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Vectors - Lesson 1 & 2
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The three representations of the Position vector of point A
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.
are:
2
a,
3
⃗ or O A
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 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.
A B ⃗
4 3
)
- as two letters letters with an overhead overhead arrow - as a column matrix (column vector)
Example 1
O P
Points P(3, 2) and and Q(-1, -3) have position vectors vectors relative to the origin O. and O Q
as column vectors OQ
1.
Express
O P
and
2.
Express
P Q
as a column vector
3.
Find the the len leng gth of P Q
m - as a lowercase letter
The size ( modulus, modulus, magnitude, magnitude, length 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
⃗= O P
3 2
= −1 OQ −3
a
Point A(2,3) 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
⃗ ⃗ OD ⃗ BO
b+c
AD =
= =
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
+
⃗ , but what Note carefully that we have the vector O P ⃗ . we need is the vector P O This is however easily found however, since the vector ⃗ is simply the reverse of the vector O P ⃗ P O i.e
⃗
2) so :
PQ =
PO
⃗
=
⃗ = −OP
−
3 2
)
−
3 2
) )
−1 −3
+
Vector c is parallel to vector b if : c = kb, and k is a constant (scalar ).
−4 −5
=
)
so if a vector is a constant (scalar) multiple of another vector, then they are parallel. Example:
= − 4 2 −5 2 3. Length (magnitude) of 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
4
c = 4 ×b
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. ⃗ =d So for example: M C
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
Use this diagram to complete the table. (Note the direction of the arrows) © cxcDirect Institute - 876 469-2775 mail:
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CxcDirect Institute
-
Vectors - Lesson 1 & 2
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Lesson 2 Example1 -Vectors
Collinear vectors ( on a straight line) C
Two points A and B have position vectors = −2 ; O B = 4 , 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 A G
B
A collinear vectors
Two vectors are collinear if one vector is a scalar ⃗ and B C ⃗ are multiple of the other vector. If A B ⃗ = k.A B ⃗ … collinear (on a straight line), then B C 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 **************************************************
⃗ A B
1) Finding
Solution:
⃗ is the vector going from A to B Now A B That is: First go from A to O and then from O to B C
⃗ = A O
⃗ so: A B B
but
⃗ A O
so
= A B
⃗ −O A
=
A
− −2 5
o
now
⃗ A G
1
=
3
so we need to find the two displacement vectors A B and then establish the relationship between and B C them.
⃗ +O B ⃗ A O
⃗ A B
=
⃗ B C
⃗ + O C ⃗ = B O
but
6 3
=
1.5
4 2
)
= − =
−
) ) ) +
1
2 3
2 3
+
8
6
=
− −2 5
)
) +
4 2
6
=
−3
⃗ 2) Finding A G
We need to prove that : ⃗ = k.A B ⃗ …. ( the condition for collinearit y) B C
−2
⃗ O B
+
=
=
4 2
6 3
3) Finding
Now
⃗ A B
= 1/3
6
−3
=
2
−1
OG
OG
= =
+ A G O A 2 −2 + −1 5
=
0 4
**************************************************
so
⃗ = k.A B ⃗ B C
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
,
⃗ = O B a
Express in the form
3 4
⃗ B C
⃗ , B A
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:
3a
= =
⃗ = D C
⇒
, and O C ⃗ = 12 −2
, vectors
b
⃗ 3D A ⃗ D A
giving so:
a
a + ½ ( b – 3a) a + ½ b - 1.5 a ½b-½a ½ ( b – a) ⃗ ⃗ D O +O X ⃗ +O X ⃗ −O D − 2a + 2b = 2 (b − a )
= = = = Finding D X = =
Two Geometrical Relationships: ii) DX and DC are parallel, (2) DX = 4 DC iii) D, C and X are collinear ( on a straight line)
