CxcDirect Institute -
Vectors - Lesson 1 & 2
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Vectors Introduction- Lesson1 To solve vector problems at the CSEC level, you will need to understand the following terms: • Position vector, • Displacement vector, • Collinear vectors, • Equal vectors, • Parallel vectors and resultant vectors.
The three representations of the Position vector of point A 2 are: a , or O ⃗ A 3
()
Displacement Vector
An example will be used to illustrate each point Vector Definition:
A vector is a quantity that has a magnitude (size) and a direction. (shown by an arrow) In the diagram above O ⃗ A and O ⃗ B are called position vectors because their starting points are taken relative to the origin (O). The starting point of vector A ⃗ B is not the origin, so the term displacement vector is used to differentiate between this vector and the position vector.
Vector Representation Vector AB is the vector going from A to B. Three different representations of the vector AB are: 1. 2. 3.
A⃗ B
() 4 3
- as two letters with an overhead arrow - as a column matrix (column vector)
Example 1 O P
Points P(3, 2) and Q(-1, -3) have position vectors relative to the origin O. and O Q
as column vectors OQ
1.
Express
O P
and
2.
Express
PQ
as a column vector
3.
Find the length of
PQ
m - as a lowercase letter
The size ( modulus, magnitude, length ) of vector AB is found using Pythagoras: 4 2 3 2 = 5
Solution: The position vectors can be found directly from the coordinates of P and Q:
1)
Position vector
()
O⃗ P=
3 2
= −1 OQ −3
a
Point A(2,3) can be viewed as being displaced from the origin O, by a vector called the position vector where: © cxcDirect Institute - 876 469-2775 mail:
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CxcDirect Institute -
Vectors - Lesson 1 & 2
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⃗ Finding P Q The displacement vector ⃗ PQ is the vector going from P to Q where : so how do we find this vector?
Find
Vector equation
⃗ = AC
b+c
⃗ AD = ⃗ OD = ⃗ BO = ⃗ BM =
⃗ MA =
Imagine that your starting position is point P and you wish to get to point Q. Note that the only know path or course is to travel first from P to O, and then from O to Q. This means that we can get from P to Q using the two vectors that we already know. i.e
⃗ PQ
⃗ PO
=
Parallel Vectors and Equal Vectors
⃗ OQ
+
Note carefully that we have the vector O ⃗ P , but what ⃗ . we need is the vector P O This is however easily found however, since the vector ⃗ is simply the reverse of the vector O ⃗ PO P i.e
⃗ PO
2) so :
⃗ PQ =
=
⃗ = −OP
()
−
3 2
+
() ( )
−
Vector c is parallel to vector b if : c = kb, and k is a constant (scalar ).
3 2
−1 −3
so if a vector is a constant (scalar) multiple of another vector, then they are parallel.
( ) −4 −5
=
Example: 3. Length (magnitude) of
−42 −5 2 P Q=
= 6.4
()
Given: a b=
c=
() 12 8
Solution:
D f
Now
M d e
A
;
Prove that the vectors are parallel.
