4.3 Addition and Subtraction of Vectors


4.3.2 Subtraction of Vectors
The subtraction of the vector b ˜ from the vector  a ˜ is written as  a ˜ b ˜ . This operation can be considered as the addition of the vector a ˜ with the negative vector of b ˜ . Therefore a ˜ b ˜ = a ˜ + ( b ˜ ) .  


Example 1:

In the diagram above, vector OP = p ˜ ,   OR = r ˜ and Q divides PR in the ratio of 2 : 3. Find the following vectors in terms of p ˜  and  r ˜
( a )  PR ( b )  OQ ( c )  QM  if M is the midpoint of OR.

Solution:
(a)
P R = P O + O R = O P + O R = p ˜ + r ˜

(b)
O Q = O P + P Q = O P + 2 5 P R = p ˜ + 2 5 ( p ˜ + r ˜ ) = p ˜ 2 5 p ˜ + 2 5 r ˜ = 3 5 p ˜ + 2 5 r ˜

(c)
Q M = Q O + O M = O Q + O M = O Q + 1 2 O R = ( 3 5 p ˜ + 2 5 r ˜ ) + 1 2 r ˜ = 3 5 p ˜ 2 5 r ˜ + 1 2 r ˜ = 3 5 p ˜ + 1 10 r ˜

Long Question 5


Question 5:
Diagram below shows a triangle KLM.


It is given that KP:PL=1:2, LR:RM=2:1,  KP =2 x ˜ ,  KM =3 y ˜ . (a) Express in terms of  x ˜  and  y ˜ , (i)  MP (ii)  MR (b) Given  x ˜ =2 i ˜  and  y ˜ = i ˜ +4 j ˜ , find  | MR | . (c) Given  MQ =h MP  and  QR =n KR , where h and n are constants,    find the value of h and of n.


Solution:
(a)(i)
MP = MK + KP   =3 y ˜ +2 x ˜   =2 x ˜ 3 y ˜

(a)(ii)
MR = 1 3 ML   = 1 3 ( MK + KL )   = 1 3 ( 3 y ˜ +6 x ˜ )   =2 x ˜ y ˜

(b)
MR =2( 2 i ˜ )( i ˜ +4 j ˜ )   =4 i ˜ + i ˜ 4 j ˜   =5 i ˜ 4 j ˜ | MR |= 5 2 + ( 4 ) 2    = 41  units

(c)
MQ + QR = MR h MP +n KR = MR h( 2 x ˜ 3 y ˜ )+n( KM + MR )=2 x ˜ y ˜ h( 2 x ˜ 3 y ˜ )+n( 3 y ˜ +2 x ˜ y ˜ )=2 x ˜ y ˜ 2h x ˜ 3h y ˜ +2n x ˜ +2n y ˜ =2 x ˜ y ˜ ( 2h+2n ) x ˜ +( 3h+2n ) y ˜ =2 x ˜ y ˜ 2h+2n=2..........(1) 3h+2n=1..........(2) ( 1 )( 2 ):5h=3  h= 3 5 From ( 1 ):h+n=1 3 5 +n=1    n=1 3 5    n= 2 5

Short Question 4 & 5


Question 4:
Diagram below shows a parallelogram ABCD with BED as a straight line.


Given that  AB =7 p ˜ ,  AD =5 q ˜  and DE=3EB, express, in terms of  p ˜  and  q ˜ . (a)  BD (b)  EC

Solution:

(a)
Note: for parallelogram, A B = D C = 7 p ˜ , A D = B C = 5 q ˜ . B D = B A + A D B D = 7 p ˜ + 5 q ˜  


(b)

D E =3 E B E B D E = 1 3 E B : D E = 1 : 3 E B = 1 4 D B = 1 4 ( B D ) = 1 4 [ ( 7 p ˜ + 5 q ˜ ) ] From (a) = 7 4 p ˜ + 5 4 q ˜ E C = E B + B C E C = 7 4 p ˜ + 5 4 q ˜ + 5 q ˜ E C = 7 4 p ˜ + 25 4 q ˜




Question 5:


Use the above information to find the values of h and k when r = 2p – 3q.

Solution:
r = 2 p 3 q ( h 1 ) a ˜ + ( h + k ) b ˜ = 2 ( 5 a ˜ 7 b ˜ ) 3 ( 2 a ˜ + 3 b ˜ ) ( h 1 ) a ˜ + ( h + k ) b ˜ = 10 a ˜ 14 b ˜ + 6 a ˜ 9 b ˜ ( h 1 ) a ˜ + ( h + k ) b ˜ = 16 a ˜ 23 b ˜ Comparing vector: h 1 = 16 h = 17 h + k = 23 17 + k = 23 k = 40

Long Question 2


Question 2:
Given that   A B = ( 10 14 ) , O B = ( 4 6 ) and C D = ( m 7 ) , find
(a) the coordinates of A,
(b) the unit vector in the direction of O A .
(c) the value of m if CD is parallel to AB .

