Vector
Quantities (anything that can be measured)
– 2 types
• Scalar
• Vector
Scalar Quantity: The quantities which can be
described by magnitude only, the direction is not
required, are called a scalar quantity. Ex- Length,
mass, speed, etc.
Vector Quantity: The quantities which require both
magnitude and direction for their description are called
vector quantities. Ex- displacement, velocity, force
weight, acceleration, etc.
Scalar
Magnitude
Vector
Magnitude
Direction
Obey the law of
vector addition
Represented vector quantity
vectors geometrically, we use a line segment
with an arrow at one end. The length of
the line represents the magnitude of the vector
and the arrow represents the direction of the
vector.
Some important topic on vector
Keep the magnitude of a vector unchanged it can
be transferred in a position parallel to its previous
position. This is known as the parallel shifting of a
vector. In this case, the vector remains unchanged.
The smallest included angle between two or
more vectors when they are joined tail-to-tail is
the required included angle of vectors.
Addition of Vectors:
 Geometric Method
 Analytical method
Resultant: The sum of two or more vectors is a new
vector. This new vector is called the resultant of the
other vectors.
Geometric Method:
1)Law of Triangle
2)Law of parallelogram
3)Law of Polygon
Triangle Law: If two vectors of the same kind
are represented as the two adjacent sides of a
triangle in a similar order then the third side of
the triangle denotes the resultant of the two
vectors in the opposite order
Law of Parallelogram: If two adjacent arms drawn
from an angular point of parallelogram represents
the magnitude and direction of the two vectors of
the same kind acting on a particle at a time, then the
diagonal parallelogram drawn from the point will
represent the magnitude and direction of the
resultant of two vectors.
Analytical method
Determining the magnitude of the resultant:
In ΔBCE,
Cos α=BE/BC
∴BE=Cos α. BC
∴
Sin α= CE/BC
CE=BC Sin α
In ΔACE,
𝐴𝐶 2 = 𝐴𝐸 2 + 𝐶𝐸 2
𝑅2 = (𝐴𝐵 + 𝐵𝐸)2 +𝐶𝐸 2
𝑅2 = 𝐴𝐵2 + 2. 𝐴𝐵. 𝐵𝐸 + 𝐵𝐸 2 + 𝐶𝐸 2
𝑅2 = 𝑃2 + 2. 𝑃𝑄. 𝐶𝑜𝑠 𝛼 + 𝑄2
∴𝑅=
𝑃2 + 2. 𝑃𝑄. 𝐶𝑜𝑠 𝛼 + 𝑄2
• Q-1: Two vectors of 4N and 6N are acting at 60°
angle. Determine the magnitude and direction of the
resultant. Ans: 8.72N, 𝜃 = 36.59
• Q-2: If two forces act perpendicularly on an
object, their resultant becomes 2 13𝑁 and their
maximum resultant is 10N. If the forces act at 120°
then determine their resultant. Ans: 2 7𝑁
Q-3: Two forces act on an object. The resultant
being perpendicular on the smaller force is 12N. If
the maximum resultant of an object is 18N, find the
two forces. Ans: 5N, 13N
Q-4: Two forces act in opposite direction and
their resultant is 10N. If two forces act
perpendicular to each other than resultant force
is 50N. Find the two forces. Ans: 30N, 40N
Q-5: If two forces act on an object then
maximum resultant is 28N and minimum
resultant is 4N. If The two forces act on the
object in such a way that the resultant is
perpendicular to the smaller force, Find their
included angle. Ans: 138.59°
• Q-6: The maximum and minimum resultant of
two forces acting on a point is 10 and 6
respectively. If each force is increased by 3N
then the forces create an angle of 90° with each
other. Determine the resultant and direction of
the new forces. Ans: 12.1N, 24.44°
• example 7: included angle between P and Q is
a and resultant magnitude is 5 𝑝2 + 𝑄2 | if
included angle between P and Q is (90° - a)
and resultant magnitude is 3 𝑝2 + 𝑄2 prove
that tan a =1/3
• Example 8: prove that the resultant of two
same vector bisect the angle between two
vectors
বিবিময় সুত্র commutative law
সংয োগ সুত্র Associative law
িণ্টি সুত্র distributive
Triangle law
𝑎
𝑏
𝑐
=
=
𝑆𝑖𝑛𝐴
𝑆𝑖𝑛𝐵
𝑆𝑖𝑛𝑐
Ex 9: The resultant of P and Q
force are 3Q , and which make 30 angle
with P. So prove that P=Q and P=2Q
• If 𝐴 + 𝐵 = 𝐴 + 𝐵 then find the angle
between them
• If 𝐴 + 𝐵 = 𝐴 − 𝐵 then find the angle
between them
9. two forces are 10N and 15N acting on a point
at time making respectively 30 and 60 angle with
x axis. Find the resultant of two force.
10. A force 14N act on object east of north 30 .
At the same time another force of 12N act on the
object 30 south of east. What is the resultant of
two force
Q-11: Amit drives a car 250km in direction East
of South at 30° angle. After that, he drives
another 250km in direction west of south at 30°
angle. Find the resultant and direction. Ans:
433km, 30°
Component
• Component rule and resolution of
vector: The process of dividing one vector
into two or more vectors is called resolution
of vector
• Perpendicular resolution : If we can find
components of R in perpendicular direction to
each other, then angle between components is
90
Use of component
