Download Vector Algebra

Lecture programme
1 Contents of lectures
1. General introductions. Scalars and vectors. Cartesian components. Direction cosines.
Engineering Maths 1.2
Geometric representation. Modulus of a vector. Unit vectors. Parallel vectors.
Vector Algebra
2. Addition of vectors: parallelogram rule; triangle law; polygon law; vector laws for addition.
Dr P. A. Sleigh
3. Multiplication of vectors: the scalar product; work done by a force; angle between vectors;
Email: [email protected]
components of forces in given directions.
4. Vector products: moments of forces; areas of triangles; parallel vectors.
Table of contents
1
2
Contents of lectures ................................................................................................................2
Scalars and Vectors ................................................................................................................3
2.1 Cartesian Co-ordinates .......................................................................................................4
2.2 Direction Cosines ................................................................................................................5
2.2.1 Examples......................................................................................................................7
2.3 A Geometric Representation ...............................................................................................8
2.4 Vector notation found in text books.....................................................................................9
2.5 Magnitude (or Modulus) ......................................................................................................9
2.6 The Unit Vector ...................................................................................................................9
2.6.1 Example: ....................................................................................................................10
2.7 Equal Vectors ....................................................................................................................10
2.8 Parallel Vectors .................................................................................................................11
3 Vector Addition......................................................................................................................12
3.1 The boat problem ..............................................................................................................12
3.2 Geometric addition laws ....................................................................................................12
3.3 Vector Laws for addition and subtraction ..........................................................................14
3.4 Vector Addition Examples .................................................................................................16
4 Vector Multiplication ..............................................................................................................18
4.1 The Scalar Product............................................................................................................18
4.2 Rules for the scalar (or dot) product..................................................................................20
4.3 Scalar Product Examples ..................................................................................................21
4.4 The Vector (or Cross) Product ..........................................................................................23
4.5 The Moment of force, F. ....................................................................................................24
4.6 Area calculation .................................................................................................................24
4.7 Laws for the Vector (or Cross) products ...........................................................................25
4.8 Parallel vectors. .................................................................................................................26
4.9 Cartesian (component) form. ............................................................................................26
4.10 Examples .......................................................................................................................27
4.11 Triple products of Vectors..............................................................................................30
5 Vector Equations of Lines and Planes..................................................................................32
5.1 The Vector Equation of a Line...........................................................................................32
5.2 Equation of line examples: ................................................................................................33
5.3 The Vector Equation of a Plane ........................................................................................35
5.4 Examples involving both lines and planes ........................................................................36
Engineering Mathematics 1.2: Vectors Algebra
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5. Triple vector products and triple scalar products: volume of parallelepiped; co-planar vectors.
6. Vector treatment of lines: vector and Cartesian forms; intersection of lines.
7. Vector treatment of planes: equation of a plane from 3 co-ordinate points; vector and
Cartesian forms.
8. Further examples: intersection of a line with a plane; perpendicular distance from a point to a
plane.
Engineering Mathematics 1.2: Vectors Algebra
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a) A temperature of 100o C
2 Scalars and Vectors
b) An acceleration of 9.8142 m/s towards earth
In engineering much of the work in both analysis and design involves forces. You will be familiar
c) The weight of a 2 kg mass
with forces in structural members in space frames, and know that a force acts in a given
d) The sum of £500.00
direction with a given magnitude. A force is a three dimensional quality and one of the most
e) A North easterly wind of 20 knots
common examples of a vector. Two other common vector quantities are acceleration and
Answers: a) Scalar b) Vector c) Vector d) Scalar e) Vector
velocity.
Note: To qualify as a vector the quantity must also satisfy some other basic rules of combination
In this section of the module we will introduce some formal mathematical notation and rules for
(addition, multiplication etc.) which we will see later in this course. For example, angular
the manipulation (multiplication, addition etc.) of these vector quantities. In the early stages it
will be easy to recognise the geometric meaning of vector addition and multiplication, but as the
problems become more complex (notably, when they are three-dimensional) the formal rules
displacement is a quantity that has both direction and magnitude but it does NOT obey the
addition rule - so it is NOT a vector
become more important.
2.1
Cartesian Co-ordinates
The general definitions of scalars and vectors are given below:
The theory of vectors is associated very closely with co-ordinate geometry so we shall start by
introducing the co-ordinate system.
A Scalar quantity is one that is defined by a single number with appropriate units.
We use a rectangular Cartesian co-ordinate system with axes O x y z or O x1 x2 x3 , as shown below.
Some examples are length, area, volume, mass and time.
