k and b =i + 3j2k.

2- Find a unit vector u in the direction of the vector from A(1 , 0 , 1) to B(3 , 2 , 0). Hence,find a vector 6 units long in that direction.


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a a a

b

5.3 The Scalar Product

A) The Scalar Product (Dot Product)

The scalar product of two vectors is a = a1i + a2j + a3k and b = b1i + b2j + b3k is theoperation which is written a.b and defined

a . b = a bcos , is angle between a and b .

Definition

The scalar product between a and b is also defined as :-

a. b =1 1 2 2a b a b or a. b = 332211 bababa

a . b > 0 a . b = 0 a . b < 0

Example 3

1- Evaluate a . b ifa = 3i7k , b =2j + 3k and the angle between a and b is 750 .2- Evaluate a) (2ij) . (3i + 4k)

b) (3j2k) . (i + 2j7k )

3- Simplify a) (ab) . (a + b)b) (a + b) .c(a + c) .b

4- Given that a = 3i + tj2k , b = (1t)i3j + 4k , find tif a is perpendicular to b.5- Given that a = 2i + 3j5k and b = 4ij3k.

Find a) a . 2a b) (2a3b) . (a + 2b)

bb


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B) Algebraic properties of the scalar product

1. a.a = a2

2. a .b = b. a3. a . (b + c) = a.b + a . c

4. (ma) .b = m (a.b) = a.(m b)

5. 0.a = 0

6. a.b = a b if and only ifa parallel to b

a.b = a b if and only ifa and b in opposite direction.

7. a .b = 0 if and only ifa is perpendicular to b with a 0, b 08. For unit vectors i, j and k we have :-

i.i = j.j = k. k = 1 and

i . j = j . k = k.j = 0

C) The Angle Between Two Vectors

If a = a1i +a2j + a3k and b = b1i + b2j + b3k are two vectors and is the angle between them.From the definition ofa.b ,

a .b = a bcos

cos =ba

b.a

The angle = cos1

ba

b.a

Example 4

1- If a = 4 , b= 3 and a.b = 7 , find the angle between a and b .2- Find the angle between 3ij and4i + 6j .3- The angle between vectors a = i+j + 2k and b = 2i + 3j + k is cos-1

84

1. Find .

4- Find the interior angles of the triangle ABC whose vertices are A(1,3,5), B(-2,0,3) and(3,1,-2).


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D) Direction Cosine Of A Vector

Consider the vector

OP where P is the point (a , b , c).

z Then

OP = ai + bj + ck and

OP = 222 cba P(a , b, c)

If

OP makes angles of , and with thex ,y andz-axis c respectively, then cos , cos and cos are known as its

direction cosines andO y

cos =

OP

a, cos =

OP

b, cos =

OP

c

x where cos2 + cos 2 + cos 2 = 1.

Notice that the unit vector

OP =

OP

OP =

OP

ai +

OP

bj +

OP

ck

= cos i + cos j + cos k

Example 5

1- Find the direction cosine of the vector

OP where P is the point (3 , -6 , 2)

2- Find the direction cosines and direction angles ofa) a = 2i + 3j - k

b) b = 4i2j + 3k

3- A vector v is inclined at 60 to they-axis and at 60 to thez-axis. Find its inclination tothe x- axis. If the magnitude ofv is 8 unit, find the vector v .


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5.4 The Vector Product

A) The Vector Product

If is the angle between vector a and b, then

a b = absin

u

where

u is a unit vector in the direction of a b

or

u =ba

ba

To determine the direction of a b , use the right hand, where the fingers turn from a to b andthe thumb points in the direction of a b .

a b

b

a

Note k

ij = k , j i =kjk = i , kj =iki = j , i k =j

j i

The vector product of a = a 1i + a 2j + a 3k and b = b 1i + b 2j + b 3k is defined in terms of

the expansion of the symbolic determinant ;

a b =1 2 3

1 2 3

a a a

b b b

i j k

= (a 2 b 3 a 3 b 2)i(a 1 b 3 a 3 b 1)j + (a 1 b 2 a 2 b 1)k


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Example 6

1- Given a = 2i +3j2k and b = 4i2j + 3k , find a b.2- Given a = i + 2j + 3k and b =i + 3jk.

a) Find a bb) Prove that a b is a vector which is perpendicular to the vector a.

