This belongs to Scalars and Vectors

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1

2. This is a REGULAR CLASS. Please make a
note by writing down what you have
seen here.
Make a separate note for CURRENT
ELECTRICITY.

3. UDAY KHANAL
Department of Physics
CCRC
3

4. Addition of two vectors
Vectors are not added like scalars but they are added
graphically using special rules – Vector algebra.

5. N
E
𝐴 𝐵 𝐴 + 𝐵
2 m/s 3 m/s 5 m/s
𝐴
5 m/s
𝐵
2 m/s
3 m/s Direction of bigger
preserves
Addition In one Dimensions
In same direction
In opposite direction

6. Note:
Vectors can be added, subtracted and multiplied
according to the vectors algebra but can not be
divided.

7. 𝑨 and 𝑩 are in the same order
𝐴
𝐵
𝐴
𝐵
𝑨 and 𝑩 are in the opposite order
Same and opposite order
H
T
H
H

8. In two dimension
• Triangle rule
• Parallelogram rule

9. Triangle law of vector addition
Statement:
If two non zero vectors 𝑨 and 𝑩 are represented by two sides of a triangle taken in the same
order then the closing side of the triangle taken in the opposite order represents the resultant 𝑅
of these two vectors.
𝐴
𝐵

10. Let OP represents the magnitude and direction of the 𝐴
and PQ that makes an angle 𝜃 with OP represents the
magnitude and the direction of the 𝐵 .
Magnitude of Resultant:
Draw QN perpendicular to OP
produced.
In triangle OQN
OQ2 = ON2 + QN2
= (OP + PN)2 + QN2
= OP2 + 2OP PN + PN2 + QN2
= OP2 + 2OP PN + PQ2
=OP2 + 2OP PQ cos𝜃 + PQ2
or R2 = A2 + 2 AB cos𝜃 + B2
= A2 + B2 + 2AB cos𝜃
∴ 𝑹 = A2 + B2 + 2AB cos𝜽
∵
𝑃𝑁
𝑃𝑄
= 𝑐𝑜𝑠𝜃,
∵ PN2 + QN2 = PQ2
∵ OP = |𝑨|= A, PQ = | 𝑩 | = B
Analytical treatment
𝐴
𝐵
- - - - - - - -
-
-
-
-
-
-
-
-
--
--
-
O P N
Q
𝛼 𝜃
Complete the triangle OPQ.
Then the closing side OQ of the triangle
represents the magnitude and direction of the
resultant 𝑅 of 𝐴 and 𝐵 as shown in the figure.

11. 𝜽 is the angle
between 𝑨 and 𝑩.
𝐴
𝐵
- - - - - -
-
-
-
-
-
-
-
-
--
-
O P N
Q
𝛼 𝜃
𝑄𝑁
𝑃𝑄
= 𝑠𝑖𝑛𝜃
𝑃𝑁
𝑃𝑄
= 𝑐𝑜𝑠𝜃,
Direction of Resultant

12. Special cases:
∴ 𝑹 = A2 + B2 + 2AB cos𝜽

13. Parallelogram law of vector addition
Statement:
If two non zero vectors 𝑨 and 𝑩 are represented by two adjacent sides of a parallelogram, then
the diagonal of the parallelogram passing through the point of intersection of these two vectors
represents the resultant 𝑅 of 𝐴 and 𝐵 .
𝐴
𝐵

14. Analytical treatment
𝐴
𝐵
- - - - - - - -
-
-
-
-
-
-
-
-
--
--
-
O P N
Q
S
𝛼
𝛽
𝜃
Let OP represents the magnitude and direction
of 𝐴 and OS that makes an angle ( 𝛼 + 𝛽 = 𝜃)
with OP represents the magnitude and direction
of 𝐵.
Complete the parallelogram OPQS.
Then the diagonal OQ represents the magnitude
and direction of the resultant 𝑅 of 𝐴 and 𝐵 as
shown in the figure.
Magnitude of Resultant :
Draw QN perpendicular to OP
produced.
In triangle OQN
OQ2 = ON2 + QN2
= (OP + PN)2 + QN2
= OP2 + 2OPPN + PN2 + QN2
= OP2 + 2OPPQ cos𝜃 + PQ2
or R2 = A2 + 2 AB cos𝜃 + B2
= A2 + B2 + 2AB cos𝜃
∴ 𝑹 = A2 + B2 + 2AB cos𝜽
∵
𝑃𝑁
𝑃𝑄
= 𝑐𝑜𝑠𝜃,
∵ PN2 + QN2 = PQ2
∵ OP = |𝑨|= A, PQ = | 𝑩 | = B

