1. vectors

Scalar and Vector quantity

There are many physical quantities in Physics. Like pressure, density, force, speed, temperature, acceleration, etc. These quantities can be classified into two categories- scalar and vector quantities.

Scalar quantities are those which have only magnitude. For example, density, temperature, distance, speed, pressure, etc and vector quantities have both magnitude and direction associated with them like force, acceleration, velocity, etc.

This section will learn about vectors and their properties in detail.

Position and displacement vector

Position vector

A position vector is defined as a vector that symbolizes either the position or the location of any given point with reference to the origin. The direction of the position vector always points from the origin of the vector towards the given point.

For example, Suppose an object is placed at coordinate (2, 3,4 ) in a rectangular coordinate system. The position of this point can be expressed as position vector     OP= 2 i +3 j+4k

Displacement vector

The change in the position vector of an object is known as the displacement vector.

Suppose an object is at position A at time t=0 and after some time ‘t’ it reaches position B. Let the position of A is (x1, y1, z1)  and B is


Position vector of A  = OA= (x1 i+y1 j+z1 k)

Position vector of B= OB= (xi+ y2 j+z2 k)

So the displacement of AB is given by displacement vector AB

AB= OB-OA = ( (x2-x1)i +(y2-y1)j+(z2-z1)k)

The displacement of an object can be defined as the vector distance between the initial point and the final point.

Suppose an object travels from point A to point B via curve path in pink. The displacement of the particle would be the vector line AB headed in the direction A to B. The direction of the displacement vector is always from the initial to the final position.

General vector and their notations

A vector has both magnitude and direction. The length of the vector represents its magnitude and the tip of the vector specifies its direction

In cartesian coordinates, we generally express a vector in 3-D in terms of its projection along the x, y and z-axis.

In the figure given above Vector A=Ax i+ Ay j+Az k.

Here Ax, Ay and Az is the projection of vector A in direction x, y and z.

Equality of vector

Two vectors are equal if all the components of the vector are equal. In other words, if both the magnitude and direction of the vector is equal then only two vectors are equal.

Negative of the vector: The vectors and their negatives have the same magnitude but opposite directions.

Vector addition: Always add vectors from tip to tail. Place the tip of the first vector next to the tail of the second vector. The Result is the sum of the first two vectors.

Vector subtraction: When we have to subtract two vectors : A- B = A + (-B) . Then first take the negative of the second vector and then add them according to the rule of vector addition.

Vector addition and subtraction in 1 dimension

Vector  addition subtraction in 2 dimension

2. resolution of vectors

Relative velocity

Relative velocity is the velocity of one object with respect to another object.

Every motion is relative as it has to be observed with respect to an observer. Relative velocity is a measurement of the velocity of an object with respect to another observer. He is defined as the time rate of change of relative position of one object with respect to another.

Most of you must have traveled on a train someday, you would have probably noticed that the other train crossing your train from the opposite side seems to be moving at a very high speed, It is because the relative velocity of the other train is the sum of your train speed and its actual speed.

Unit vectors

Unit vector is a vector with unit magnitude. It is used to specify direction.  i, j and k are the unit vectors along x, y and z-direction respectively.

  • We can define the unit vector of any vector. Unit vector of a vector is of unit length and along with the same direction as that of the vector

  • If we have a vector A= Ax i + Ay j+ Az k

And the magnitude of the vector A is given by |A|= (Ax^2+Ay^2+Az^2 )

Then unit vector of A=  A/(|A|)

Below is one the example

Resolution of a vector in a plane

Consider a vector in x-y plane  a^→= ax i+ ay j

Here ax and ay is the x and y component of Vector A.

Suppose vector A makes an angle θ with the horizontal then

ax= |a| cosθ    and  ay= |a| sinθ  where |a|= magnitude of vector a

We have two ways to specify a vector

  1.  When we have the magnitude of the vector A and angle θ made with the horizontal.

A= Acosθ i+ Asinθ j

  1. When we have its horizontal and vertical components Ax and Ay.

A= Ax i + Ay j

If there are two ways to represent the same vector, there must be some relation connecting the two representation

The magnitude of vector A=(Ax^2+Ay^2 )

tanθ= Ay/Ax    ;  θ= tan^(-1) (Ay/Ax)

Similarly, we can resolve any vector quantity in its horizontal and vector component as shown above.

In the figure below the velocity vector with magnitude 50m/s which is at 60 degrees from the horizontal is resolved.

Another example of the resolution of force. Suppose a dog is held by its neckband with a force of 60 N acting at an angle of 40 degrees with the horizontal.

Vector Addition by rectangular components.

Finding the resultant of the vectors by using the component method.

Suppose we have two vectors given below and we need to find the resultant of these two vectors.

  • First we will resolve the two vectors in their components and write them in vector notation.
  • Vector A makes an angle of 70 with the positive horizontal direction( +x axis) and has a magnitude A= 3.6 m.

Ax= A cos 70  = 3.6 cos70= 1.23

Ay= A sin 70 = 3.6 sin 70= 3.38

        So vector A= 1.23 i + 3.38 j

  • Similarly, vector B has a magnitude of 2.4 and makes an angle of 30 with the negative x-axis. When we resolve vector B, the components Bx and By are along the -x and -y-axis as this vector is in the third quadrant.

Bx= B cos 30= 2.4 cos 30=2.078

By= B sin 30=2.4 sin 30= 1.2

vector B= 2.078 (-i) + 1.2 (-j)= -2.078 i -1.2 j

  • Now we have vector A and Vector B, we will now do the vector addition. We will add ith component of the A with ith component of B and Jth component of A with J^th component of B.