A
Class Activity 2 The Position vectors of R and J are :
D C
= O R O
B
−2
and
3
= O J
1
−1
X
in the form R J
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. ⃗ = 3a and The vectors a and b are such that: O A ⃗ =b O B Express the following in terms of a ; A C ; DC and D X A B
and
2. Find the length
R J
a b
3. Given that another point is such that
= R T
8
2
Find the coordinates of T
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
OT *********************************************************
Class Activity 3 ABCD is a quadrilateral such that: Finding A B = now A B = = Finding
⃗= O A
⃗ + O B ⃗ A O
⃗ +O B ⃗ = −O A
−8 4
)
;
⃗= O B
−5 7
)
;
⃗ = OC
1 4
)
−3a + b
b −3a
OM is a point on OA such that the ratio OM:OA = 1:4 Prove that ABCM is a parallelogram
A C
now: C is mid point AB = ½ A B ⇒ A C
= ½ ( b – 3a)
Solution: *********************************************************
Finding now
D C = D A A C , D 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
⃗ + D A ⃗ O D ⃗ + D A ⃗ 2 D A
= 3a = 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
P M
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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Vector lesson - Solutions to activity questions
Activity 2 T
Activity 1 R T ⃗
R(-2,3)
B(3,4)
O T ⃗
A(6,2
O J(1, -1)
O C(12, -2)
⃗ R J ⃗ = B A ⃗ = B C so:
⃗ B C
⃗ + O A ⃗ B O
⃗ + O C ⃗ = BO ⃗ 3 B A
=
−
=
3
+
4
3 − 4
)
+
6 2
12 −2
=
3
=
)
⃗ + O J ⃗ = R O
−2 =
( geometrical relationship)
9 −6
)
−
−2 3
)
1 −1
+
√ ( 3 +(−4 ) ) 2
length =
2
)
3 −4
=
)
=5
⃗ OT
Position vector of T =
=
−2
=
3
B(3,4)
)
⃗ O R
+
⃗ R T
)
=
8
+
2
6
)
5
so coordinates = T (6, 5)
M A(6,2
*********************************************
Activity 3
O
B(-5,7)
C(12, -2)
C(1,4)
⃗ if M = mid AB; then B M =
1
3
2
−2
)
=
1
=
2
⃗ B A
A(-8,4)
)
1.5 −1
M O
Now the coordinates of M can be found from the ⃗ position vector O M where
⃗ O M
= =
⇒
⃗ + B M ⃗ O B 3 4
+
1.5
−1
=
4.5 3
The objective is to prove that :
(1)
Now:
= M C ; and (2) A B ⃗ = A O ⃗ + O B ⃗ A B
coordinates = M( 4.5, 3)
= Also:
⇒ so :
)
− −8 4
+
⃗ = M O ⃗ + O C ⃗ M C
=¼ O M
O A
= O M
−2
⃗ Giving M O © cxcDirect Institute - 876 469-2775 mail:
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= C B M A
=
1
= ¼
)
−5 7
)
=
)
3 3
but OM:OA = 1:4
−8 4
=
−2 1
− −2 1
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CxcDirect Institute
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Vectors - Lesson 1 & 2
1 − −2 4 1
= M C
so:
3 3
=
= M C A B
⇒
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⇒
1 1 a b− a 3 2
= P M
….... (1)
1
=
6
3b − a
**********************************************
= C B We now need to prove that : M A
Points P, M and N are collinear if:
= k P M M N now:
⃗ = M A
also
⃗ C B
=
3 4
⃗ O A
=
3
−8
4
4
−6
=
3
)
so we need to find M N
⃗ +O B ⃗ C O
= −
1
now
−5
+
4
7
−6
=
…
⃗ M N
where: M B
3
M N
so
= C B M A
⇒
)
( k is a constant)
(2)
⃗ + B N ⃗ = M B =
½ b − a ; and
=
½ b −a
=
+b
½ 3b − a
⃗ now note that M N
From (1) and (2), ABCM is a parallelogram.
=b B N
⃗ 3 P M
=
⇒ k=3
therfore P , M and N are on a stright line ( collinear)
Activity 4 : consider triangle OAN: To find the length of A N
N
b
B
⇒
=O A A N O N
⇒
=O N – O A = A N
⇒
A N
M A
=2
b 1/3 (a)
P
= 2/3 (a)
2b − a
1 2
2 4
-
-
6 2
6 2
=
−4 2
O
Length =
Finding:
⃗ + O B ⃗ A O
= A B
−a + b
=
Finding
now:
= P A
1 3
2
=
4.47
************************************************
b−a
a
=
⃗ + A M ⃗ P A
= now A M = MB =½ A B
⃗ P M
=
⃗ + O B ⃗ − A O
2
⃗ P A
⃗ Finding P M
so
=
−4 2
=
1 3
a
+
1 2
½ b −a
( a− b)
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