The position of any point, P say, is given by co-ordinates, or components (x,y,z) or (x1,x2,x3).
x -axis
3
x
A Vector quantity is defined completely when we know both its magnitude (with units) and its
3
direction of application.
Some examples are force, velocity and acceleration.
r
P(x ,x ,x )
1
o
2
3
x
2
x -axis
2
Two very simple (and common) examples demonstrating the differences between scalars and
vectors are speed - a scalar, and velocity - a vector.
x
1
x -axis
A speed of 10 km/h is a scalar quantity
1
A velocity of 10 km/h at 20 degrees is a vector quantity.
x, y, z, axis system
Examples:
What are these quantities, vectors or scalars?
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Z-axis
Z-axis
z
z
P(x,y,z)
r
P(x,y,z)
y
o
r
Y-axis
y
β
o
Y-axis
x
x
X-axis
X-axis
x1 , x2 , x3 axis system
The order and direction of axes are assumed to be 'right handed'. This definition comes from
Z-axis
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
z
and MUST be adhered to.
P(x,y,z)
The distance from point O to point P, r, can be calculated using the Pythagoras theorem to give
(
r = x2 + y2 + z2
)
1/ 2
o
2.2 Direction Cosines
We will see later that it is useful to know the angle the line OP makes with each axis.
γ
r
y
Y-axis
x
Fortunately this can be easily calculated. The shaded areas of the figures below each form
right-angled triangles with the axes.
X-axis
So, knowing the length and the appropriate co-ordinate, each angle can be calculated.
Z-axis
The angle ∠POx is given by
z
P(x,y,z)
α
x
r
m = cos β =
y
r
n = cos γ =
z
r
The angle ∠POy is given by
r
o
l = cos α =
y
Y-axis
The angle ∠POz is given by
x
X-axis
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The cosines of these angles, α , β , γ are (often written) l , m, n and these are known as the
Calculate the direction cosines:
x
= cos α = cos 60 = 0.5
r
y
m = = cos β = cos 50 = 0.6428
r
direction cosines of the line OP.
l=
It can easily be shown that
l 2 + m2 + n2 =
x2 y2 z2 x2 + y2 + z2
+
+
=
=1
r2 r2 r2
r2
We need the third direction cosine.
Using l 2 + m 2 + n 2 = 1
2.2.1
Examples
1)
If P has co-ordinates (5, 4, -9). What is the length OP and its direction cosines.
z
= cos γ = 1 − l 2 − m 2
r
= 0.58035
m=
γ = 54.52 o
OP 2 = 25 + 16 + 81
The length OT, (r) can be calculated as we know the x co-ordinate, i.e. from l
OP = 122
x
= 0.5
r
7.0
r=
= 14.0
0.5
l=
Direction cosines:
5
l=
m=
n=
2)
122
4
From the other direction cosines
y = 0.6428 × 14.0 = 9.0
z = 0.58035 × 14.0 = 8.125
122
−9
Co-ordinates of the corner are: ( x, y, z ) = (7.0,9.0,8.125)
122
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
2.3 A Geometric Representation
We represent vectors geometrically by line segments in space. The length of the line
and the following angles were measured: TOx = 70o and TOy = 50o .
representing the magnitude and the direction of the line the direction of the vector.
Calculate the co-ordinates of the corner relative to the position of he theodolite.
From this definition the starting point is irrelevant.
T
What is the difference between these two vectors?
A
z
A’
y
a
x
7m
Engineering Mathematics 1.2: Vectors Algebra
O
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Engineering Mathematics 1.2: Vectors Algebra
a
O’
8
Their length is the same.
The unit vectors i, j, k
The unit vectors in the co-ordinate directions are denoted i, j, and k.
Their direction is the same.
i = (1,0,0),
(The arrows indicate the direction along the line)
These two vectors are equivalent (equal).
j = (01,0)
and
k = (0,0,1)
Any vector r can be expressed in terms of its component x, y, z with respect to the unit vector
(i,j,k) by
r = xi + yj + zk
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
The notation (x,y,z) is interpreted as (xi, yj, zk).
space).
2.6.1
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
(
The modulus of a = a = 3 2 , (− 1) ,4 2
used either a geometrical or a co-ordinate (component) one.
2.4
Example:
A vector a = (3,−1,4 ) = 3i − j + 4k
2
)
1/ 2
= 26
So the unit vector in the direction of a is
Vector notation found in text books
aˆ =
A vector can be written down in many ways. Some of the more common (and acceptable) ways
1
4 
a  3
=
i−
j+
k 
a  26
26
26 
you will come across are:
a
OA
(a1 , a 2 , a3 )
a
2.7
Equal Vectors
Two vectors a and b are equal if they have the same magnitude and direction i.e.
