Vector Part 1

8/16/2019 Vector Part 1

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Vector Algebra: Part 1 Page 1

Chapter 3: Vector Algebra

3.1 Introduction to Vector Analysis

In the world of engineering, physical quantities can be divided mainly into scalar and vector.These quantities can be represented by numbers alone (i.e., magnitude only), with the

appropriate units, and they are called scalars. Another physical quantities with magnitude

and direction are called vectors. Scalars and vectors are the underlying elements in vector

analysis.

3.1.1 Scalar vs Vector

Scalar Vector

Example Mass; length;

temperature; voltage

Displacement;

velocity; force;

accelerationUnit of quantities kg; m; degree; Volt m; ms-1; N; ms-1

Direction No Yes

Symbol/Notation a ; b ; A ; B ; PQ a ; b ; OA ; OB ; PQ

3.2 Basic Concept

3.2.1 Properties of vector

(a) Equality of vector

1. Vectors are used to represent quantities that have both a magnitude and a direction.

2. They don’t impart any information about where the quantity is applied.

Let’s consider force for an example. A force of 5N (i.e., magnitude) that is applied toward

West (i.e. a particular direction) is a vector – (Fact 1); and it can be applied at any point in

space as shown in sketch below –(Fact 2).

Think: Are a , b , c , d the same vector?


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In a graphical sense, vectors can be represented by line segments. Length of the segment is

the magnitude and the direction of vector is pointed by the line segment. – (Fact 1); Vector

don’t impart any information about where the quantity is applied, thus we can said they are

all the same vector (i.e., = = =a b c d ) – (Fact 2)

Noted that vector a , b , c , d can be represented by its initial and terminal point: =a OA ,

=b PQ , =c ST , =d UV ; Since = = =a b c d , we get = = =OA PQ ST UV .

From this outcome we can said vector is not determined by its position – (Fact 2) i.e., not

determined by its initial point and terminal point. Vectors with two different initial point and

terminal point can be equal.

(b) Negative of vectors

A negative vector is a vector that has the opposite direction to the reference positive direction.

Negative of vector (i.e.,-OA ) has negative sign on its vector. This means the vector has

direction opposite to its reference positive vector OA and has same magnitude with vector

OA .

TS , AO and -OA are same vectors because they are equal. TS and AO are negative

vectors of reference positive vector OA because they have opposite direction to the

reference positive direction of vector OA (i.e., = = -TS AO OA as shown in diagram). They

have same magnitude (i.e., = = -TS AO OA )

Exercise 3.2.1 (b)(i) Find the vectors which have the same

magnitude.

(ii) Find the vectors which have the same

magnitude and same direction.

(iii) Find the vectors which have the same

magnitude and opposite direction.

(iv) Find the vectors which are not equal.


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3.2.2 Definition of Vector

(a) The Zero Vector, 0

A zero vector is not the same as zero.

Zero, 0 Zero Vector, 0 It is a number that has magnitude of zero. It

can be used to represent the origin of space.

- It is a vector that has magnitude of zero

and no direction.

- The magnitude of zero vector, 0 is zero,

0

(b) Position vector

Let O be the origin and A be any point. From the diagram, it shows point ( , , ) A x y z in 3D

Cartesian coordinate system.

- Position vector A , known as OA or a .

- It is vector with initial point at originO and terminal point A .

- = = ( + + )a OA xi yj zk or , , x y z for 3D system where i , j and

k are unit vector in , , x y z respectively.

Exercise 3.2.2 (a)

Draw a position vector = (5 + 0 + 7 )OB i j k , = -2, -5, 0c and = 0,1, 0d in a 2D and 3D

Cartesian coordinate system.

Position vector vs Coordinate:

Coordinate Position Vector Vector/ Arbitrary vector

Show location of a

point ( , , ) x y z in

coordinate system

Represents the position of any

point in space in relation to a

reference origin O

- Represents the position of

any point in space.

- Not necessary in relation to a

reference origin O

Example: Point(3,4,8) A

Example: = (3 + 4 + 8 )OA i j k or

3,4,8

Example: = (3 + 4 + 8 ) BC i j k

or 3,4,8


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Be careful to distinguish vector notation , , x y z , from the notation we use to represent

coordinates of points ( ), , x y z . Don’t mix the notations up!

