Someone recently bought our

students are currently browsing our notes.

X

Vectors Notes

Natural Sciences Notes > Mathematics for Natural Sciences Notes

This is an extract of our Vectors document, which we sell as part of our Mathematics for Natural Sciences Notes collection written by the top tier of Cambridge University students.

The following is a more accessble plain text extract of the PDF sample above, taken from our Mathematics for Natural Sciences Notes. Due to the challenges of extracting text from PDFs, it will have odd formatting:

Mathematics for NST Part IA Cambridge University, 2012-2013

Notes for Vectors Basic vector algebra: Commutivity of vector addition: The geometry of a parallelogram can be used to prove that vector addition is commutative:

B

C c

a
A

C

can be reached either through

B

or

B ' , since opposing sides of

the parallelogram have the same length and direction:

B'

b

AC =
AB+
BC
[?] c=a+b=b+ a

Non-commutivity of vector subtraction: The geometry of a triangle can be used to prove that vector subtraction is non-commutative: It can be seen that:

-c

a
b

b -a =c

a )=-c a -b =-( b-

-b
-a

c

[?] c=b-a=-(a-b) Associativity of vector addition: The geometry of a parallelepiped can be used to prove that vector addition is associative:

B

c

b
a

a

C

a
a
c

b

X'

c

d
b

c P X

b

R

A

X'

can be reached in one

of six ways, since opposing sides of the parallelogram have the same length and

XX '=
XA+
AC +
CX ' Q

XX '=
XA+
AQ+
QX '

XX '=
XB +
BR+
RX '

XX '=
XB+
BC +
CX '

XX '=
XP+
PR+
RX '

Mathematics for NST Part IA Cambridge University, 2012-2013 The ratio theorem: The ratio theorem is a method for solving geometric problems using vectors, by expressing the vectors as a ratio of a third vector:

A

l

AX=
OX -
OA

X m

XB=
OX +
OB B

By the ratio theorem:

|
AX| l
=
|
XB| m

O Therefore,

OX

can be expressed in terms of

OA ,
OB ,

l and

m :

AX l l
= [?] OX -OA = ( OX + OB )

m
XB m
[?] m (
OX -
OA ) =l (
OX +
OB ) [?] ( m-l )
OX =m
OA + l
OB

[?]
OX =

m
OA + l
OB ( m- l )

AB

Where, if the total length of

is taken to be 1, then

be expressed in terms of

and

such that

OX

can

l only.

Using the ratio theorem to prove that the diagonals of a parallelogram bisect:

A

C 1-l

m

For a parallelogram:

OA =
BC

X l

OB=
AC 1-m

O

OA =
OX +
XA= l
OC-m
AB

BC =
BX +
XC =( 1- l )
O C-(1-m)
AB

OA =
BC

OA ,
OB

m=1-l

B

Prove that

l=m=

1 2

Mathematics for NST Part IA Cambridge University, 2012-2013

[?] l
OC -m
AB= (1-l )
OC-(1-m)
AB Comparing coefficients:

l=1-l , m=1-m

1 [?] 2l=1, 2m=1 [?] l=m= [?]

2 Mathematics for NST Part IA Cambridge University, 2012-2013

Vectors in kinematics: The kinematic quantities of displacement, velocity and acceleration have both magnitude and direction, hence are vector quantities. They are related by vector calculus as follows: Displacement: For a moving object, displacement is a function of time,

r (t) .

Velocity: Velocity is the rate of change of displacement:

v ( t )=lim
dt -0

(

r (t +dt ) -r ( t )
dr dr
=lim
= =r (t )
dt dt
dt -0 dt

)

[?] v ( t )=r ( t) Acceleration: Acceleration is the rate of change of velocity:

a ( t )=lim

dt -0

( v ( t+ dtdt)-v ( t ) )= lim ddtv = ddtv =v( t)=r ( t)
dt - 0

[?] a ( t )=v (t)=r (t ) Since the differentiation of vector functions may change their direction, the displacement, velocity and acceleration of a moving object are not necessarily parallel. Scalar quantities in kinematics: The kinematic quantities of distance and speed are the cumulative scalar analogues to displacement and velocity. The scalar equivalent to acceleration can be found using the scalar product, and will therefore be explored later: Distance: Since distance is a cumulative quantity, it cannot be directly related to the displacement function Since

d r

r (t) ; only to an incremental change in r (t) , denoted d r .