2.5
Magnitude (or Modulus)
a=b
The magnitude, modulus or length of the vector a is written as a or OA and given in
Using the component definition
component form by
a = (a1 , a 2 , a 3 )
(
a = a12 + a 22 + a
)
2 1/ 2
3
then
Geometrically this is the length of the line OA
2.6
b = (b1 , b2 , b3 )
a1 = b1
a 2 = b2
a3 = b3
The Unit Vector
A vector whose modulus is 1 is called a unit vector, sometimes written as â and
aˆ =
Engineering Mathematics 1.2: Vectors Algebra
a
a
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2.8
Parallel Vectors
3 Vector Addition
If λ is a scalar and a = λb
3.1
then
The boat problem
A boat steams at 4 knots due East for one hour. The tide is running North-North-East at 3 knots.
if λ>0, the vector a is in the same direction as b and has magnitude λb
Where will the boat be after one hour?
if λ<0, the vector a is in the opposite direction to b and has magnitude λb
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.
In component form
a1 = λb1
a 2 = λb2
NNE
a 3 = λb3
North
B
C
This can be summarised as the vectors a and b are
Parallel if λ > 0
a+b
b
b
Anti-parallel if λ < 0.
a
O
East
A
The net velocity of the boat is represented by the line OC which is the sum of a 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.
B
b
O
C
a+b
a
A
In component form
a + b = (a1 + b1 , a 2 + b2 , a3 + b3 )
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3.3
Vector Laws for addition and subtraction
Commutative law
Because the vector a is the same as any other vector which is parallel and in the same
direction, a or b could be "moved" in the figure above to form a triangle. This leads to the
a+b=b+a
triangle law.
Order is NOT important
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
a+b
side.
b
a
a
b
a+b
a+b
b
a
In component form
The triangle rule can be made more general to apply to any geometrical shape - or polygon.
(a1 + b1 , a2 + b2 , a3 + b3 ) = (b1 + a1 , b2 + a2 , b3 + a3 )
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,
Associative law
and e. Then the closing side represents the vector r and r is the sum of the vectors a, b, c, d,
(a + b) + c = a + (b + c)
and e.
B
D
d
b
C
e
E
E
c
B
r
e
a
r
b
O
a
This follows from the component definition and geometrically - from the triangle or polygon law.
A
O
A
D
In component form
c
d
C
c
a+(b+c)
b+c
(a+b)+c
r =a+b+c+d+e
a+b
b
a
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(a1 + b1 ) + c1 = a1 + (b1 + c1)
B
(a2 + b2 ) + c2 = a2 + (b2 + c2)
(a3 + b3 ) + c3 = a3 + (b3 + c3)
a-b
b
a
O
A
Distributive law
-b
λ (a + b ) = λa + λb
This follows from the component definition or geometrically from similar triangles
Note BA = OA − OB
In component form
This is an important result as it shows that the vector represented by the line joining two points
(λ (a1 + b1 ), λ (a2 + b2 ), λ (a3 + b3 )) = (λa1 + λb1 , λa 2 + λb2 , λa3 + λb3 )
is given by taking the vector giving the position of the first point from that for the second
B’
BA = OA − OB = BO + OA = −b + a = a − b
λ(a+b) = λ a+λb
λb
3.4
B
a+b
O
(1)
b
a
Vector Addition Examples
A
A’
λa
a = (2,1,0 )
b = (− 1,2,3) c = (1,2,1)
a + b = (1,3,3)
2a − b = (2 × 2 + 1, 2 × 1 − 2, 2 × 0 − 3) = (5,0,−3)
a + b − c = (0,1,2 )
O’
Subtraction
Modulus of c
a - b = a + (-b)
(
c = 12 ,2 2 ,12
In component form:
)
1/ 2
= 6
The unit vector in the direction of c
a − b = (a1 − b1 , a 2 − b2 , a 3 − b3 )
cˆ =
c  1 2 1 
=

,
,
c  6 6 6 
Not that the modulus of cˆ = cˆ = 1
Geometrically:
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2) From analysis of the forces in a new bridge it is calculated that the following forces are
4 Vector Multiplication
imposed at one of the bridge supports (in Newtons).
We have seen how vectors can be added, another natural operation in maths is to multiply
F1 = (1,1,1)
quantities together. The idea of multiplication of vectors becomes very useful in engineering. In
F2 is 6N and acts in the direction (1,2,-2)
particular we will show how to use them to calculate moments and work-done with great ease.
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
Two types of multiplication exist. They are called the vector and scalar products.
support must impose on the bridge to reduce the resultant force to zero.