(c) Parallel vector vs Normal/Perpendicular/Orthogonal vector

Parallel vector Normal/Perpendicular/Orthogonal vector

Two nonzero vectors AB and CD are said

parallel if AB are in same direction

( = 0°)θ or in exactly opposite direction

( = 180°)θ with CD .

Two nonzero vectors AB and CD are said

perpendicular if AB are 90°normal to CD .

AB CD denote vector AB parallel to CD AB CD denote vector AB normal to CD

Note: We will discuss this further in scalar multiplication and dot product sections.

(d) Unit Vector, û

Unit vector, û is any vector, u with magnitude of 1, i.e., ˆ =1u

Unit vector, ˆ =u

u

u

…………………………………………………………………………………………………………...(1)

(e) Standard Basis Vectors

Unit vector can be used as standard basis vector in a particular direction. What are the

standard basis vector in x , y and z - axis for a 3D space?

Let vectors i , j and k to be standard basis vectors for x , y and z - axis respectively.

In 2D space, 2

there are two standard basis vectors:

= 1,0i = 0,1 j

In 3D space,3 there are three standard basis vectors:

= 1,0,0i = 0,1, 0 j = 0,0,1k

Note that standard basis vectors are also unit vectors where2 2 2

1,0,0ˆ = = 1, 0, 0

1 + 0 + 0i


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Precaution: In actual, vectors can exist in n -dimensional space instead of 2D or 3D space. The

general notation for a n -dimensional space:

1 2 3= , , , , nv a a a a where 1 2 3, , , , na a a a are the components of the vector.

This study will focus on 2D and 3D cases as it is easier to visualize things in these dimensions

and also more practical.

3.2.3 Vector Algebra/ Vector Arithmetic

(a) Vector addition

Let a and b be two vectors. Then the vector addition is the sum of a and b , denoted by

+a b . The vector addition can be done by using Graphical Method according to the following

laws:

(i) Triangle Law (ii) Parallelogram Law

Definition If a and b are represented by

two sides of a triangle, then the

vector addition, +a b is at the

3rd sided of the triangle.

If a and b are represented by

two sides of a parallelogram, then

the vector addition, +a b is at

the diagonal of the parallelogram.

Geometric

interpretation

Calculation( + ) = = +a b OC OA AC

Where = AC OB

( + ) = = +a b OC OA OB

Remarks Used head-to-tail method

Head-to-tail method : The sum or resultant of vector addition is a vector drawn from the tail

of the first vector to the head of the last vector. It does not matter in which order you add

them.


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From the figure below, the n vectors 1 2 3( + + + + )na a a a together with vector -v form a

closed polygon. Hence, for sums of more than three vectors,

1 2 3

- -

= ( + + + + )n

Head to tail

v a a a a

Exercise 3.2.3 (a) :

Given points (3,4,8) A and (-2,-3,-5) B . Find the vector addition between position vector A and B .

(b) Vector subtraction

Vector subtraction is very similar to vector addition. Let a and b be two vectors. Then the

vector subtraction is denoted by -a b .

(a) Head-to-tail method

(For vector subtraction)

(b) Parallelogram Law

Definition The sum or resultant of

vector addition is a vector

drawn from the tail of the

first vector to the head of the

last vector.

If a and -b are represented

by two sides of a

parallelogram, then the

vector addition, +( - )a b is at

the diagonal of the

parallelogram.

Geometric

interpretation

Calculation

- - -

+ =

Resultant Head to tail vector

b BA a

- = = -a b BA OA OB

( + (- )) = + (- )a b a b

- = -a b OA OB

Remarks Remember that to construct a vector from starting point B to

ending point A , we have to subtract the position vector of

starting point OB from the ending point OA .


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Exercise 3.2.3 (b) :

(i) Compute the vector from point (2,-7,0) A to (1,-3,-5) B

(ii) By using the figure below, find + - + -c f g h d . Express the vector e as the sum of two

and three different vectors separately.

(c) Scalar Multiplication

Let α is a scalar and 1 2 3= , ,v v v v is a nonzero vector, then any number α the scalar

multiplication is:

1 2 3 1 2 3= , , = , ,αv α v v v αv αv αv

It shows that scalar multiplication is the operation where we multiply all the components of

v by the constant α .