Q1. why the handle of luggage is kept longer?
Q2. Pulling lawn roller is easier than pushing
Q3. Pulling a boat :
Q1. Two forces A1 and A2 acts on a point at a
time. Magnitude of A1 is 4 unit and it produces
30 angle with horizontal. Magnitude of A2 is 3
unit and it acts along the vertically. Find
horizontal and vertically component of resultant
Q2. Magnitude of two vector quantities P
and Q are 4 units and 3 units respectively. P
is acting along x-axis and Q is inclined with
x-axis at 60°. Find resultant and direction of
the two vectors. Also explain- if you can find
horizontal and vertical component of the
resultant
Q3. Magnitude of two vector quantities P
and Q are 10 units and 4 units respectively.
and Q is inclined with x-axis at 30° and P is
acting 60 angle with x-axis . Find resultant
and direction of the two vectors. Also
explain- if you can find horizontal and
vertical component of the resultant
Q4. A man rows a boat with velocity 20ms-1 at
angle 30 of north of west. Shoreline makes angle
of 15 south of west. Find the component of the
velocity of the boat along the shoreline and
perpendicular to the shoreline
Q5. Find the
resultant force
Boat river problem
The velocity of a boat and current is 8 kmh-1 and 4
kmh-1. The width of the river is 2 km
B
C
B
A
A
Figure 1
Figure: 2
C
B
C
A
Figure: 3
D
The velocity of a boat and
current is 8 kmh-1 and 4 kmh-1.
The width of the river is 2 km
B
A
Figure:1
• A. find the resultant velocity
• B, find AC distance
• C. find the required time to reach c point
C
B
C
The velocity of a boat and
current is 8 kmh-1 and 4 kmh-1.
The width of the river is 2 km
A
Figure 2
a. What should the boatman do to cross the river
exactly the opposite side
b. what will be the result in velocity?
c. find the required time to reach B point
B
The velocity of a boat and
current is 8 kmh-1 and 4 kmh-1.
The width of the river is 2 km.
𝛼 = 60
A
• Figure 3
a. what will be the resultant velocity?
b. Find the distance of AD
c. find the required time to reach D point
C
The velocity of a boat and
current is 12 kmh-1 and 4 kmh1. The width of the river is 4
km
B
A
Figure:1
• A. find the resultant velocity
• B, find AC distance
• C. find the required time to reach c point
C
The velocity of a boat and
current is 12 kmh-1 and 4 kmh1. The width of the river is 4
km
B
A
C
Figure 2
a. What should the boatman do to cross the river
exactly the opposite side
b. what will be the result in velocity?
c. find the required time to reach B point
The velocity of a boat and current
is 12 kmh-1 and 4 kmh-1. The width
of the river is 4 km. 𝛼 = 30
B
A
• Figure 3
a. what will be the resultant velocity?
b. Find the distance of AD
c. find the required time to reach D point
C
D
• The river velocity 2km/h and boat velocity is
4km/h. the river width 2km
a. What should the boat man do to cross the
minimum distance
b. Find the minimum time to cross the river
D
B
C
1.5km
30
30
A
Q2: The width of the river is 1.5 km and the velocity of the
current is 4 km/h. The boatman started along the path AB and
reached a long AC. The velocity of the boat is 3 km/h.
1) Determine the time to reach AC .
2) Is it possible to reach the point B, if the boat starts rowing
along the path AD?
3) What would be the distance and required time to reach the
other side, if the boat starts rowing along the path AC?
• A thief running along the river bed with
same direction of current, after seeing a
police boat opposite side of river. The river
velocity is 4ms-1 and boat is running with
velocity of 5ms-1. boat is making angle 60
with stream. What is velocity of man, if
police catch the thief
Unit vector: The vector whose magnitude is one
unit. The vector represents the direction. Except
for zero vector if any other vector quantity is
divided by its magnitude only then the unit vector
is achieved.
Rectangular unit vector : In three dimensional
cartesian co-ordinate system, the three axes X, Y
and Z mutually perpendicular, represent three unit
vectors which are called rectangular unit vector.
• Ex1: find vector of 5 unit along with x axis
• Ex2: find vector of 7 unit along with y axis
• Ex3: find vector of 5 and 7 unit along with x
and y axis
• Position vector : The vector which represents a
vector from origin to a point is called position
vector of that point
• Displacement : Change in position vector of
any particle is called displacement
Magnitude and unit vector
Q1. r=4i-6j+12k then find its magnitude and
parallel unit vector
Ex2: the coordinate of A(2,1,5) and B(-1,5,-2).
a. Determine 𝑂𝐴, 𝑂𝐵
b. Find the magnitude and unit vector of 𝑂𝐴, 𝑂𝐵
c. Find the magnitude and unit vector of 𝐴𝐵
Q3 : A = 2i − j + 3k , B = 3i − 2j − 2k
Determine the unit vector which is parallel to the
resultant vector
Q4. |A| = 5 and |B| = 6 find the 𝐴 + 𝐵