B) Properties of cross product

If a and b is a vector , m is a scalar , then

1. a b =ba

2. (ma) x b = m(a b) = a (m b)

3. a (b + c) = (a b) + (a c)

4. ( a + b) x c = a x c + b x c

5. a.(b x c ) = ( a x b ) . c

6. a x ( b x c ) = ( a.c)b(a.b)c

7. a b = 0 if a is parallel tob

Example 7

1- Find all vectors of length 11 unit which are perpendicular to both a = i + 2jk andb = i3k

2- Find a unit vector of the normal to the plane containing points A( 1 , -1 , 2 ) ,B( 3 , 1 , 0 ) and C( -1 , 2 , -3 )


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C) Area of Parallelogram

Example 8

1- A plane contains points A(1,1,1), B(3,2,-1) and C(1,-4,2) and D. If A, B,C and Dforms a parallelogram, find

a) the coordinates of D

b) the area of the parallelogram ABCD

2- A plane contains points A(1,1,0) , B(3,-2,1) and C(5,7,2). Finda) a vector normal to the planeb) the area of triangle ABC

Area of parallelogram = a b

Area of triangle =1

2 a b


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5.5 Application Of Vectors In Geometry

A) Lines In Space

A line in space is a straight line which continues indefinitely in both directions and contains acontinuous infinite set of points.

Suppose that R(x , y, z) is a point which is free to move on a line containing a fixed point

A(x1, y1, z1) . If v= ai + bj + ck is a direction vector of the line, it is clear that a line consists

precisely of those points for which the vector AR is parallel to v, that is

AR = tv for some scalar t

OR

OA

= tvOR =

OA + tv

or r = a +tv (1)

In terms of components, it can be written as

xi +yj +zk = (x1i +y1j +z1k) + t(ai + bj + ck)

So that x = x1 + tay = y1 + tb (2)

z = z1+ tc

Isolating tin each of these equations gives

a

xx1

=

b

yy1

=

c

zz1

(3)

So there are three ways of expressing the line in space :

(1) is the vector equation of the line,

(2) are the parametric equations of the line,(3) are the Cartesian equations of the line.

R

A

v

y

x

z

0

ra


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Summary :

If a straight line passes through A(x1, y1, z1) and parallel to ai + bj + ck ,

Its vector equation is r = (x1i +y1j +z1k) + t(ai + bj + ck)

Its parametric equations are x = x1 + ta y = y1 + tb

z = z1 + tc

t is called theparameterand can take any real value.

Its Cartesian equations are 111c

z-z

b

y-y

a

x-x .

Example 9

1- Find a vector equation of the line that contains points A(1,2,3) and B(-2,1,3)2- Find the vector equation , the parametric equations and the cartesian equations of the line

through (1,2, 3) in the direction 4i + 5j 6k

3- A line has Cartesian equations3

1x=

4

2y=

5

3z. Find a vector equation for a

parallel line passing through the point with position vector 5 i 2j 4k and find thecoordinates of the point on this line wherey = 0.

4- Finda. parametric equations for the line l passing through the points P1 (2 , 4 ,1) and

P2 (5 , 0 , 7).

b. Where does the line intersect thexy-plane?


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B) The Angle Between Two Straight Lines

Suppose two straight lines vector equation are

r1 = a1 + tv1

r2 = a2 + sv2

With t, s are any scalar and is angle between two straight lines.

r1 =a1 + tv1

v2

v1r2 = a2 + sv2

If is angle between two straight lines, its also between v1 and v2 . because of the lines are

parallel.

v1 . v2 = cosvv 21

21

21

vv

.vvcos

Hence, the angle between two straight lines given by

= 1cos

21

21

vv

vv

Two straight lines are perpendicular ifv1 . v2 = 0 and the straight lines are parallel if

v1 = kv2 for kscalar.


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Example 10

1- The vector two straight lines equation are given byp = ( 2i + 2j - 4k ) + t( i + 3j - 3k )

and q = ( i +j + k) + s( i + 2j - 4k )

with tand s scalar. Find the acute angle between that two straight lines.

2- Line 1 has equationsx =1 + 2s ,y = 12s andz = 1 + 4s.Line 2 has equationsx = 1t,y = t andz = 32t.

Line 3 has equations3

41

2

1

zy

x.

a) Show that line 1 and line 2 are parallel.

b) Show that line 2 and line 3 intersect and find the angle between them.