Exercise 3.2.3 (c) : Graphical interpretation and the effect of scalar multiplication

(i) For the position vector = 2, 4a , compute 3a ,1

2a and -2a . Sketch all four vectors on

the same axis system. Discuss the effect of scalar multiplication on the magnitude and

direction of the original vector.

An application of scalar multiplication to identify parallel vector.

Let a and b are parallel vectors. If they are parallel then there must be a scalar number α sothat =a αb

Previous example shows that we can create multiple parallel vectors by using the scalar

multiplication:1

= 2, 4 3 = 6,12 = 1, 2 -2 = -4, -82

a a a a

Exercise 3.2.3 (c)

(ii) Determine if the sets of vectors are parallel or not.

(a) = 2, 4,-1a , = -6,12, -3b ; (b) = 4,10a , = 2,-9b


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Standard Basis Vectors Revisited (Additional knowledge)

In the previous section we introduced the idea of standard basis vectors without really

discussing why they were important. With the knowledge of scalar multiplication. We can

now do that.

Question: Why an arbitrary vector in a 3D space is defined as

1 2 3 1 2 3= , , = + +a a a a a i a j a k ?

Using scalar multiplication, we can get

1 2 3 1 2 3= , , = 1, 0, 0 + 0,1, 0 + 0, 0,1

i j k

a a a a a a a

This shows that the vector a is the sum of scalar multiplication between the its component

1a with unit vector i , its component

2a with unit vector j and its component

3a with unit

vector k .

In other words, magnitude in the x-direction (i.e., scalar1a ) multiply the unit vector along the

x-axis (i.e., unit vector i ) will retrieve the x-component of vector a .

Remarks: Scalar multiplication can be used to find unit vector of an arbitrary vector.

Exercise 3.23 (c)

(iii) Find unit vector that has the same direction as = -5, 2,1u ,

(d) Basic Properties of Vector Addition and Scalar Multiplication

Let a , b and c be three vectors and let α and β be a scalar. Then,

(i) + = +a b b a (commutative law)

(ii) ( + ) + = + ( + )a b c a b c (associative law)

(iii) + =a 0 a

(iv) ( ) = ( )α βa αβ a

(v) ( + ) = +α β a αa βa

(vi) ( + ) = +b c β βb βc


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Application in Geometry – Ratio Formula

Let A and B be two points and C is the point on the line segment AB which divides AB

into two segments which are in the ratio :m n . Then1

= ( + )+

c na mbm n

.

Proof: = = ?c OC

= - = - AB OB OA b a

= = ( - )+ +

m m AC AB b a

m n m n

= - AC OC OA >>

= + = + ( - )+

m

OC OA AC a b am n

+ = + ( ) - ( )

+ + +

m n m ma b a

m n m n m n

= + ( )+ +

n ma b

m n m n

1 = ( + )

+ na mb

m n -------------------------------------------------------------------(1)

Remarks: If point C be the mid point of the line segment AB , : = 1:1m n . Then we get

Midpoint formula,1

= ( + )2

OC a b

Exercise 3.2.3 (d)

(i) Let (1,3,5) A and (4,6,2) B . Find the point C so that it is located on the line segment

AB which divides AB into two segments which are in the ratio 1: 3 .


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3.3 Cartesian domain and Polar domain of 2D and 3D vectors

3.3.1 Cartesian coordinates of a vector in 2D space (Plane) and its polar expression

(a) Definition of a 2D vector

Let O be the origin and let x

O and yO be two mutually perpendicular coordinate axes.

Then, the plane containing x

O and yO is called the xy-plane or the xy-coordinate system) and

xO is called the x axis and yO is y axis.

The vector i is the vector from the origin O to the point

(1,0) .

The vector j is the vector from the origin O to the point

(0,1) .

Note: i and j are unit vectors and also position vectors.

Any vector v in xy-plane can be represented by = +v ai bj or = ,v a b where a and b are

scalars. The scalars a and b are called the components of the vector v with respect to that

coordinate system.

The vector ai and vector bj are called the vector components in the direction of i and j ,

respectively.Notation:

(i) The vector = +v ai bj can be denoted by = ,v a b

(ii) The point P at ( , )a b can be denoted by ( , )a b , ( , ) P a b or = ( , ) P a b

(iii) Note that ( , ) ,a b a b to avoid confusion. ( , )a b represent coordinates of a

point. ,a b represent components of a vector.

