Transcript Vectors

Chapter 46
Vectors
ch46 Vectors
by Chtan FYKulai
1
A VECTOR?
□ Describes the motion of an object
□ A Vector comprises
□ Direction
□ Magnitude
Size
□ We will consider
□ Column Vectors
□ General Vectors
□ Vector Geometry
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Column Vectors
NOTE!
Vector a
Label is in BOLD.
2 up
a
When handwritten,
draw a wavy line
under the label
a
i.e. ~
4 RIGHT
 4
 
 2
COLUMN Vector
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Column Vectors
Vector b
2 up
b
3 LEFT
 3 
 2
 
COLUMN Vector?
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Column Vectors
Vector u
2 down
n
4 LEFT
 4 
 2 
 
COLUMN Vector?
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Describe these vectors
4
1
 
a
1
 3
 
b
c
 2 
 3
 
d
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 4 
 3 
 
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Alternative labelling
B
D
EF
E
AB
F
CD
G
C
A
GH
H
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General Vectors
A Vector has BOTH a Length & a Direction
All 4 Vectors here are EQUAL in Length and
Travel in SAME Direction. All called k
k
k
k
k
k can be in any position
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General Vectors
Line CD is Parallel to AB
B
A
CD is TWICE length of AB
k
D
2k
Line EF is Parallel to AB
E
C
-k
EF is equal in length to AB
EF is opposite direction to AB
F
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Write these Vectors in terms of k
B
k
D
2k
½k
1½k
F
G
E
A
C
-2k
H
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Combining Column Vectors
 2
k   
1
AB
k
B
AB  3k
D
CD  2k
 2
AB  3  
1
A
 2
CD  2  
1
C
6
AB   
 3
AB
 4
CD   
 2
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Simple combinations
4
AB   
1 
C
1 
BC  
3

 
5
AC =  
4
B
A
a  c   a  c 
      

 b   d  b  d 
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Vector Geometry
Consider this parallelogram
Q
OR  b  PQ
P
R
a
b
O
OP  a  RQ
Opposite sides are Parallel
OQ  OP  PQ
 a +b
OQ  OR  RQ
 b +a
a +b  b + a
OQ is known as the resultant of a and b
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Resultant of Two Vectors
□ Is the same, no matter which route is
followed
□ Use this to find vectors in geometrical
figures
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e.g.1
S is the Midpoint of PQ.
Work out the vector
.
Q
S
P
OS
OS  OP  ½PQ
= a + ½b
R
a
b
O
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Alternatively
S is the Midpoint of PQ.
.
Q
S
P
Work out the vector
OS
OS  OR  RQ  QS
R
a
b
O
= b + a - ½b
= ½b + a
= a + ½b
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e.g.2
C
AC= p, AB = q
p
A
M
q
Find BC
M is the Midpoint of BC
B
BC = BA + AC
= -q + p
=p-q
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e.g.3
C
AC= p, AB = q
p
A
M
q
Find BM
M is the Midpoint of BC
B
BM = ½BC
= ½(p – q)
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e.g.4
C
AC= p, AB = q
p
A
M is the Midpoint of BC
M
q
Find AM
B
AM = AB
+ ½BC
= q + ½(p – q)
= q +½p - ½q
= ½q +½p
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= ½(q + p)
by Chtan FYKulai
= ½(p + q)
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Alternatively
C
AC= p, AB = q
p
A
M is the Midpoint of BC
M
q
Find AM
B
AM = AC + ½CB
= p + ½(q – p)
= p +½q - ½p
= ½p +½q
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= ½(p + q)
by Chtan FYKulai
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Distribution’s law :
The scalar multiplication of a vector :
𝑘 𝒂 + 𝒃 = 𝑘𝒂 + 𝑘𝒃
𝑘 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
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𝑘 > 0 𝑜𝑟 𝑘 < 0
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Other important facts :
ℎ𝑘 𝒂 = ℎ𝑘 𝒂
ℎ + 𝑘 𝒂 = ℎ𝒂 + 𝑘𝒂
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A vector with the starting
point from the origin point
is called position vector.
位置向量
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Every vector can be
expressed in terms of
position vector.
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e.g.5
2
−2
Given that 𝒂 =
,𝒃=
5
3
10
and also 𝑘𝒂 + 𝑙𝒃 =
. Find
1
the values of 𝑘 𝑎𝑛𝑑 𝑙.
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e.g.6
Given that 𝒂 = 𝑚𝑖 − 4𝑗, 𝒃 =
3𝑖 − 2𝑗, and 𝒂 𝒂𝒏𝒅 𝒃 are
parallel. Find the value of m.
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e.g.7
2
3
𝐴𝐵=
, 𝐵𝐶 =
, a point
5
−2
𝐶 1,4 . Find the coordinates
of 𝐴 𝑎𝑛𝑑 𝐵, then express
point 𝐶 in terms of 𝒊 𝑎𝑛𝑑 𝒋 .
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e.g.8
5
If 𝑃 3,5 , 𝑃𝑄 =
, find the
−7
coordinates of 𝑄.
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e.g.9
Given that 𝒂 = 2𝑖 + 𝑝𝑗, 𝒃 =
7 + 𝑝 𝑖 + 4𝑗, and 𝒂 𝒂𝒏𝒅 𝒃
are parallel. Find the value
of 𝑝.
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by Chtan FYKulai
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Magnitude of a vector
𝐴 𝑖𝑠 𝑥1 , 𝑦1 , 𝐵 𝑖𝑠 𝑥2 , 𝑦2 .
𝒂 = 𝑨𝑩
=
𝒙𝟐 − 𝒙 𝟏
𝟐
+ 𝒚𝟐 − 𝒚𝟏
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by Chtan FYKulai
𝟐
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𝒙, 𝒚
𝒂
0
𝑦
𝒂 =
𝟐
𝒙
+
𝟐
𝒚
𝑥
Unit vector :
𝟏
𝒂=
∙𝒂
𝒂
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e.g.10
Find the magnitude of
the vectors :
−𝟐
𝒂 𝒑=
𝟓
(b) 𝒓 = 𝟗𝒊 − 𝟏𝟐𝒋
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e.g.11
Find the unit vectors in
e.g. 10 :
−𝟐
𝒂 𝒑=
𝟓
(b) 𝒓 = 𝟗𝒊 − 𝟏𝟐𝒋
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Ratio theorem
𝒚
A
P
1
a  b
B p
1 
𝒙
𝟎
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by Chtan FYKulai
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e.g.12
M is the midpoint of AB,
find b in terms of a, m .
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by Chtan FYKulai
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e.g.13
𝑨
2 𝑷
3
𝑩
6b
4a
P divides AB
into 2:3. Find
OP in terms
of a, b .
𝑶
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by Chtan FYKulai
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Application of vector in plane geometry
e.g.14
A
M
X
C
In the diagram, CB=4CN,
NA=5NX, M is the midpoint
of AB.
B
CN  u , BM  v
N
(a) Express the following vectors in terms
of u and v ; (i) NB (ii) NA
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2
(b) Show that CX  4u  v 
5
(c) Calculate the value of
(i) CX
(ii)
Area ACX
CM
Area ACM
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Soln:
(a) (i)
CB  CN  NB
NB  CB  CN  4CN  CN  3CN  3u
(ii)
NA  NB  BA  3u  2v
(b)
1
CX  CN  NX  CN  NA
5
1
8
2
2
 u  3u  2v   u  v  4u  v 
5
5
5
5
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by Chtan FYKulai
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(c) (i) CM  CB  BM  4u  v
2
CX  CM
5
CX
2