4.1
The first force is given in the usual vector form. The second two are in valid forms but need to
The Scalar Product
(also known as the dot product or the inner product)
be converted so that we can perform the usual vector addition.
First calculate the unit vectors in the direction of F2 and F3
A
F
(1,2,−2) = 1 (1,2,−2)
Fˆ 2 = 2 =
F2
1+ 4 + 4 3
F
(3,−4,0) = 1 (3,−4,0)
Fˆ 3 = 3 =
F2
9 + 16 + 0 5
a
So the in vector for the two forces are
θ
P
O
N
The component of the vector a in the direction of OP is easily calculated as equivalent to the
 1 2 −2
F2 = 6 , ,
 = (2,4,−4 )
3 3 3 
3 −4 
F2 = 10 ,
,0  = (6,−8,0 )
5 5 
length ON by
ON = a cosθ
The resultant force F is found by vector addition
F = F1 + F1 + F1 = (1,1,1) + (2,4,−4 ) + (6,−8,0 ) = (9,−3,−3)
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 vector a, then we can calculate the work done by the
The force the support must impose on the bridge is equal and in the opposite direction as the
force. (Remember: work done is the component of the force in the direction of movement
resultant i.e.
multiplied by distance moved by the point of application.)
Reaction force = (− 9,3,3)
F
F
θ
O
a
A
The component of F along OA is F cosθ . This force is moved the distance a , giving,
work done = F a cosθ
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This is the scalar product of the two vectors F and a and can be seen as a geometric
definition.
(3) Use of the "dot" is essential to indicate that the calculation is a scalar product.
An equivalent component definition can be written.
4.2 Rules for the scalar (or dot) product.
Commutative Law
Scalar product definition:
The scalar product of two vectors a = (a1 , a 2 , a 3 ) and b = (b1 , b2 , b3 ) is written with a dot (⋅)
a⋅b = b⋅a
a b cosθ = b a cosθ
between the to vectors.
It is defined in component form as
a ⋅ b = a1b1 + a 2 b2 + a 3b3
Associative Law
You cannot have a scalar product of three vectors as 'dotting' the first two gives a scalar.
And in geometrical form
Distributive Law (for a scalar multiplier)
a ⋅ b = b a cosθ
Brackets are multiplied out in the usual way.
a ⋅ (λb ) = (aλ ) ⋅ b = λ (a ⋅ b )
Where θ is the angle between the two vectors and 0 ≤ θ ≤ π .
Distributive Law (for vector addition)
The two definitions can be proved to be equivalent by the cosine rule for a triangle,
a ⋅ (b + c ) = a ⋅ b + a ⋅ c
B
Powers of vectors
a ⋅ a = a12 + a 22 + a32
b
= a a cos 0
=a
θ
O
a
AB 2 = OA 2 + OB 2 − 2( AO )(OB ) cosθ
2
No other powers of vectors are possible other than 2.
A
Note that for unit vectors i, j, k
j2 = j ⋅ j = 1
i2 = i ⋅i = 1
k2 = k ⋅k =1
Which can be expanded to show that
a1b1 + a 2 b2 + a 3b3 = b a cosθ
Perpendicular Vectors
If a and b are perpendicular then θ =
Three important points about a scalar product:
π
2
and cosθ = cos
π
2
=0
Hence a ⋅ b = 0
BUT a ⋅ b = 0 does not imply that a and b are perpendicular, because a could be zero or b could
(1) The scalar product of two vectors gives a number (a scalar)
be zero.
Note: Suppose a ⋅ b = a ⋅ c
(2) The scalar product is a product of vectors
This does not mean that b = c, since a could be zero or a perpendicular to b-c.
(it cannot be of two scalars nor a vector and a scalar)
Engineering Mathematics 1.2: Vectors Algebra
[ a ⋅ (b − c ) = 0 ]
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(3) Find the work done by the force F = (3, 1, 5) in moving a particle from P to Q where position
i.e. vectors cannot be cancelled in the same way as scalars.
vectors of P and Q are (1, 2, 3) and (2, 4, 4) respectively.
For unit vectors which are perpendicular
r = PQ = OQ − OP
= (1,2,1)
i ⋅ j = j ⋅k = k ⋅i = 0
Work done by the force F = F ⋅ r
(3,1,5) ⋅ (1,2,1) = 3 + 2 + 5 = 10
Perpendicularity is important in engineering, for example pressure acts normal to a surface, so
units of work.
force per unit area is pn̂ , where p is the pressure and n̂ the unit normal vector.
We often have to find a vector that is normal to another vector.