Y
A
60

B
X
Multiplication of vector
Multiplication of vector can be done in two
ways:
• scalar multiple/dot product
• vector multiple/cross product
• Scalar Multiplication:
the product is scalar
represented by ( . ) between two vectors
• 𝐴.𝐵. = |A| |B| cos  = AB cos 
Special case:𝜗 = 0,180,90,
Type-1:
• 𝐴.𝐵. = 𝑨𝒙𝑩𝒙 + 𝑨𝒚𝑩𝒚 + 𝑨𝒛𝑩𝒛
Type-2: .  =90
Type-3: Included angle between vectors,
Type-4: Projection
Type2
Ex 1: prove that two vectors A=2i+3j+k and
B=4i-2j-2k are perpendicular to each other
Ex 1: if two vectors A=ai-2j+k and B=2ai-aj-4k
are perpendicular to each other, find the value of
a
Type 3
Q1. Find the angle between the vector
A=2i+2j+k and B=2i+10j-11k
Q2. if A=6i-6j+5k and B=6i+j-6k find the angle
between A and B

 
A– B
B

Q1. find the value of
𝛼 𝑎𝑛𝑑 𝜎


 
A
Q2. a. find the angle produced by
C with axis Y
b. Verify mathematical whether
the vector B and A perpendicular
to each other or not
A+ B
w
Pò - 1
w
Pò - 2
^
^
^

A = 2i + 2j – k

^
^
B = 6i – 3j + 2k
Y
L(2,–1,1)

A
M(1,1,–2)

