stpm mathematics t / a level - vectors · pdf file06.09.2012 · stpm mathematics t...
TRANSCRIPT
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.
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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
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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
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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.
M.K.Lim STPM Mathematics T / A Level
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
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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.
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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.
M.K.Lim STPM Mathematics T / A Level
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.
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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
M.K.Lim STPM Mathematics T / A Level
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
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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.
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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
M.K.Lim STPM Mathematics T / A Level
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
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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
M.K.Lim STPM Mathematics T / A Level
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
)
M.K.Lim STPM Mathematics T / A Level
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
M.K.Lim STPM Mathematics T / A Level
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
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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
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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
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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
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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
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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
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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
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Vector Equation of a Line
• The equation of a line can be expressed in two forms
• Vector form
• Cartesian form
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Vector Equation of a Line
• 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
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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
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Vector Equation of a Line
A
P(x , y , z)
b
O
r
a
x
y
r = a + λb
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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
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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
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Cartesian Equation of a Line
A
P(x , y , z)
x i + bj + ck
r
a
O x
y
z
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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
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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.
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Equations of a Plane
• Two types namely, Vector and Cartesian form
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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
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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.
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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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