(4) Find the component of the vector F = (2, -2, 3) in
Component of a vector
(i)
the i direction
(ii)
the direction (3, 1, -2)
The component of a vector, F say, in a given direction a is given by F cos θ .
(i) The i direction is the vector (1,0,0) so the component in the i direction is
And we can write
F ⋅ (1,0,0 ) = (2,−2,3) ⋅ (1,0,0 ) = 2
F ⋅ aˆ = F aˆ cosθ = F cosθ
(iii)
to give the component of F in the direction on a.
4.3
(9 + 1 + 4)1 / 2 =
Unit vector =
Scalar Product Examples
14 Therefore it is not a unit vector.
(3,1,−2)
14
Component of F in direction (3, 1, -2) is
(1) Find the angle between the two vectors a = (2, 0, 3) and b = (3, 2, 4)
F⋅
a ⋅ b = b a cosθ and a ⋅ b = a1b1 + a 2 b2 + a 3b3
Using
Is the vector (3, 1, -2) a unit vector?
(3,1,−2) = (2,−2,3) ⋅ (3,1,−2) /
14
14 =
−2
14
a ⋅ b = 6 + 0 + 12 = 18
a = (2, 0, 3) = (4 + 0 + 9 )
1/ 2
b = (3, 2, 4 ) = (9 + 4 + 16 )
= 13
1/ 2
= 29
18 = 13 29 cosθ
cosθ =
18
13 29
θ = 22.02°
= 0.927
(2) For a = (1, 2, -2),
i)
a ⋅c
b = (0, 2, 2)
ii)
(a + b ) ⋅ c
i)
a ⋅ c = (3 + 4 − 2 ) = 5
ii)
(a + b ) ⋅ c = (1,4,0) ⋅ (3,2,1) = (3 + 8 + 0) = 11
iii)
(a ⋅ b )c = (0 + 4 − 4)(3,2,1) = 0(3,2,1) = (0,0,0)
c = (3, 2, 1) find
iii)
(a ⋅ b )c
Note that a and b are perpendicular since neither a or b are zero.
Remember that a ⋅ b is just a number (not a vector).
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4.4 The Vector (or Cross) Product
The main practical use of the vector product is to calculate the moment of a force in three
4.5
The Moment of force, F.
dimensions. It is of very limited use in two dimensions.
n
For two vectors a and b, the vector product is defined as:
F
(a × b ) = a b sin θnˆ
r
where θ is the angle between the vectors a and b, ( 0 ≤ θ ≤ π ) and n̂ is the unit vector normal to
O
θ
P
both a and b.
a , b and n̂ are three vectors which form a right-handed set. (Remember how a right handed
axes system was defined earlier.)
If the force F passes through the point P and OP = r , then the moment of the force about O is
defined as
It is very important to realise that the result of a vector product is itself a vector.
M = r×F
M is a vector in the direction of the normal n̂ .
n
So, as with all vectors, moments add by the parallelogram rule.
a
b
θ
θ
The use of calculating moments by the vector (cross) product comes into itself when used in
b
a
situations (particularly in three dimensions) which are very difficult to visualise.
n
4.6
Area calculation
D
Note how from this definition order of multiplication matters:
The vector given by (b × a ) is in the opposite direction to (a × b ) .
C
h
θ
(This follows from the right-hand screw rule), so
(a × b ) = −(b × a )
A
B
The area of the parallelogram ABCD is given by
Area = h AB = AD sin θ AB = AD × AB
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The area of the triangle PQR is given by
Distributive law with addition
a × (b + c ) = (a × b ) + (a × c )
Q
4.8 Parallel vectors.
From the definition of the vector product, if a and b are parallel, then
θ =0
and
R
h
a×b = 0
θ
So we can say that if a × b = 0 then
a=0
P
or
1
Area = h PR
2
1
= PQ sinθ PR
2
1
= PQ × PR
2
b=0
or
a and b are parallel.
If a × b = a × c it cannot be deduced that b = c. You must first show that a ≠0 and that a is not
parallel to b - c.
4.7
Laws for the Vector (or Cross) products
4.9 Cartesian (component) form.
For unit vectors i,j,k
Anti-Commutative
(asθ
i×i = j× j = k ×k = 0
(a × b ) = −(b × a )
This follows from the definition that includes the 'right-handed axes', n̂ changes direction when
the multiplication is reversed.