Vector Equation Exercise:
a
3 2
so
C
O
⇒
c
b
() 12 8
can be written as
c=4×b
4
() 3 2
(take 4 as a factor)
( k = 4 a constant)
c = 4b ( so b and c are parallel)
B
In the diagram above, the points are A, B, C , M , and O, and the vectors are a,b,c,d,e, and f. ⃗ So for example: M C=d Use this diagram to complete the table. (Note the direction of the arrows) © cxcDirect Institute - 876 469-2775 mail:
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Equal Vectors
Vector e is equal to vector h if they both have the same magnitude and direction. ( so e = h) It follows that equal vectors are also parallel vectors
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CxcDirect Institute -
Vectors - Lesson 1 & 2
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Collinear vectors ( on a straight line) 6
7
Lesson 2 Example1 -Vectors
8 C
Two points A and B have position vectors −2 4 ; O B = , where O is the O A = 5 2 origin ( 0,0). The point G lies on the line AB such that 1 x AG = AB . Express in the form , 3 y ; position vector O G A B ; AG
B
A collinear vectors
Two vectors are collinear if one vector is a scalar ⃗ are multiple of the other vector. If A ⃗B and B C ⃗ ⃗ collinear (on a straight line), then B C = k.A B … where k is a scalar (constant) (Note that this is the same condition for parallel vectors)
A(- 2, 5)
G
B(4,2)
Example: Given Points A(-2,1); B(2,3) and C(8,6) . Use a vector method to prove that the points are collinear
0
**************************************************
A⃗ B
1) Finding
Solution:
Now A ⃗ B is the vector going from A to B That is: First go from A to O and then from O to B
C B
so:
A⃗ B
=
but
⃗ AO
=
so
A B =
⃗ AO
O⃗ B
+
−O ⃗A
=
A
( )
−
o
2) Finding
We need to prove that : ⃗ = k.A ⃗ BC B …. ( the condition for collinearity)
now
−2 5
⃗ +O ⃗ AO B
=
⃗ BC
⃗ ⃗ = B O+O C
( ) () ()
−2 2 = − + 1 3 2 8 = − + 3 6
=
=
4 2 6 3
1 ⃗ AB 3
⃗ = AG
3) Finding Now
−2 5
4 2
6 −3
⃗ AG
so we need to find the two displacement vectors A B and then establish the relationship between and B C them. A⃗ B
+
( ) = −
= 1/3
6 −3
=
2 −1
OG
OG
= =
O A + AG −2 2 + 5 −1
= 0 4
**************************************************
but
6 3
=
()
4 1.5 2
so
⃗ = k.A ⃗ BC B
k = 1.5
so the three points are on a straight line (collinear.)
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CxcDirect Institute -
Vectors - Lesson 1 & 2
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Class Activity 1 The position vectors of points A, B and C are: O ⃗A =
() 6 2
,
O⃗ B=
Express in the form
() a b
() 3 4
( )
, and O C ⃗ = 12 −2
, vectors
B⃗ A ,
3D ⃗ A ⃗ DA
giving so:
⃗ BC
⃗ = DC
a + ½ ( b – 3a) = a + ½ b - 1.5 a = ½b-½a = ½ ( b – a) = DO ⃗ +O X ⃗ DX ⃗ +O X ⃗ = −O D = −2a+2b = 2 (b−a )
⇒
Finding
2) State one geometrical relationship between BA and BC 3) If Point M is the mid point of AB; Find the coordinates of M. Example 2: A
3a a
= =
Two Geometrical Relationships: ii) DX and DC are parallel, (2) DX = 4 DC iii) D, C and X are collinear ( on a straight line)
Class Activity 2 The Position vectors of R and J are :
D C
O
O R = −2 3 B
and
O J=
1 −1
X
1. Express
In the diagram above: C is the mid point of AB and B is the mid point of OX, and D is such that OD = 2DA. The vectors a and b are such that: O ⃗A=3a and O⃗ B =b
R J in the form
2. Find the length
a b
R J
3. Given that another point is such that
R T = 8 2
Find the coordinates of T
Express the following in terms of ; DC and D X A B ; AC
a
and
b *********************************************************
State two geometrical relationship between DX and DC State one geometrical relationship between the points D, C and X **************************************************
Hint: The coordinates of T comes from the position vector O T *********************************************************
Class Activity 3 ABCD is a quadrilateral such that: Finding A B now
=
⃗ +O ⃗ AO B −O ⃗ A+O ⃗ B =
=
b−3a
A B =
Finding
O⃗ A=
−8 4
;
O⃗ B=
( ) −5 7
;
()
⃗= OC
1 4
−3a +b
OM is a point on OA such that the ratio OM:OA = 1:4 Prove that ABCM is a parallelogram
AC
now: C is mid point AB = ½ A B ⇒ AC
( )
= ½ ( b – 3a)
Solution: *********************************************************
Finding now
DC =D , DC A A C
where OD = 2DA
Hint: The opposite sides of a parallelogram are equal., so this question is testing that you know how to prove that two vectors are equal. *********************************************************
and so
⃗ ⃗ = 3a O D+D A 2D⃗ A+D ⃗ A = 3a
( given)
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CxcDirect Institute -
Vectors - Lesson 1 & 2
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Class Avtivity 4 The position vectors A and B relative to the origin are a and b respectively. The point P is on OA such that OP = 2PA The point M is on BA such that BM = MA A
B
b a
O
OB is produced to N such that OB = ON Express in terms of a and b, the vectors: 1.