(b) Vector Algebra of a 2D vector

Let 1 1 1= +v a i b j and 2 2 2= +v a i b j be two vectors. Then(i)

1 2= >>v v Then,

1 2=a a ;

1 2=b b

(ii) 1 2

+ >>v v Then, 1 2 1 2( + ) + ( + )a a i b b j

(iii) 1 2

- >>v v Then, 1 2 1 2( - ) + ( - )a a i b b j

(iv) Let α is a scalar, then 1 1 1= ( ) + ( )αv αa i αb j

Proof Exercise


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(c) Theorem of an arbitrary vector in 2D space

Let P and Q be the points 1 1( , )a b and 2 2( , )a b respectively. Then, the vector PQ is given

by 2 1 2 1= - = ( - ), ( - ) PQ OQ OP a a b b

Proof:

Let1 1

= ,OP a b ,2 2

= ,OQ a b

2 1 2 1= - = ( - ), ( - ) PQ OQ OP a a b b

(Follow head-to-tail method in section 3.2.3 (a) & (b))

(d) Magnitude & Angle of a vector in 2D space

Let= +

v ai bj be a 2D vector.

(i) The magnitude of v is defined as 2 2= +v a b

(ii) The angle between v and a line parallel to the x-axis is defined as -1= tanb

θ a

Hint: Identify the quadrant; θ is positive if it is measured in the direction of anti-clockwise;

θ is negative if it is measured in the direction of clockwise.

(e) Transformation of Cartesian form of a 2D vector to polar form

By using magnitude and angle of a vector, the Cartesian form of a vector (i.e., = +v ai bj ) can

be transformed into polar form (i.e., = (cos( ) + sin( ) )angle angle

magnitude

v v θ i θ j )

Thus, we have = + = (cos( ) +sin( ) )

Cartesian domain Polar domain

v ai bj v θ i θ j

Exercise 3.3.1

(i) Let u , v and w be position vectors of the points (2,3)U , (1,5)V and (3,-4)W ,

respectively. Find

(a) = - 2 +3 z u v w

(b) the magnitude of z

(c) the angle between z and x

O

(d) transform the vector z from Cartesian domain into Polar domain

(e) compare the result in (a) and (d), explain your finding and relate this in the application

of engineering.


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(ii) Determine the unit vector in the direction of = 2 - 3u i j

(iii) Find the unit vector from the point (1,4) P to the point (3,-5)Q

(iv) Find a vector of magnitude 3 in the direction of = - + 3v i j

3.3.2 Cartesian coordinates of a vector in 3D space (Volume) and its polar expression

(a) Definition of a 3D vector

Let Obe the origin and let x

O , yO and z O be three mutually perpendicular coordinate axes.

Then, the plane containing x

O , yO and z O is called the xyz-plane or the xyz-coordinate

system (Follow right hand rule) and x

O is called the x axis, yO is y axis and z O is z axis.

The vector i is the vector from the origin O to thepoint (1,0,0) .

The vector j is the vector from the origin O to the

point (0,1,0) .

The vector k is the vector from the origin O to the

point (0,0,1) .

Note: i , j and k are unit vectors and also position

vectors.

Any vector v in xyz-plane can be represented by = + +v ai bj ck or = , ,v a b c where a , b

and c are scalars. The scalars a , b and c are called the components of the vector v with

respect to that coordinate system.

The vector ai , vector bj and vector ck are called the vector components in the direction of

i , j and k respectively.

Notation :

(i) The vector = + +v ai bj ck can be denoted by = , ,v a b c

(ii) The point P at ( , , )a b c can be denoted by ( , , )a b c , ( , , ) P a b c or = ( , , ) P a b c

(iii) Note that ( , , ) , ,a b c a b c to avoid confusion. ( , , )a b c represent coordinates of

a point. , ,a b c represent components of a vector.