Non-associative
= 0, sin θ = 0 )
Because of perpendiculatiry
i × j = k,
j × k = i,
k ×i = j
j × i = −k ,
k × j = −i ,
i×k = − j
Component form of the vector product:
(The triple vector product)
a = (a1 , a 2 , a3 ) = a1i + a 2 j + a3 k
a × (b × c ) ≠ (a × b ) × c
b = (b1 , b2 , b3 ) = b1i + b2 j + b3 k
The vector b × c is perpendicular to both b and c and the plane containing b and c.
a × b = (a1 , a 2 , a3 ) × (b1 , b2 , b3 )
Similarly, (a × b ) × c is in the plane of a and b, so a × (b × c ) and (a × b ) × c are different vectors.
= ( a1i + a 2 j + a3 k ) × (b1i + b2 j + b3 k )
= a1b1 (i × i ) + a1b2 (i × j ) + a1b3 (i × k )
Brackets must always be used for more than two vectors in a vector product.
+ a 2 b1 ( j × i ) + a 2 b2 ( j × j ) + a 2 b3 ( j × k )
+ a 3b1 (k × i ) + a 3b2 (k × j ) + a3 b3 (k × k )
Distributive law with multiplication by a scalar
= a1b2 k + a1b3 (− j ) + a 2 b1 (− k )
a × (λb ) = (λa ) × b
+ a 2 b3 i + a3 b1 j + a3 b2 (− i )
So, how do you remember this!!?
Engineering Mathematics 1.2: Vectors Algebra
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Engineering Mathematics 1.2: Vectors Algebra
26
There are several ways, the determinant form is shown here
i
a × b = a1
b1
=i
a2
b2
j
a2
b2
It can be shown that in general,
(a × b ) × c = (a ⋅ c )b − (b ⋅ c )a
k
a3
b3
a3
a
−j 1
b3
b1
It can also be shown that
a × (b × c ) = (a ⋅ c )b − (a ⋅ b )c
a3
a
+k 1
b3
b1
a2
b2
(2) Find the area of the triangle having vertices at
P= (3, 2, 2), Q= (1, -1, 2), R = (2, 1, 1)
a × b = (a 2 b3 − a 3b2 )i − (a1b3 − a3 b1 ) j + (a1b2 − a 2b1)k
Q
4.10 Examples
(1) Given the three vectors a = (3, 1, 1), b = (2, -1, 1), c = (0, 1, 2)
i
j k
a×b = 3 1 1
2 −1 1
=i
R
θ
1 1
3 1
3 1
−j
+k
2 1
2 −1
−1 1
P
Area PQR =
= (2,−1,−5)
1
PQ × PR
2
PQ = (− 2,−3,0 )
This vector is perpendicular to the plane containing a and b.
i
j
PQ × PR = − 2 − 3
i
j
k
(a × b ) × c = 2 − 1 − 5
0 1
2
=i
PR = (− 1,−1,−1)
−1
=i
2 −5
2 −1
−1 − 5
−j
+k
1
2
0 2
0 1
−3
k
0
−1 −1
0
−1 −1
−j
−2
0
−1 −1
+k
−2 −3
−1
−1
= (3,−2,−1)
= (3,−4,−2 )
Area =
This lies in the plane containing a and b.
1
1
PQ × PR =
2
2
(9 + 4 + 1) =
14
2
(a ⋅ c )b − (b ⋅ c )a = 3b − 1a
= (6,−3,3) − (3,1,1)
= (3,−4,2 )
This is the same as the solution for (a × b ) × c above.
Engineering Mathematics 1.2: Vectors Algebra
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Engineering Mathematics 1.2: Vectors Algebra
28
(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).
4.11 Triple products of Vectors
Definitions of the two triple products are
F
P
θ
Q
(a × b ) × c
is the Triple Vector product.
(a × b ) ⋅ c
is the Triple Scalar product.
The unit vector in the direction of the force is
The triple scalar product has an interesting geometrical meaning:
We know that
4i + j + 2k
1
2 
 4
=
,
,

16 + 1 + 4  21 21 21 
(a × b ) = a b sin θnˆ
= (area of the parallelogram defined by a and b )
So the force F, has components
1
2 
 4
F = 6
,
,

 21 21 21 
Thus
The position vector of P relative to Q is
(a × b ) ⋅ c = (area of the parallelogram)nˆ ⋅ c
= (area of the parallelogram ) nˆ c cos φ
QP = (1,1,0 )
i.e. r
So the moment of the force about Q is
M = QP × F =
6
21
i j k
1 1 0
h
4 1 2
φ
c
B
b
i j k
1 0
1 0
1 1
−j
+k
1 1 0 =i
1 2
4 2
4 1
4 1 2
θ
O
= (2,−2,−3)
C
a
A
But c cos θ = h =height of the parallelepiped normal to the plane containing a and b . ( φ is the
 12 − 12 − 18 
,
,
M=

 21 21 21 
angle between c and n̂ ).
