STPM Mathematics T / A Level - Vectors · PDF file06.09.2012 · STPM Mathematics T / A Level Vectors M.K.Lim September 6, 2012 M.K.LimSTPM Mathematics T / A Level

STPM Mathematics T / A LevelVectors

M.K.Lim

September 6, 2012

M.K.Lim STPM Mathematics T / A Level

Representation of Vectors

B

A A vector is a quantity which has magnitude andspecific direction in space. A quantity with magnitude but nodirection is called a scalar.We write as ~AB to show the displacement from A to B.Displacement is move from A to B as ~AB can be called vector aNote the arrowhead points(direction) from A towards B.


Equal Vectors

• Two vectors with same magnitude and same direction areequal. a = b.

• It follows that they can be represented by any line of rightlength and direction.

• In this case, both vectors has same direction and length.

a

b


Negative vectors

• If two vectors a and b ,have the same magnitude but oppositedirections, we say a = −b

• An other useful notation is valid such as ~CD = − ~DC

a

b


Subtraction of Vectors

• If we wish to subtract two vectors a and b, we can expressedas

• a− b =a + (-b)

We say the subraction of vectors can be considered as the additionof a reversed vector b.It is easier to add a reversed vector form of b.


Modulus of a Vector

• The modulus of a vector is its magnitude.

• It is written as |a|. This is the length of the line represented.

• Given vector a = 3i + 4j + 5k ,

• Then modulus |a| is√

32 + 42 + 52


Scalar Multiplication of a Vector

• If λ is positive real number , then λ is a vector in the samedirection as a and of magnitude λa.

• It is natural that −λa is in opposite direction.


Example of Scalar Multiplication

A

B

D

C

• Vector ~CD is twice as long as vector ~AB.

• Represent it by ~CD = 2 ~AB

• We say λ is 2.

• Scalar means magnitude is involved, direction is not.


The Addition of Vectors - Triangle Law

This law is important for solving problems.

Consider 4 ABC.

CB

A

p

q

p + q

• Vector for p for side AB and Vector for q for side BC

• Resultant Vector is p + q represented by side AC

• Note that the arrow point towards C for resultant vectorp + q.


Addition Law Triangle Law Contd...

• Its the head-to-tail story...

• ~AB + ~BC = ~AC

• If side AB represents vector p

• Side BC represented by vector q

• Then side AC is the resultant, as p + q going from tail of p tohead of q.

• Note :The tail of vector q follows the head of vector p


Addition Law Using Parallelogram ABCD

B

D

C

A

a a

b

b

• Parallel sides AB and DC represented by vector a

• Similarly, parallel sides BC and AD represented by vector b

• In the triangle ABC, resultant ~AC = a + b

• In triangle ADC , ~AC = a+ b

• Therefore a + b = b + a

• Since AC is the common between 4 ABC and 4 ACD


Diagonals of a Parallelogram

P

RQ

O

b

a

a + ba− b

Given ~OP = a, ~OR = b, then ~OQ = a + bLooking at 4 PQR: ~RP = ~RQ + ~QP = a + (−b) = a− b fromsubtraction of Triangle Law.Also ~PR = - ~RP = - (a- b) = b - a .These are important to be remembered:

• (a− b) is the vector from endpoint of b to to theendpoint of a.

• (b− a) is the vector from endpoint of a to the endpointof b.


Diagonals in a Parallelogram

Consider 4 OPR, we have

• Solid black line is vector (a− b)

• Dashed black line is vector (b− a)

b

a

a− b


Vectors Illustrated in Cartesian coordinates

i

j

0 1 2 3 4

1

2

3

4

5

A

C

• Vector a is ~OA = 3i + 4j and vector b is ~OC = i + 2j• Resultant vector arrow :Aligning head of vector a with the tail

of vector b.• So it becomes a + b = a + b• ~OA + ~OC = ~CA


Area of a Parallelogram Using Vectors

i

j

1 2 3 4

1

2

3

4

5

6

B

CD

A

h

~AB

~AD

θ


Determinant Method to compute Cross product

• Use determinant method to solve 2D matrix

• Area by determinant method should yield the same answer of8 units squared (

2 02 4

)


Angle between two vectors

• Angle between two vectors is unique labelled as θ.

• Two vectors a and b are shown with angle in between.

• It is the angle between the directions when the both linesconverge or diverge from a point shown as a blue dot. It isonly angle θ and not any other.

a

θb


Unit Vector

• Given a is a vector

• The unit vector is written as a

• A unit vector is a vector whose length is 1, so magnitude of ais 1

• Definition: a =a

|a|• A unit vector is in the direction of v is vector over its

magnitude

• Applied to Cartesian coordinates, i is the unit vector in Oxdirection and j is the unit vector in Oy direction


Scalar Product (Dot)

• The scalar product of two vectors a and b is defined asab cos θ

• where θ is the angle between them

• a.b = ab cos θ

• Sometimes Scalar Product is also known as Dot Product


Vector Product (Cross)

• The vector product of two vectors a and b is defined asab sin θ where θ is the angle between them