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(b) Vector Algebra of a 3D vector

Let1 1 1 1

= + +v a i b j c k and2 2 2 2

= + +v a i b j c k be two vectors. Then

(i) 1 2

= >>v v Then,1 2

=a a ; 1 2=b b ; 1 2=c c

(ii) 1 2

+ >>v v Then, 1 2 1 2 1 2( + ) + ( + ) + ( + )a a i b b j c c k

(iii) 1 2

- >>v v Then, 1 2 1 2 1 2( - ) + ( - ) + ( - )a a i b b j c c k

(iv) Let α is a scalar, then 1 1 1 1= ( ) + ( ) + ( )αv αa i αb j αc k

Proof Exercise

(c) Theorem of an arbitrary vector in 3D space

Let P and Q be the points 1 1 1( , , )a b c and 2 2 2( , , )a b c respectively. Then, the vector PQ is

given by 2 1 2 1 2 1= - = ( - ), ( - ), ( - ) PQ OQ OP a a b b c c

Proof:

Let1 1 1= , ,OP a b c , 2 2 2= , ,OQ a b c

2 1 2 1 2 1= - = ( - ), ( - ), ( - ) PQ OQ OP a a b b c c

(Follow head-to-tail method in section 3.2.3 (a)

& (b))

(d) Magnitude & Angle of a vector in 3D space

Let = + +v ai bj ck be a 3D vector and let α , , β and γ be the direction angles of

= + +v ai bj ck

The magnitude and angle that define vector v can be obtained as following:

(i) The magnitude of v is defined as2 2 2

= + +v a b c


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(ii) The angle between v and a line parallel to the x-axis is defined as-1

= cosa

αv

;

The angle between v and a line parallel to the y-axis is defined as-1

= cosb

β v

;

The angle between v and a line parallel to the z-axis is defined as-1

= cosc

γv

.

(e) Transformation of Cartesian form of a 3D vector to polar form

By using magnitude and angle of a vector, the Cartesian form of a vector = + +v ai bj ck can

be transformed into polar form = (cos + cos + cos ).v v αi βj γk

Thus, we have = + + = (cos + cos + cos )Cartesian Domain Polar Domain

v ai bj ck v αi βj γk

(f) Important remarks for polar form of a 3D vector

(i) The unit vector v̂ is =v

v

(cos + cos + cos )αi βj γk or cos , cos , cosα β γ

(ii) Magnitude of a unit vector, v̂ is 1. Thus, we get 2 2 2cos + cos + cos = 1α β γ

(iii) The direction angles of negative vector, - v are -π α , -π β , -π γ

(iv) Have a clear definition for the following term:

Direction angles Direction cosines Direction ratio

α , , β and γ are called

the direction angles of v

cosα , cos , β and cosγ are

called the direction cosines ofv

The ratios : :a b c is

called the direction ratio

of v

For polar coordinate

= (cos + cos + cos )v v αi βj γk

For Cartesian coordinate

= + +v ai bj ck

Additional remarks:(i) If 2 2 2cos + cos + cos 1α β γ , then there does exist a unit vector with the direction

cosines cos +cos +cosα β γ .

Note: because unit vector has magnitude of 1.

(ii) Two vectors u and v are have the same direction cosines if and only if they have

the same direction.

Note: Different direction cosines shows different directions.

(iii) Two vectors u and v are have the same direction ratios if and only if they are

parallel (i.e. u and v are in the same direction or in opposite directions).Note: As explained by the scalar multiplication and parallel vector.


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Exercise 3.3.2

Let u , v and w be position vectors of the points (2,3,1)U , (0,-5,1)V and (-3,0,0)W ,

respectively. Find

(i) = - 2 +3 z u v w

(ii) transform z from Cartesian domain (i.e., + +ai bj ck ) to Polar domain (i.e.,

(cos + cos + cos )r αi βj γk where r is its magnitude.

(iii) the angle between z and x

O

(iv) direction cosines of z in three direction i , j and k .

(v) unit vector of z

(vi) If given direction angle as following, can you identify whether the vector with the

following direction cosine (cos + cos + cos )αi βj γk is exist or not?

------ vector m has direction angle α , , β and γ of ( / 4,2 / 3, / 3)π π π .

------ vector n has direction angle α , , β and γ of ( / 2, / 3, / 3)π π π .

(vii) Find the direction cosines of negative vector - z . Then find the relationship between the

direction cosines of vector z and - z .

3.4 Applications in geometry: Equations of lines in 2D and 3D spaces

Problems in space (in 2D and 3D space)

1. Find the equation of a line passing through a given point and parallel to a given vector

(all in 2D & 3D space).

2. Determine whether two lines intersect in three dimension space and find the point

of intersection if they intersect.