So
(a × b ) ⋅ c
= area of the parallelepiped defined by a, b and c.
It follows then that:
(a) If any two vectors are parallel (a × b ) ⋅ c = 0 (zero volume)
(b) If the three vectors a co-planar then (a × b ) ⋅ c = 0 (zero volume)
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Engineering Mathematics 1.2: Vectors Algebra
30
(c) If (a × b ) ⋅ c = 0 then either
i)
a = 0, or
ii)
b = 0 or
5 Vector Equations of Lines and Planes
iii)
c =0 or
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
iv)
two of the vectors are parallel or
respectively and a point P, which has position vector r, and lies on this line as shown
v)
the three vectors are co-planar
A
(d) (a × b ) ⋅ c = a ⋅ (b × c ) = b ⋅ (c × a ) = the same volume.
P
a
B
r
b
O
OA = a
OB = b
OP = r
From this diagram we can see that
OP = OA + AP
r = a + AP
If t is some multiple of AB such that
r = a + t AB
as a + AB = b , then
r = a + t (b − a )
So the equation of the line AB is:
r = (1 − t )a + tb
for − ∞ < t < ∞ .
In component form
r = ( x, y , z )
a = (a1 , a 2 , a 3 )
b = (b1 , b2 , b3 )
( x, y, z ) = (1 − t )(a1 , a 2 , a3 ) + t (b1 , b2 , b3 )
= (a1 , a 2 , a 3 ) + t [(b1 , b2 , b3 ) − (a1 , a 2 , a3 )]
If we equate components e.g. those of x gives
Engineering Mathematics 1.2: Vectors Algebra
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Engineering Mathematics 1.2: Vectors Algebra
32
x = (1 − t )a1 + tb1
In Cartesian form
x − a1
t=
b1 − a1
x −1 y −1
z
=
=
=s
4
−3
−7
(3) Do r1 and r2 intersect?
Similarly for y and z
t=
y − a2
b2 − a 2
t=
For the lines to intersect then t and s should be such that:
z − a3
b3 − a3
3t = 1 − 4 s
1 + 3t = 1 − 3s
− 2 + 5s = −7 s
Hence the Cartesian equation of a line passing through the points (a1,a2,a3) and (b1,b2,b3).is
z − a3
x − a1
y − a2
=
=
=t
b1 − a1 b2 − a 2 b3 − a3
are all satisfied.
Solving the first two simultaneously s = 1 and t = -1.
Note: For any equation in Cartesian form we can readily express the equation in vector form.
Substituting these values into the third gives
e.g.
-7 = -7
x y −9 z −2
=
=
=t
3
1
−1
i.e. the three equations are satisfied, so the two lines intersect.
In vector form
r = a + t (b − a )
r = (0,9,2) + t (3,−1,1)
Substituting the values for s or t 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)
5.2
Equation of line examples:
c ⋅ d = c d cosθ
r = (1 − t )a + tb = a + t (b − a )
c ⋅ d = −12 − 9 − 35 = −56
(1) Find the equation of the line r1 through the points (0,1,-2) and (3,4,3).
r = (1 − t )a + tb
= (0,1 − t ,−2 + 2t ) + (3t ,4t ,3t )
= (3t ,1 + 3t ,−2 + 5t )
i.e. x = 3t
y = 1 + 3t
c = (9 + 9 + 25)
1/ 2
d = (16 + 9 + 49 )
= 43
1/ 2
= 74
z = −2 + 5t
cosθ =
In Cartesian form
− 56
43 74
θ = 173.1°
x − 0 y −1 z + 2
=
=
=t
3 − 0 3 −1 3 + 2
x − 0 y −1 z + 2
=
=
=t
3
2
5
= −0.9927
(2) Find the equation of the line r2 through the points (1,1,0) and (-3,-2,-7)
r = (1 − s )a + sb
= (1 − s,1 − s,0) + (− 3s,−2s,−7 s )
= (1 − 4s,1 − 3s,−7 s )
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Engineering Mathematics 1.2: Vectors Algebra
34
5.3
5.4 Examples involving both lines and planes
(1) Find the equation of the plane through the points
The Vector Equation of a Plane
a = (3,2,0)
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.
b = (1,3,-1)
c = (0,-2,3)
a - b lies in the plane
n and AP are perpendicular.
a - b = (2, -1, 1)
a - c lies in the plane
a - c = (3, 4, -3)
n
A normal, n, to the plane is
n = (a − b ) × (a − c ) = (2,−1,1) × (3,4,−3)
P
A
i
j
k
= 2 −1 1
3 4 −3
=i
O
= ((3 − 4 ), − (− 6 − 3), (8 + 3))
n = (− 1, 9, 11)
The vector n is normal to the plane, a is the position vector of A, r is the position vector of P (r =
(x, y, z)).