• |a× b| = ab sin θ

• Sometimes the Vector Product is also known as the CrossProduct

• This product acts in a direction perpendicular to both a and b


Vector Product of two vectors a and b

• Two vectors are parallel, θ = 0◦, then |a× b| = 0

• Two vectors are perpendicular θ = 90◦, then |a× b| = ab


Parallel Vectors

Properties of Scalar Product

a

b

π

• Two vectors a and b are parallel, then ab = ab cosπ

• Then a.b = - a.b since cos 180◦= -1

• For like parallel vectors a.b = ab

• For unlike parallel vectors a.b = - ab

• when a = b, then a.b = a.a = a2

• In Cartesian unit vectors i,j,k we have i.i = j.j = k.k = 1


Perpendicular Vectors

• When two vectors a and b are perpendicular, then the dotproduct of them is a.b = 0

• Because cos 90◦ = 0

• For unit vectors, we have i.j = j.k = k.i = 0

a

b


Cartesian Unit Vectors

• Now if i is the unit vector in direction of Ox

• Now if j is the unit vector in direction of Oy

• Now if k is the unit vector in direction of Oz

x

y

z


Vector Equation of a Line

• The equation of a line can be expressed in two forms

• Vector form

• Cartesian form



• We want to find the vector equation of the blue line shown asAP.

• This line is parallel to a direction vector b which shows thedirection

• Recall the straight line equation y = mx + c

• Similarly, we can use vectors to find the equation of a line

• Consider a line parallel to vector b which passes through afixed point A with position vector a

• Vector b is the direction vector for the line

• We shall see the development of r = a + λb


Vector Equation of a Line Contd...

• If r is the position vector ~OP then ~AP = λb

• where λ is a scalar parameter. Relationships of length isshown by λ

• Now ~OP = ~OA + ~AP

• Therefore we have r = a + λb since r is ~OP

• This equation gives the position of one point on the line

• That is P is on the line ⇔ r = a + λb

r = a + λb



A

P(x , y , z)

b

O

r

a

x

y

r = a + λb


Example of Vector equation of a Line

A line whose vector equation is r = (5i − 2j + 4k) +λ(2i − j + 3k) is parallel to vector 2i − j + 3k and is passesthrough the point whose position vector is 5i − 2j + 4k .r is the position vector of any point P.

x i + y j + zk = 5i − 2j + 4k + λ(2i − j + 3k)

= (5 + 2λ)i + (−2− λ)j + (4 + 3λ)k

Equating coefficients from above, we have ∴ λ =

x = 5 + 2λ

y = −2− λz = 4 + 3λ


Example of Vector equation of a Line Contd ...

∴ λ =x − 5

2

∴ λ =y + 2

−1

∴ λ =z − 4

3

So, the Cartesian form of a vector equation of a line is

x − 5

2=

y + 2

−1=

z − 4

3


Cartesian Equation of a Line

A

P(x , y , z)

x i + bj + ck

r

a

O x

y

z


General Vector Equation of a Line

If a line passes through A(x1, y1, z1) and is parallel to ai + bj + ckits equation may be written as

r = (x1 i + y1 j + z1k) + λ(ai + bj + ck)

x = x1 + λa

y = y1 + λb

z = z1 + λc


V.Equation Cartesian form

In Cartesian format, it is shown as

λ =x − x1

a=

y − y1b

=z − z1

c

λ =

x = x1 + λa

y = y1 + λb

z = z1 + λc

Note that the point A (x1, y1, z1) is one of the infinite set of pointson the line. Hence the equations representing a given line is notunique.


Equations of a Plane

• Two types namely, Vector and Cartesian form


Vector Equation of a Plane

Plane ( green ) is defined as distance d from origin O and isperpendicular to unit vector n shown.

N

P

O

d

n

r

x


Standard form of Vector equation of a Plane

If line ON is perpendicular to the plane then, for any point P onthe plane , NP is perpendicular to ON.If r is position vector of P,then ~ON = d n.Since P is on the plane,itmeans that ~NP. ~ON = 0The equation is called the scalar product form of the vectorequation of a plane.If r is a position vector,then ~NP = r − d n.Therefore it becomes (r - dn).dn = 0This implies that r.n− dn.n = 0But n.n = 1,So that means

r.n = d

The equation is the standard form of a vector of a plane.


Cartesian form of Plane

Using r.n = d idea and n = lx + my + nzNow, if P is a point (x , y , z) on this plane, its position vectorr = xi + yj + zk , satisfies the equation r.n = d.So that (x i + y j + zk).(lx + my + nz) = d

lx + my + nz = d

Example:Find the Cartesian equation of this plane r.(2i+3j-4k) = 1Solution:Comparing with r.n = d(2i + 3j − 4k) means n= lx + my + nz = 1.here d = 1. so l = 2,m = 3, n = −4 therefore2x + 3y − 4z = 1


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STPM Mathematics T / A Level - Vectors · PDF file06.09.2012 · STPM Mathematics T / A Level Vectors M.K.Lim September 6, 2012 M.K.LimSTPM Mathematics T / A Level