CM
5
(ii)
1
CX h 
Area ACX
CX
2
2



1
Area ACM
CM
5
CM h 
2
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e.g.15
M
B
A
M and N are
midpoints of AB, AC.
N Prove that
C
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1
MN  BC and MN // BC
2
by Chtan FYKulai
41
e.g.16
A
2a
1
6a
B
1
K
In the diagram K divides AD into
1:l, and divides BC into 1:k .
l
k
O
2b
C
6b
D
Express position vector OK in 2 formats.
Find the values of k and l.
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by Chtan FYKulai
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More exercises on
this topic :
高级数学高二下册
Pg 33 Ex10g
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by Chtan FYKulai
43
Scalar product of two vectors
If a and b are two non-zero
vectors, θ is the angle between
the vectors. Then ,
a  b  a b cos 
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by Chtan FYKulai
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Scalar product of vectors satisfying :
Commutative law :
a b  b  a
Associative law :
k a   b  a  k b   k a  b 
Distributive law :
a  b  c   a  b  a  c
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by Chtan FYKulai
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e.g.17
Find the scalar product of the
following 2 vectors :
a  6 , b  5 ,  between is 60
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by Chtan FYKulai
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e.g.18
(a) If a  b  a b , find the angle
between them.
(b) If a  1, b  2, a  k b and a  k b
are perpendicular, find k.
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Scalar product (special cases)
1. Two perpendicular vectors
a  0, b  0,
a  b  a b  0
N.B.
Unit vector for y-axis
i  j  j i  0
Unit vector for x-axis
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2. Two parallel vectors
a  0, b  0,
a // b  a  b   a b
N.B.
i i  1  j  j
 
i   i   1  j   j
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e.g.19
Given a  3, b  8, a  b  2 14 ,
Find
a b
.
Ans:[17/2]
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Scalar product (dot product)
The dot product can also be defined
as the sum of the products of the
components of each vector as :
 x1 
 x2 
a   , b   
 y1 
 y2 
 a  b  x1 x2  y1 y2
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e.g.20
Given that
 3
7
a   ; b   
 4
1
Find (a) a  b
(b) angle between a and b .
Ans: (a) 25 (b) 45°
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Applications of Scalar product
高级数学高二下册
Pg 42 to pg43
Eg30 to eg 33
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More exercises on
this topic :
高级数学高二下册
Pg 44 Ex10i
Misc 10
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Miscellaneous
Examples
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e.g.21
Given that D, E, F are three
midpoints of BC, CA, AB of a
triangle ABC. Prove that AD, BE
and CF are concurrent at a
point G and
AG BG CG


2 .
GD GE GF
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A
Soln:
From ratio theorem
1
d  b  c 
2
1
e  a  c 
2
1
f  a  b 
2
ch46 Vectors
B
by Chtan FYKulai
F
G
D
E
C
57
We select a point G on AD such
𝑨𝑮
that
= 𝟐.
𝑮𝑫
From ratio theorem,
1
2 1
1
g  a   b  c   a  b  c 
3
3 2
3
Similarly,
We select a G1 point on BE such that
𝑩𝑮𝟏
= 𝟐.
𝑮𝟏 𝑬
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1
g 1  a  b  c 
3
Similarly,
We select a G2 point on CF such that
𝑪𝑮𝟐
= 𝟐.
𝑮𝟐 𝑭
1
g 2  a  b  c 
3
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Because g1, g2, g are the same,
G, G1, G2 are the same point G!
G is on AD, BE and CF, hence
AD, BE and CF intersect at G.
𝑨𝑮
𝑮𝑫
And also
=
established.
𝑩𝑮
𝑮𝑬
ch46 Vectors
=
𝑪𝑮
𝑮𝑭
by Chtan FYKulai
= 𝟐 is
60
Centroid of a ∆
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The end
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