2 1
2 −1
−1 1
−j
+k
4 −3
3 −3
3 4
Equation of the plane is r ⋅ n = a ⋅ n
The vector
r ⋅ (− 1,9,11) = (3,2,0 ) ⋅ (− 1,9,11)
= −3 + 18 + 0
= 15
AP = OP − OA = r − a
In component form this is
Now r - a is perpendicular to n if
( x, y, z ) ⋅ (− 1,9,11) = 15
(r − a ) ⋅ n = 0
− x + 9 y + 11z = 15
or
r ⋅n = a⋅n
Check
or
a:
b:
c:
r ⋅n = p
The above two are the general form of the vector equation of a plane.
−3 + 18 + 0 = 15
−1 + 27 − 11 = 15
0 − 18 + 33 = 15
All are consistent.
If we can take n = (α, β, γ) the equation in Cartesian form is
αx + βy + γz = p
since r = (x, y, z)
Engineering Mathematics 1.2: Vectors Algebra
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Engineering Mathematics 1.2: Vectors Algebra
36
(2) (a) Find the point where the plane r ⋅ (1,2,1) = 4 (or x + 2 y + z = 4 ) meets the line
r = (2, 1, -1) + t(1, 1, 2) or (
(3) Find the perpendicular distance from the point P (2,-3,4) to the plane x+2y+2z=13
The equation of the plane in vector form is
x − 2 y −1 z +1
=
=
=t)
1
1
2
r ⋅ (1,2,2) = 13
(b) Find the angle that the line makes with the plane.
A vector normal to the plane is
n = (1,2,2 )
[Remember the equations of a line
Hence the equation of a line perpendicular to the plane and passing through P is
r = a + t (b − a )
r = a + t (b − a )
= (2,−3,4 ) + t (1,2,2 )
z − a3
x − a1
y − a2
=
=
=t
b1 − a1 b2 − a 2 b3 − a3
This line meets the plane when
r ⋅ (1,2,2 ) = (2,−3,4 ) ⋅ (1,2,2 ) + t (1,2,2 ) ⋅ (1,2,2 ) = 13
]
i.e.
The point of intersection must satisfy both the equation of the plane AND the equations of the
(2 − 6 + 8) + t (1 + 4 + 4) = 13
line.
4 + 9t = 13
t =1
i.e (using the vector form)
[a + t (b − a )] ⋅ (1,2,1) = 4
(2 + 2 − 1) + t (1 + 2 + 2) = 4
Thus the line meets the plane at N with co-ordinates
(2,−3,4) + 1(1,2,2) = (3,−1,6)
3 + 5t = 4
t = 1/ 5
Hence the perpendicular distance PN
[(3 − 2 ) + (− 1 + 3)
2
Substituting into the equation of the line gives the point of intersection
2
+ (6 − 4 )
]
2 1/ 2
= 3 units
r = a + t (b − a )
1
(1,1,2)
5
 11 6 3 
=  , ,− 
 5 5 5
(3) Find the equation of the line of intersection of the two planes given by:
= (2,1,−1) +
x+y+z=5
The normal to the plane is (1, 2, 1) = n.
The line of intersection must be perpendicular to both the vectors (1,1,1) and (4,1,2).
A vector in the direction of the line is (1, 1, 2) = b - a
Hence a vector in the direction of the line must be in the direction
(1,1,1)×(4,1,2) = (1,2,-3)
(b − a ) ⋅ n = b − a n cosθ
(b − a ) ⋅ n = 1 + 2 + 2 = 5
1/ 2
n = (1 + 4 + 1)
1/ 2
4x + y + 2z = 15
r ⋅ (1,1,1) = 5
r ⋅ (4,1,2) = 15
(b) Find the angle that the line makes with the plane.
b − a = (1 + 1 + 4 )
and
In vector from these planes can be written:
To complete the equation of the line we need to find any one point on the line. Choosing x=0
= 6
then, from the Cartesian equations of the two planes,
= 6
y+z=5
5
cosθ = = 0.833
6
θ = 33.56°
y + 2z = 15
Hence for x=0, y=-5 and z = 10
And the point (0, -5, 10) is a point on the line.
The equation of the line can then be written
Engineering Mathematics 1.2: Vectors Algebra
37
Engineering Mathematics 1.2: Vectors Algebra
r = (0, -5, 10) + t (1, 2, -3)
38