A B;
P A;
PM
2.
use a vector method to Prove that Points P, M and N are collinear
3.
Calculate the length AN given that : a=
6 2
and b =
1 2
***************************************************
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CxcDirect Institute -
Vectors - Lesson 1 & 2
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Activity 2
Vector lesson - Solutions to activity questions
T
Activity 1
⃗ RT
R(-2,3)
B(3,4)
⃗ OT
A(6,2
O
O
J(1, -1)
C(12, -2)
R⃗ J
() ( ) () ( ) ( )
⃗ ⃗A = B O+O
3 − 4
⃗ = B O+O ⃗ ⃗ = BC C
3 − 4
B⃗ A =
so:
⃗ BC
3B⃗ A
=
+
6 2
+
12 −2
3 −2
=
=
( geometrical relationship)
9 −6
=
⃗ +O ⃗ RO J
=
−
( ) −2 3
( )=( ) 1 −1
+
√( 3 +(−4 ) ) 2
length =
2
=5 O T⃗
Position vector of T =
−2 3
B(3,4)
O⃗ R
+
() 8 2
+
R T⃗
=
() 6 5
so coordinates = T (6, 5)
M A(6,2
*********************************************
Activity 3
O
B(-5,7)
C(12, -2)
if M = mid AB; then =
⃗ BM
C(1,4)
1 ⃗ BA 2
=
A(-8,4)
( )=( )
1 3 2 −2
1.5 −1
M O
Now the coordinates of M can be found from the ⃗ position vector O M ⃗ OM
= =
⇒
=
( )
=
where
3 −4
⃗ O⃗ B+ B M
() + ( ) = ( ) 3 4
1.5 −1
4.5 3
The objective is to prove that: (1) Now:
; and (2) A B =M C
⃗ ⃗ A⃗ B = A O+O B
coordinates = M( 4.5, 3)
= Also: ⇒
so : Giving © cxcDirect Institute - 876 469-2775 mail:
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M A= C B
( )
−
−8 4
+
⃗ = MO ⃗ +O C ⃗ MC OM
=¼ O A = ¼
= OM ⃗ MO
=
( ) () −5 7
=
3 3
but OM:OA = 1:4 −8 −2 = 4 1
( ) ( ) −2 1
−
−2 1
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Vectors - Lesson 1 & 2
1 − −2 4 1
= MC
so:
3 3
=
A B =M C
⇒
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⇒
1 1 a b− a 3 2
= PM
….... (1)
1 3b− a 6
=
********************************************** Points P, M and N are collinear if:
We now need to prove that : M A= C B
=k P M M N
( )
3 ⃗ OA = 4
now:
M⃗ A =
also
⃗ +O ⃗ C⃗ B = CO B
()
= −
1 4
+
3 −8 4 4
−6 3
( ) ( ) −5 7
…
−6 3
=
M N
so we need to find
⃗ M N
=
⃗ M⃗ B+B N
where:
M B
=
½ b−a ; and
so
= M N
now
M A= C B
⇒
( )
=
( k is a constant)
(2)
now note that
=b
½ b −a + b
=
From (1) and (2), ABCM is a parallelogram.
BN
½ 3b− a
⃗ M N
⃗ 3PM
=
⇒ k=3
therfore P , M and N are on a stright line ( collinear)
Activity 4 N
: consider triangle OAN: AN
To find the length of
b
B
M
A
b
⇒
=O ON A A N
⇒
=O N – O A = AN
⇒
AN
=2
1/3 (a)
P
=
2/3 (a)
2b− a
1 2
2 4
-
-
6 2
6 2
=
−4 2
O
Length =
Finding:
A B =
= Finding now:
⃗ +O ⃗ AO B
= =
−a+b
⃗ ⃗ −A O+O B
2
−4 2
2
=
4.47
************************************************
b−a
P⃗ A
1 P A= a 3
⃗ Finding P M
=
⃗ P⃗ A+ A M
now A M = MB =½ A B =
⃗ so P M
=
½ b −a
1 1 a + ( a−b) 3 2
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