is very small, it can be approximated as a straight line, hence the cumulative

nature of distance is not significant:

distance=|d r|
Speed: Speed is the rate of change of distance:

speed =lim
dt -0

(|ddtr|)= lim (|dtdtv|)= lim (|v|)=|v|

[?] speed=|v|

dt - 0

dt- 0

Mathematics for NST Part IA Cambridge University, 2012-2013

Geometric vector equations: Through the use of vectors, it is possible to generate equations to represent any line, plane or sphere, at any point in three-dimensional space. Vector equation of a line: A line is one-dimensional and is therefore defined along only one direction in threedimensional space. Thus, the vector equation of a line contains only one direction vector, along with a point vector to specify its position:

r=a+ l( b-a) The position vector second vector

a defines the position of the line at a point A in space, while the

l(b-a) defines an arbitrary distance

l along the vector joining points

A and B, therefore specifies the direction of the line.

r

is therefore the position vector of any point on the line from the origin.

Vector equation of a plane: A plane is two-dimensional and is therefore defined along two directions in three-dimensional space. Thus, the vector equation of a plane contains two direction vectors, along with a point vector to specify its position:

r=a+ l ( b-a ) + m(c-a) The position vector vector

l( b-a)

a defines the position of the plane at a point A in space; the second

defines an arbitrary distance

and the third vector

l along the vector joining points A and B,

m(c-a) defines an arbitrary distance

points A and C. The vectors

(b-a) and (c-a)

m along the vector joining

therefore span a two-dimensional area

containing the points A, B and C, specifying the direction of the plane.

r

is therefore the position vector of any point on the plane from the origin.

Vector equation of a sphere: A sphere is three-dimensional; defined in every direction, at a fixed radius, from a single, fixed point. Thus, the vector equation of a sphere is defined in terms of a vector of length

R which radiates from a fixed point A in any direction: R= r -a[?]

The vector

(r-a) defines a distance

R

from the centre of the sphere at A to a point

on its surface at R, in an arbitrary direction. Thus, it spans a spherical surface of radius around the central point, defining a sphere in terms of the position vectors of R and A.

R

Mathematics for NST Part IA Cambridge University, 2012-2013

a is the position vector of the sphere's centre from the origin, and r vector of any point on the sphere's surface, from the origin.

is the position

Mathematics for NST Part IA Cambridge University, 2012-2013 Geometric problems using vector algebra: Many geometric problems can be solved using the vector definitions of shapes such as lines, planes and spheres, and manipulating these vectors through vector algebra: Distance between a point and a line: In general, the point B is given by some position vector

b and the line is given by

r=a+ l l , where a is the position vector of A, and

d=r -b=a+ l l-b

another point

A
th

is the

l

l l
b -a

d

d [?] l [?] d [?] l=0

l=0
[?] ( a+ l l-b ) [?] l=0
[?] a [?] l-b
[?] l+

unit

[?] l=l [?](b-a)

B

a
b

O

[?] d =a-b+ l[?](b-a) l

vector which specifies the direction of the line:

Distance between a point and a plane: In general, the point A is given by some position vector

a and the plane is given by

r [?] n =l , where n is a unit vector normal to the plane and l is a scalar. d defined as the vector joining the point to the plane:

d [?] n [?] d=a+ l n

Thus at the point where

d

meets the plane:

r=a+ l n [?] ( a+ l n ) [?] n =l [?] a [?] n + l=l
[?] l=l-a[?] n

[?] d -a=(l-a[?] n ) n
[?]|d-a|=|( l-a[?] n ) n|
Minimum distance between skew lines:

is

Buy the full version of these notes or essay plans and more in our Mathematics for Natural Sciences Notes.