3.4.1 Definition (Equation of a line in 2D or 3D space)

Let L be a straight line passing through the point A and is parallel to a given vector v (i.e.,

( AR v ). Suppose that ( , ) R x y or ( , , ) R x y z is any point on L . Find the vector equation,parametric equation and Cartesian equation of line L .


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(a) Vector equation of line

(i) Let ( =OA a and ( =OR r be the position vectors of A and R respectively.

(ii) Since ( AR v , then ( = AR tv , where t .

Note: as t changes, we have all the points on the line L .

(iii) Now following head-to-tail method , +=OR OA AR

Then, we get = +r a tv -----------------------------------------------------------------(1)

This is called the vector equation of the line L .

The vector v is called a direction vector of the line L .

Note: The Eqn. (1) can be applied for problem in 2D or 3D space. Since the computation and

derivation for 2D and 3D space are similar. We will give the example in 3D one for the

demonstration.

(b) Parametric equation of a line in 2D and 3D space

Now, let position vector of an arbitrary point on line L , + +=r xi yj zk , position vector of a

point passing through line L ,1 2 3

+= +a a i a j a k and direction vector parallel to line L

1 2 3+= +v v i v j v k

From Eqn. (1) we have

= +r a tv >>1 2 3 1 2 3

, , , , ,= + , x y z a a a v v vt

>>

1 1

2 2

3 3 Point at time t Initial Point Direction Vector

x a v

y a t v

z a v

(Matrix form of vector equation of line)

By comparing the components of i , j , and k , we have

= , ,r x y z 1 1

2 2

3 3

= +

= + },

= +

x a v

y a v

t

t

z a vt

t ……………………………………………………………………………………………………………(2)

This system of equations (2) is called the parametric equations of the line L .

The variable/scalar t is called the parameter of the system of equations.

Note: The Eqn. (2) can be applied for problem in 2D or 3D space in the same manner.


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(c) Cartesian equation of a line in 2D and 3D space

By equal the parameter t in the Eq. (2), we have

31 2

1 2 3

-- -= = =

z a x a y at

v v v……………………………………….………………………………………………..………..(3)

The Eqn. (3) is called the Cartesian equation or symmetric form of the line L .

Precaution Zero denominator leads to zero numerator as shown in Eqn. (3).

Proof: From Eqn. (2), if1

= 0v , then1

- = 0 x a . This also apply to2v and

3v .

3.4.2 Intersection between a line to a plane or between two lines

(a) Intersection of a line to a planeIf we know the information of a plane and the line equation are given either in format of

Vector Eqn./Parametric Eqn./Cartesian Eqn., for example:

Information of a plane

(i) The line intersect at a specific plane, i.e., yz-plane at coordinate ( , , ) = (0, , ) x y z y z

Lets say Parametric Eqn. have been derived from a given information of point A

passing through the line, and the vector direction of the line L , v are given as

well. Then, we can use Parametric Eqn.

1 1

2 2

3 3

= +

= + },= +

x a v

y a v

t

t z a vt

t to solve for t , y and z .

Then the point of intersection at yz-plane, (0, , ) y z can be obtained.

(ii) The plane of eqn. of the intersection plane is given, i.e., + + =ax by cz k

The step is similar to procedure above. Given the a , b , c and k , substitute the

Parametric Eqn. into the eqn. of intersection to solve for t . Then, you can get the

point of intersection ( , , ) x y z by substitute t into Parametric Eqn.


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Precaution: There are three possibilities of the intersection: (i) line intersects the

plane in a point; (ii) line is parallel to the plane (no point of intersection); (iii) line

is in the plane

Note: You will know how to derive the equation of plane after you learn about product of vectors.