B

C
O
Z
^
N(1,–2,3)
X
• A. show the vector format of F1 and F2
• B. show that 𝐹1 +𝐹2 and 𝐹1 - 𝐹2 are
perpendicular to each other
The projection of b on a is b cos 𝜃
𝑎𝑏
𝑏 cos 𝜃 =
𝑎
The projection of a on b is 𝑎. cos 𝜃
𝑎𝑏
𝑎 cos 𝜃 =
𝑏
• A=i+2j+2k B=2i-3j+6k, then find the project
of the vector B upon the vector A
• A=2i+3j+2k B=4i+3j+ k, then find the
project of the vector B upon the vector A
• A=i+2j+2k B=2i-3j+6k, then find the project
of the vector A upon the vector B
• A=2i+3j+2k B=4i+3j+ k, , then find the project
of the vector A upon the vector B
Cross multiplication
 The product of cross multiplication is a vector
quantity.
C = | A x B | = AB sin𝜃
If A and B are two vectors and  is their included
angle, then the cross product of A and B is another
vector which is perpendicular to the same plane, when
(0 ≤ 𝜃 ≤ 180°) The direction of the product is found
from the right hand screw the rule.
Rectangular unit vector
𝑖 × 𝑖 = 𝑘 × 𝑘 = 𝑗 × 𝑗=0
𝑖×𝑗=𝑘 , 𝑗×𝑘 =𝑖 , 𝑘×𝑖=𝑗
𝑗 × 𝑖 = −𝑘 , 𝑘 × 𝑗 = −𝑖 , 𝑖 × 𝑘 = −𝑗
CROSS PRODUCT
𝐴=(a1 i + a2 j + a3 k ) 𝐵=(𝑏1 i + b2 j + b3 k )
𝐴 × 𝐵 = (a1 i + a2 j + a3 k ) × (𝑏1 i + b2 j + b3 k )
=a1 i × 𝑏1 i + b2 j + b3 k + a2 j × 𝑏1 i + b2 j + b3 k + a3 k ×
𝑏1 i + b2 j + b3 k
=a1 b2 k − a1 b3 j − a2 b1 k + a2 b3 i + a3 b1 j − a3 b2 i
=i (a2 b3 -a3 b2 ) − j(a1 b3 − a3 b1 )+ k(a1 b2 − a2 b1 )
i
𝐴 × 𝐵 = a1
𝑏1
j
a2
b2
k
a3
b3
EX.1 𝐴=2i − 4j + 3k and 𝐵=(i − 3j + 4k ) ,
find the 𝐴 × 𝐵
EX2.
𝐴 = 3i + 7j + 3k and
𝐵 = (i + 2j −
7k ) find the 𝐴 × 𝐵
EX3: 𝐴=i − 2j − 2k and 𝐵=(6i + 3j + 2k ) find
the 𝐴 × 𝐵 ans: 425
𝑣𝑒𝑐𝑡𝑜𝑟 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
Unit vector =
𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒
Unit vector=
Unit vector
𝐴×𝐵
𝐴×𝐵
Example: 𝐴 =2i + 4j − 5k and 𝐵=(i + 2j + 3k )
find the unit vector perpendicular to the plane
formed by two vectors
Two vectors parallel
If the two vectors are parallel, their cross
product is zero। conversely if 𝜃 = 0° , then | A x
B | = AB sin 0° = 0
ex:1 show that 𝐴=i + j + 2k and 𝐵=(3i + 3j +
6k ) are parallel to each other.
ex:2 find the value of a, if 𝐴=5i + 2j − 3k and
𝐵=(15i + aj − 9k ) are parallel to each other.
Area of parallelogram:
,Area of parallelogram= base × height
= 𝑂𝑃 × 𝑅𝑆
= 𝑂𝑃 × 𝑂𝑅 sin θ
= | 𝑂𝑃 × 𝑂𝑅 |
= |𝐴 × 𝐵|
Ex3: the two adjacent side of two vectors
𝐴 = 4i − 4j + k and 𝐵 = (2i − 2j − k ) formed a
parallelogram.
Determine
the
area
of
parallelogram. । ans: 6 2
Area of triangle
1
area of triangle = × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚
2
1
= × |𝐴 × 𝐵|
2
Example: the three vertex of a triangle are
respectively A(1,-2,-3), B(2,1,-1), C(1,3,-2). Find
the area of triangle. ans: 2.5 3
• a=a1i+a2j+a3k, b=b1i+b2j+b3k, c=c1i+c2j+c3k
• If the a,b,c be the three coplanar vector, then
(a× 𝑏). 𝑐 = 0
coplanner
The prove that A , B and
C are same plane
Y
L(2,–1,1)

A
M(1,1,–2)

B

C
O
Z
N(1,–2,3)
X

Y
A
60

B
Q2. |A| = 5 and |B| = 6
a. find the 𝐴 + 𝐵
b. Show that 𝐴 × 𝐵 is perpendicular to 𝐴 + 𝐵
X
Prove that, (𝐴. 𝐵)2 + 𝐴 × 𝐵 2 = A2B2
Subtraction of Vector:
Vectors cannot be subtracted but can only be
added. If two vectors are 𝐴 and 𝐵.
But if the negative vector of 𝐵 is added with 𝐴
,then we get a subtraction vector 𝐴 and 𝐵
That is, 𝐴 + −𝐵 = 𝐴 − 𝐵
Q1. if A=5N and B=6N, the included angle A
and B is 6০। determine ( A B or resultant
Q2. |A| = 5 and |B| = 6
Determine ( A B )