Solution:

(a)
A B = A O + O B ( 10 14 ) = ( x y ) + ( 4 6 ) ( x y ) = ( 10 14 ) ( 4 6 ) A O = ( 6 8 ) O A = ( 6 8 ) A = ( 6 , 8 )


(b)
| OA |= ( 6 ) 2 + ( 8 ) 2 | OA |= 100 =10 the unit vector in the direction of  OA = OA | OA | = ( 6 8 ) 10 = 1 10 ( 6 8 ) =( 3 5 4 5 )


(c)
Given  CD  parallel  AB   CD =k AB ( m 7 )=k( 10 14 ) ( m 7 )=( 10k 14k ) 7=14k k= 1 2 m=10k=10( 1 2 )=5

Vectors In Cartesian Plane

(A) Vectors in Cartesian Coordinates
1. A unit vector is a vector whose magnitude is one unit.
2. A unit vector that is parallel to the x-axis is denoted by  i ˜ while a unit vector that is parallel to the y-axis is denoted by j ˜ .
3. The unit vector can be expressed in columnar form as below: OA =x i ˜ +y j ˜ =( x y )(Column Vector) 
4. The magnitudes of the unit vectors are  | i ˜ | = | j ˜ | = 1.

5. The magnitude of the vector O A can be calculated using the Pythagoras’ Theorem.
| O A | = x 2 + y 2
 
(B) Unit Vector in the Direction of a Vector
  Unit vector of  a ˜ ,  a ^ ˜ = x i ˜ +y j ˜ x 2 + y 2     

Example 1:
If   r ˜ = k i ˜ 8 j ˜ and | r ˜ | = 10 , find the values of k. Determine the unit vector in the direction of   r ˜   for each value of k.   

Solution:
r ˜ =k i ˜ 8 j ˜ Given | r ˜ |=10 x 2 + y 2 =10 k 2 + ( 8 ) 2 =10 k 2 +64=100 k=±6 Unit vector of   r ˜ ^ = x i ˜ +y j ˜ x 2 + y 2 When k=6,                                When k=6 r ˜ ^ = 6 i ˜ 8 j ˜ 10 = 3 i ˜ 4 j ˜ 5        ,         r ˜ ^ = 6 i ˜ 8 j ˜ 10 = 3 i ˜ 4 j ˜ 5 r ˜ ^ = 1 5 ( 3 i ˜ 4 j ˜ )            ,               r ˜ ^ = 1 5 ( 3 i ˜ 4 j ˜ )



Example 2:
It is given that a ˜ =( 6 3 ) and  b ˜ =( 3 7 ).  
(a) Find b ˜ a ˜  and | b ˜ a ˜ |.  
(b) Hence, find the unit vector in the direction of b ˜ a ˜ .

Solution:
(a)
b ˜ a ˜ = ( 3 7 ) ( 6 3 ) = ( 3 6 7 3 ) = ( 3 4 ) | b ˜ a ˜ | = ( 3 ) 2 + 4 2 = 9 + 16 = 25 = 5


(b)
The unit vector in the direction of  b ˜ a ˜ = 1 5 ( 3  4 ) =( 3 5   4 5 )

Short Question 1 – 3


Question 1:
Given that O (0, 0), A (3, 4) and B (9, 12), find in terms of the unit vectors,   i ˜ and   j ˜.
(a) A B
(b) the unit vector in the direction of  A B

Solution:
(a) 
A=(3,4), thus  OA =3 i ˜ +4 j ˜ B=(9,12), thus  OB =9 i ˜ +12 j ˜ AB = AO + OB AB =( 3 i ˜ +4 j ˜ )+( 9 i ˜ +12 j ˜ ) AB =3 i ˜ 4 j ˜ 9 i ˜ +12 j ˜ AB =6 i ˜ +8 j ˜

(b)
The magnitude of | AB |, | AB |= ( 6 ) 2 + ( 8 ) 2 =10 The unit vector in the direction of  AB , AB | AB | = 1 10 ( 6 i ˜ +8 j ˜ )= 3 5 i ˜ + 4 5 j ˜



Question 2:
Given that A (3, 2), B (4, 6) and C (m, n), find the value of m and of n such that    2 A B + B C = ( 12 3 )

Solution:

A=( 3 2 ), B=( 4 6 ) and C=( m n ) AB = AO + OB AB =( 3 2 )+( 4 6 )=( 7 4 ) BC = BO + OC BC =( 4 6 )+( m n )=( 4+m 6+n ) Given 2 AB + BC =( 12 3 ) 2( 7 4 )+( 4+m 6+n )=( 12 3 ) ( 144+m 86+n )=( 12 3 ) 10+m=12 m=2 2+n=3 n=5




Question 3:
Diagram below shows a rectangle OABC and the point D lies on the straight line OB.
 
It is given that OD = 3DB.
Express  OD  in terms of  x ˜  and  y ˜ .