15. 𝜽 is the angle
between 𝑨 and 𝑩.
𝐴
𝐵
- - - - - - - -
-
-
-
-
-
-
-
-
--
--
-
O P N
Q
S
𝛼
𝛽
𝜃
Similarly, 𝜷 = 𝐭𝐚𝐧
_𝟏
𝑨𝒔𝒊𝒏𝜽
𝑩+𝑨𝒄𝒐𝒔𝜽
𝑄𝑁
𝑃𝑄
= 𝑐𝑜𝑠𝜃
𝑃𝑁
𝑃𝑄
= 𝑐𝑜𝑠𝜃,
Direction of Resultant

16. Special cases

18. 1 Two vectors having equal magnitude A make an angle 𝜽 with each other. Find the magnitude and
direction of the resultant.
2 If the magnitude of two vectors are equal to the magnitude of their resultant, find the
direction of the resultant.
3 At what angle the magnitude of two vectors are equal to the magnitude of their
resultant?

21. Polygon law of vector addition
Statement:
If a number of non zero vectors are represented by n-1 sides of a n sided
polygon taken in the same order than the closing side of the polygon in the
opposite order represents the resultant 𝑹 of them.
A
B
C
D
A+B
A+B+C
R=A+B+C+D
A
B
C
D

22. Addition Properties :
1: A+B+C = C+A+B = B+C+A : commutative
2: α(A+B) = αA+ αB : Distributive
3: A+(B+C) =(A+B)+C :Associative

23. Subtraction between two vectors: Triangle Law concept
A
B
A-B = A+(-B)
B
A
A-B
B-A = B+(-A)
B
A
- - - - - - - - -
-A
-B
B-A

24. A
B -B A
B
A
A-B
-B
A
-B
A-B
-B
A
A-B
Methods: 3 Different ways (H-T)

25. Subtraction between two vectors: Parallelogram law
concept
A
B
A-B
B-A
A-B = 𝐴2 + 𝐵2 + 2𝐴𝐵𝑐𝑜𝑠(180 − 𝜃)
Or = 𝐴2 + 𝐵2 − 2𝐴𝐵𝑐𝑜𝑠𝜃
𝑡𝑎𝑛𝛼 =
𝐵𝑠𝑖𝑛 180 − 𝜃
𝐴 + 𝐵𝑐𝑜𝑠(180 − 𝜃)
Or =
𝐵𝑠𝑖𝑛𝜃
𝐴−𝐵𝑐𝑜𝑠𝜃
A-B = A+(-B)
. . . . . . . . . . . .
B
-B
A
𝜃
B-A = B+(-A)
. . . . . . . . . . . .
B
-A A
. . . . . . . . . . . .
𝜃

26. A
B
A-B
B-A
A-B = 𝐴2 + 𝐵2 + 2𝐴𝐵𝑐𝑜𝑠(180 − 𝜃)
Or = 𝐴2 + 𝐵2 − 2𝐴𝐵𝑐𝑜𝑠𝜃
𝑡𝑎𝑛𝛼 =
𝐵𝑠𝑖𝑛 180 − 𝜃
𝐴 + 𝐵𝑐𝑜𝑠(180 − 𝜃)
Or =
𝐵𝑠𝑖𝑛𝜃
𝐴−𝐵𝑐𝑜𝑠𝜃
A-B = A+(-B)
. . . . . . . . . . . .
B
-B
A
𝜃
B-A = B+(-A)
. . . . . . . . . . . .
B
−𝑽𝑨 A
. . . . . . . . . . . .
𝜃

27. A
B
A
-B
A
-B
Method: Only one way (T-T)
A-B
. . . . . . . . . . . .
B
-B
A
𝜃

28. A
B
A-B
. . . . . . . . . . . .
B
-B
A
. . . . . . . . . . . . .
A+B
-(A+B)
B-A
. . . . . . . . . . . .
B
-A
. . . . . . . . . . . . . . . . . . . . . . . .
𝜃

29. R = A-B = 𝐴2 + 𝐵2 − 2𝐴𝐵𝑐𝑜𝑠𝜃 𝛼 = tan
_1 𝐵𝑠𝑖𝑛𝜃
𝐴−𝐵𝑐𝑜𝑠𝜃
Find the value of R and 𝛼 for 𝜃 = 0°, 90°, 180°, 𝐴 = 𝐵, 𝑎𝑛𝑑 𝐴 = 𝐵 = 𝑅
Work Out

31. 𝑜𝑟, 𝛼 = tan
_1(
60
25
)
= 𝟔𝟕. 𝟑𝟖°

32. The essence of SCIENCE: ask an impertinent question, and
you are on the way to a pertinent answer.

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