Y
A
60

B
X
• Relative velocity: Velocity of a rest or
moving object with respect to another rest
or moving object is called relative velocity.
To calculate relative velocity, you will add
negative velocity of yours to the velocity
of that object. From this, you can apply
parallelogram law to find resultant velocity
Figure 1,2,3 and 4 determine the velocity B
respect to A .
Determine the relative velocity of rain respect to man
• At some days it was raining at a velocity of
30ms-1 vertically. The man is running from
west to east at a velocity of 10ms-1, then at
which direction should you hold your umbrella
so that you can be safe from rain?
• Asad was watching the rainfall standing at the door
office house in rainyday. The rain was falling vertically
with velocity 6km/h. at this moment he saw a man walking
with an umbrella placing it 33.8 angle with vertically.
Another man by running by cycle with an umbrella placing
it 53.06 angle with vertically. Both of them were safe
from rain.
• a. find the velocity of the man walking from above stem.
Ans: 4km/h
• b. analysis the cause of placing of the umbrella at
different angles of the men in stem to save themselves
from rain. ans: 7.98km/h
EXAMPLE3: a man walking with velocity 5km/h
at the east to west. He observes that rain
vertically falling on him with velocity 12km/h.
What is the real velocity of rain ? Ans: 13km/h,
22.630
• A ship is going toward east with velocity 30km/h
and another ship is going towards south with velocity
40km/h. find the relative velocity of the second with
respect to the first. Ans: 50km/h. angle 53.13
• A car driver moving with 40km/h in the east
direction and see a truck is moving with double
velocity of count by hour in the north direction.
Which direction really the moving on? 40 5 ,26.57
• A bus is rising with uniform velocity 72km/h on
the slope a hill with an angle 30. At the time rain
starts to fall downward with uniform velocity 6ms1. When the rain almost end, the wind started flow
horizontally. a. At the starting the bus driver will
see in which direction to fall rain? Explain 47.27.
b. due to flow of wind, if the bus driver saw the rain
fall vertically down ward, then analysis
mathematically, the real value of direction of the
wind
Operator:
The mathematical operation which can transform a
quantity to another quantity is called an operator
Vector Operator:
In case of vector there is a vector differential operator which
is expressed by . This is known as nabla operator. This is one
kind of mathematical operation the vector differential
operation can be written in the following way by using
various components:  ^  ^  ^ 
=i +j +k
x y z
This is used to determine gradient, diverges and curl.
Gradient
 The differentiating any scalar field or scalar function
through vector operator we get gradient
 It is a mechanism to transform a scalar field into a vector
field
 It indicates the highest / maximum rate of change of a
scalar field and its direction is towards the maximum
change


^  ^  ^  
=  i + j + k 
y
z 
 x
^  ^ 
^ 
=i
+j
+k
x
y
z
If p(x, y, z) = 3xy2-4xy , then find the gradient
Divergence
The dot product of vector operator and the vector field is
called divergence of that vector field. 𝐷𝑖𝑣 = 𝛻. 𝜎
It's acts on a vector field.
The divergence of a vector field is scalar quantity.
Divergence = positive ; then the field is diverging from a point
So, volume increases and density decreases.
Divergence =negative ; then the field is converging from a
point
So, volume decreases and density increases.
If the divergence is zero, it is known as solenoid.
Ex-2: Determine the divergence of 𝐴 = 3𝑥𝑦𝑧 3 𝑖 +
2𝑥𝑦 2 𝑗 − 𝑥 3 𝑦 2 𝑧𝑘 at point (1, -1, 1)
Ex-3: Is 𝑟 = 𝑥 + 3𝑦 𝑖 + 𝑦 − 𝑧 𝑗 + (𝑥 − 2𝑧)𝑘
solenoid?
Curl
1. The cross product of any vector field with nabla operator
is called curl . Curl of vector field is a vector quantity.
2. It explain the rotation of the quantity
Let, 𝜑=(𝑥i + 𝑦j + 𝑧k ) is any vector field. Curl = × 𝜑
× φ=
𝑖
𝑗
𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑥
𝑦
𝑧
3. if the curl is zero then the vector is irrotational and
conserved.
• The divergence of curl vector field is zero that is
V.(V× 𝑉) = 0
• The curl of linear velocity is twice of angular
velocity V× V= 2𝜔
Ex-4: Determine the curl of 𝐴 = 𝑥𝑧 2 𝑖 − 2𝑥 3 𝑦𝑧𝑗 +
3𝑦𝑧 3 𝑘 at point (1, 1, -1).
Ex-6: 𝐴 = 6𝑥𝑦 + 𝑧 3 𝑖 + 3𝑥 2 − 𝑧 𝑗 + (3𝑥𝑧 2 − 𝑦)𝑘 .
Prove that, the vector field in non-rotational.