3- Find the coordinates of the point of intersection of the two lines r = (2i + j - k) + (i +3j -4 k) and r = (-i2k)+ t(-2i + 2j - 5 k)


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C) Plane

Planes equation in 3 dimension can be measure with this criteria :

1. A plane in space has normal vector n = ai + bj + ck and that it passes through the fixed

point A (x1, y1,z1) or

2. A plane passes through 3 fixed point or3. A plane has two vectors.

Suppose a plane in space has normal vector n = ai + bj + ck and that it passes through the fixedpoint A (x1, y1,z1).

R(x , y , z) moves anypoint on the plane.

Now

AR is perpendicular to n.

n = ai + bj + ck

R(x, y, z)

A (x1, y1,z1)

AR .n = 0

( OR OA ) .n = 0

(ra) .n = 0

r. na . n = 0

r. n = a. n

r. n = p where p = a. n

Example 11

1- Find the equation of the plane with normal vectori

+ 2j

+ 3k

and containing the point(-1 , 2 , 4).

2- Find the equation of the plane through A(-1 , 2 , 0) , B(3 , 1 , 1) and C(1 , 0 ,3).

3- Find the vector equation of the plane that contains (2,-1,3) and is

a) parallel to thexy plane b) parallel to the plane with equation 3xy +z = 2


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4- Show that the line L whose vector equation is r = 2i - 2j + 3k + t ( ij + 4k ) is parallelto the plane P whose vector equation r . ( i + 5j + k ) = 5

5- Find the Cartesian equation for each of the following planes ;a) The plane

1 through A(1,2,3) and perpendicular to the line

1

2 3 1

x y z

b) The plane2

through B(1,-1,3) and perpendicular to both the planesxy + 2z = 3

and 2x +yz = 3


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D) The angle between two planes

It is important to choose the correct angle here. It is defined as the angle between 2 lines, one in

each plane, so that they are at right angles to the line of intersection of the 2 planes (like theangle between the tops of the pages of an open book).

The picture below shows part of 2 planes and the angle between them.

To find this angle, will we first have to find the equation of the line of intersection of the 2

planes, and then find 2 vectors which are in the planes and perpendicular to this?

Fortunately no! We just need to know a normal vector to each of the planes. Then we can findthe angle we want very neatly as shown in the drawing below.

The angle between the planes is the same as the angle between their 2 normal vectors (sliding

their tails together if necessary).

Now we just use n.m = |n||m| cos Aand find the angle in the same way as we did for the 2 lines.


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Example 12

1- Find the angle between the two planes

3x - 2y + 5z = 1 and 4x + 2y -z = 4

2- Calculate the angle between the (1,1,1) and (2,0,0) planes.

3- From the equations to the two given planes, 3x+4y = 5 and, x+2y+3z = 6, the normal to

Plane 1 is parallel to the vector , 3i + 4j and the normal to Plane 2 is parallel to the vector ,

i + 2j + 3k. The angle between the two normals is therefore, the angle between the twovectors3i + 4j and i + 2j + 3k

4- Find the acute angle between the two planes 5 0x y z and 2 2 3x y z .


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E) The angle between a line and a plane

The sharp angle between a line and a plane is determined by the angle between the direction

vector of the line and the normal vector of the plane..

We slide the normal vector n until its tail is at the point of intersection with the line L with the

plane P. Then n and L together define a plane which is perpendicular to plane P. The anglewhich line L makes with plane P is defined to be the red angle A in this plane.

Since A and B together make a right angle, we can find A by using the dot product ofn and thedirection vector b of line L to first find cos B.Or we can find A even more directly by using the trigonometry identity

cos B = cos (90 - A) = sin A

so n.b = |n||b| cos B so giving B

OR

n.b = |n||b| sin A so giving A

Example 13

1- From the equation to the given plane, r.( 3i + 4k) = 5, the normal to the plane is parallelto the vector , 3i + 4k. The line is parallel to the vector , 2ij . Find the angle between

the line and the plane.

2- Find the angle between the straight line r =6 4

5 0

3 1

t

and the plane 3 2x y z

3- Find the vector equation of line passing through the point ( 3 , 1 , 2 ) and perpendicular tothe plane r . ( 2ij + k ) = 4 .Find also the point of intersection of this line and the

plane.

4- Find the point where the straight line r = ( -4i +j + 9k ) + ( -2i + 4k ) intersects theplane r.( 2i + 2jk) = 5

chapter5 vectors

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