(b) Intersection between two lines

Let us have line1

L :1 1 1 1 1 2 3 1 2 3

= , , = , , + , ,r x y z a a a t v v v and line2

L :

2 2 2 2 1 2 3 1 2 3= , , = , , + , ,r x y z b b b s u u u where t and s are the parameters, 1 2 3= , ,a a a a ,

1 2 3= , ,b b b b are the position vectors specified at line

1 L and

2 L respectively.

1 2 3= , ,v v v v ,

1 2 3= , ,u u u u are the vectors parallel to line

1 L and

2 L .

If the line 1 L and line 2 L intersect each other, then:

1 2=r r

1 2 1 1 1 1= >> + = + x x a tv b su ----------------------------------------------------------------------------------(a)

1 2 2 2 2 2= >> + = + y y a tv b su ----------------------------------------------------------------------------------(b)

1 2 3 3 3 3= >> + = + z z a tv b su -----------------------------------------------------------------------------------(c)

This means that all the three Eqns. (a), (b) and (c) must be satisfied if the two lines1

L and

1 L are intersecting with each other. In the other words, if the parameter t obtained from

Eqn. (a) and parameter s obtained from Eqn. (b) will not satisfy Eqn.(c) if there is no point of

intersection.

It intersection exist, the point of intersection is as following:

1 2 1 2 1 2( = , = , = ) x x y y z z or 1 1 2 2 3 3(( + ), ( + ), ( + ))a tv a tv a tv or 1 1 2 2 3 3( + , + , + )b su b su b su .

Exercise 3.4.1 & 3.4.2

(i) Find the (a) vector equation, (b) parametric equation and (c) Cartesian equation

for the line L passing through the point (3,1,-2) P and (-2,7,-4)Q .

(ii) Where does the line L

intersect the xy -plane?

(iii) Where does the line intersect the plane: 2 +1 + =-4 4 x y z if exist? The parametric

eqns. of lines are provided as following:

(a) Line T : = x t , = 2 + 3 y t ; = z t where t

(b) Line V : =1+ x t , = 4 + 2 y t ; = z t where t

(c) Line W : = x t , = 4 + 2 y t ; = z t where t

(d) Sketch the results obtained in (iii) (a), (b) & (c) and categories them into 3

categories: unique solution; infinite solution & no solution.


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(iv) Do the following two lines intersect? If so, find the point of intersection.

1 L :

1 1 1 1= , , = 1, -6, 2 + 1, 2,1r x y z t &

2 L :

2 2 2 2= , , = 0, 4,1 + 2,1, -3r x y z s

(v) Given line1

L passes through the points (5,1,7) and (6,0,8) and the line2

L

passes through the points (3,1,3) and (-1, 3, )α . Find the value of α so that thelines

1 L and

2 L intersect and then find the point of intersection.

3.4.3 Linear Combination and Linear Dependence

(i) Linear combination

A linear combination of two or more vectors is the vector obtained by adding two or more

vectors (with different directions) which are multiplied by scalar values.

1 1 2 2 3 3= + + +... +

n nv α a α a α a α a

(ii) Linear dependent

Vector are linearly dependent if there is a linear combination of them that equals the zerovector, without the coefficients of the linear combination being zero.

1 1 2 2 3 3+ + + ... + =

n nα a α a α a α a 0 , where scalar

1 2 3, , ,...,

nα α α α

Note: The vectors are linearly dependent if the determinant of the matrix is zero, meaning

that the rank of the matrix is less than its full rank.

(Hint: In a matrix system, zero determinant helps to indicate an infinite or no solution system)

1 2 3+ + + ... + = 0

na a a a

1 2 3+ + + ... + 0

na a a a

(iii) Linear independent

Vector are linearly independent if none of them can be expressed as a combination of others

1 1 2 2 3 3+ + + ... + =

n nα a α a α a α a 0 , where scalar

1 2 3, , ,...,

nα α α α

Note: The vectors are linearly dependent if the determinant of the matrix is non-zero,

meaning that the rank of the matrix is equal to its full rank.

(Hint: In a matrix system, non-zero determinant helps to indicate an unique solution system)

……… v

1 1α a

n nα a

2 2α a

3 3α a


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Exercise 3.4.3

(v) Determine whether the vectors are linearly dependent or independent:

1 = 2,3,1a ,

2 = 1, 0,1a ,

3 = 0,3,-1a

(vi) Express the vector = 1, 2,3v as a linear combination of 1 = 2,3,1a , 2 = 1, 0,1a and 3 = 0,3,-1a

(vii) Determine whether the vectors are linearly dependent or independent:

1 = 1, 0,1a ,

2 = 1,1, 0a ,

3 = 0,1,1a

(viii) Express the vector = 1, 2,3v as a linear combination of1

= 1, 0,1a ,2

= 1,1, 0a

and3

= 0,1,1a

Vector Part 1

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Transcript of Vector Part 1