Solution:

O B = O A + A B = 3 x ˜ + 12 y ˜ O D = 3 D B O D D B = 3 1 O D : D B = 3 : 1 O D = 3 4 O B = 3 4 ( 3 x ˜ + 12 y ˜ ) = 9 4 x ˜ + 9 y ˜

4.3.1 Addition of Vectors


4.3 Addition and Subtraction of Vectors

4.3.1 Addition of Vectors
1. The addition of two vectors, u ˜  and  v ˜ , can be written as u ˜ + v ˜ . The result of this addition is a vector which is called the resultant vector.

2. When two vectors with the same direction is added up, the resultant vector has
(a) the same direction with both the vectors.
(b) a magnitude equal to the sum of the magnitudes of both the vectors.


Addition of Non-parallel Vectors
1. Addition of two non-parallel vectors, u ˜  and  v ˜ , can be shown by using two laws. 
 
(a) Triangle Law of Addition

The resultant vector u ˜ + v ˜ is represented by the third side AC.




(b) Parallelogram Law of Addition

The resultant vector u ˜ + v ˜ is represented by the third side AC.

4.1 Vector


(A) Definition of Vector and Vector Notation
1. Vector is a quantity that has both magnitude and direction.
2. Scalar is a quantity that has magnitude only.
3. A vector can be presented by a line segment with an arrow, known as a directed line segment.





(B) Equality of Two Vectors
1. Negative vector of   A B has the same magnitude as   A B but its direction is opposite to that of A B .


2. A zero vector is a vector whose magnitude is zero. It is denoted by 0 ˜ .

3. Two vectors are equal if both the vectors have the same magnitude and direction.


Long Question 4


Question 4:
Diagram below shows quadrilateral ABCD. The straight line AC intersects the straight line BD at point E.


It is given that BE:ED=2:3,  AB =10 x ˜ ,  AD =25 y ˜  and  BC = x ˜ +15 y ˜ . (a) Express in terms of  x ˜  and  y ˜ , (i)  BD (ii)  AE (b) Find the ratio AE:EC.


Solution:
(a)(i)
BD = BA + AD   = AD AB   =25 y ˜ 10 x ˜

(a)(ii)
AE = AB + BE   = AB + 2 5 BD   =10 x ˜ + 2 5 ( 25 y ˜ 10 x ˜ )   =10 x ˜ + 2 5 ( 25 y ˜ 10 x ˜ )   =10 x ˜ +10 y ˜ 4 x ˜   =6 x ˜ +10 y ˜   =2( 3 x ˜ +5 y ˜ )

(b)
EC = EB + BC   = BC BE   = BC 2 3 ED   = BC 2 3 ( EA + AD )   = x ˜ +15 y ˜ 2 3 ( 6 x ˜ 10 y ˜ +25 y ˜ )   = x ˜ +15 y ˜ 2 3 ( 6 x ˜ +15 y ˜ )   = x ˜ +15 y ˜ +4 x ˜ 10 y ˜   =3 x ˜ +5 y ˜ AE EC = 2( 3 x ˜ +5 y ˜ ) 1( 3 x ˜ +5 y ˜ ) AE:EC=2:1

4.4 Expression of a Vector as the Linear Combination of a Few Vectors

4.4 Expression of a Vector as the Linear Combination of a Few Vectors
1. Polygon Law for Vectors

P Q = P U + U T + T S + S R + R Q

2.
To prove that two vectors are parallel, we must express one of the vectors as a scalar multiple of the other vector.

For example, AB =k CD  or  CD =h AB . . 

3.
To prove that points P, Q and R are collinear, prove one of the following.

   PQ =k QR  or  QR =h PQ    PR =k PQ  or  PQ =h PR    PR =k QR  or  QR =h PR



Example:
Diagram below shows a parallelogram ABCD. Point Q lies on the straight line AB and point S lies on the straight line DC. The straight line AS is extended to the point T such that AS = 2ST.


It is given that AQ : QB = 3 : 1, DS : SC = 3 : 1, AQ =6 a ˜  and  AD = b ˜   
(a) Express, in terms of a ˜  and  b ˜ :   
 (i)  AS    (ii)  QC
(b) Show that the points Q, C and T are collinear.


Solution:
(a)(i)  AS = AD + DS  = AD + AQ AQ:QB= 3:1 and  DS:SC= 3:1 AQ = DS  = b ˜ +6 a ˜  =6 a ˜ + b ˜


(a)(ii)  QC = QB + BC   = 1 3 AQ + AD AQ:QB= 3:1 AQ QB = 3 1 QB= 1 3 AQ and for parallelogram,  BC//AD, BC=AD    = 1 3 ( 6 a ˜ )+ b ˜   =2 a ˜ + b ˜


(b)  QT = QA + AT   = QA + 3 2 AS AS=2ST AT=3ST= 3 2 AS   =6 a ˜ + 3 2 ( 6 a ˜ + b ˜ )   =3 a ˜ + 3 2 b ˜   = 3 2 ( 2 a ˜ + b ˜ )   = 3 2 QC Points Q, C and T are collinear.