A+B = (1.23 +(-2.078) )i + (3.38 + (-1.2) )j

So the resultant vector  R= -0.848 i +2.18 j

  • Here the  i^th component of the resultant vector R is negative and the J^thcomponent of the resultant vector R is positive. So this R vector lies in the 2nd quadrant.
  • Now tanθ=Ry/Rx=2.18/(-0.848)= -2.5708

θ= tan^(-1) (-2.5708)= -68.744 degree .

Therefore it makes 68.744 degrees clockwise  with the negative X -axis.

Dot product :

The dot product is one way of multiplying two or more vectors. The resultant of the dot product of the vector is a scalar quantity, That’s why the dot product is also called a scalar product.

Suppose we have two vectors  a and b with components (a1, a2, a3 ) and (b1, b2, b3)

Then the dot product of vectors a and b will be given as

ab= a1*b1+ a2*b2+a3*b3

When the magnitudes of vectors and the angle between them is given then the dot product ab= |a||b| cosθ

A Properties of dot product:

  • The dot product of two perpendicular vectors is zero. Two vectors are orthogonal only when ab=0

ab=|a||b|  cos90= 0

  • Dot product is commutative.   ab= ba
  • Dot product is distributive.     a(b+c) = ab+ac
  • Scalar multiple property :  (xa)∙(yb)= xy (ab)
  • Since i, j and k are the unit vectors along the x, y and z-axis. These unit vectors are perpendicular to each other.

So , ij=ji=ik=ki= jk=kj=0

Also , ii=jj=kk=1

Let me show you this with an example

Cross product

The Vector product of the two vectors refers to a vector that is perpendicular to both of them. In other words, we can say that the cross-product of two vectors is a vector that is orthogonal to both.

If we have two vectors  a and b and the angle between them is θ

Then  c= a×b=|a||b| sinθ   n ̂   

When the two vectors a and b have components (ax, ay, az) and

 (bx, by, bz) respectively then a×b  will be given as

Let me show a solved example

Suppose we have two vectors x1 and x2 with components (2,-3,1) and (-2,1,1) respectively.

Properties of the cross product

  • Cross product is anticommutative : a×b= -b×a.
  • Cross product is distributive : a×(b+c)= a×b+ a×c.
  • Cross product of two parallel vectors is zero.

a×b= |a||b| sinθ= |a||b|sin 0=0

  • Cross product of two vectors is equal to the area of the parallelogram formed by two vectors.

  • The direction of the cross product is given by the right-hand rule.

  • Cross product of the unit vectors i, j and k will be given as.

  • Also i×i=j×j=k×k=0
  • Two vectors are parallel if their cross product is zero and vice versa.

3. Motion in 2-D plane

Motion in a plane

Motion is a plane (2-D) that will be difficult to understand if we would treat it like a 2-D  motion. For simplicity, we can break the motion of 2-D into two independent motions along with X and Y directions ( two 1-D motions) with which we are already familiar.

We use the equation of motion in 1-D with subscripts x and y for motion along the X and Y direction.

We resolve the force in the X and Y direction and call it Fx and Fy.

Then from the force components, we find acceleration along the x and y-direction. 

 ax=Fx/m  ; and ay= Fy/m

 a= ax i + ay j

Then similarly we can resolve the initial velocity ‘u’ along the x and y direction as ux and uy.

Initial speed  u= ux i + uy j

Then we can use the equation of motion separately for x and y directions and find the final velocity and displacement along x and y (vx , vy ,x,y).

final speed   v= vx i + vy j

displacement  r= x  i + y j

Projectile Motion

Motion of an object under gravity is called projectile motion.

Projectile motion are of three types: Oblique projectile, horizontal projectile and projectile motion along an inclined plane.

In this discussion, we will discuss oblique and horizontal projectiles.

Figure given below is an example of a horizontal projectile. Here an object is thrown with velocity ‘u’ in the horizontal direction from a height of H and the object follows a projectile path.

Figure given below is the example of the oblique projectile in which an object is thrown from a ground with velocity u making some angle Ө with the horizontal.

Oblique projectile

In this type of projectile, an object is thrown with the initial velocity ‘u’ which makes an angle θ with the horizontal direction. The only force here is gravity which acts in ‘y direction’. There is no force along the x-axis.

Fx=0 , Fy=mg

Also ax=0 , ay=g  (downward)

We will discuss and calculate Horizontal distance, maximum height and time of flight T. As you can see from the projectile path shown below that the motion is symmetric about maximum height.

To calculate maximum height and time of flight we will use the ‘y - direction’ motion. And to calculate the range we will use the equation of x-motion.

Since there is a force along only the y-direction hence acceleration to velocity along with the y-axis changes while velocity along y remains the same as no force is acting on x.

Components of initial velocity ‘u’ : ux= u cos θ , uy= u sin θ

Acceleration  ax=0 , ay=g (downward)

We know that at maximum height velocity along y will be zero.

Vy=0  at t=T/2   where T= time of flight. Using this we have,

v_(y_ )=u_y+a_y t   ;  0=u sinθ -g(T/2)  ;      u sinθ= g(T/2)

Therefore we have the time of flight   T= 2 u sinθ/g

Now we will find maximum height using the 2nd equation of motion, here we will consider the first half of motion. (t=T/2) and y= ymax =H

v_y^2= u_y+2a_y y  ,    

 0= u^2 sin^2 θ-2gH  ;    2gH= u^2 sin^2 θ .

Therefore we have a maximum height of H=u^2 sin^2 θ/(2g)

To calculate the range we need to use x motion. Motion of X direction is uniform motion as ax=0  so , x= U_x*Time of flight

R= u cosθ* (2 u sinθ/g)=u^2 (2 sinθ cosθ)/g= u^2  sin 2θ/g

Therefore we have Range  R= u^2 sin 2θ/g

Equation of path of projectile motion.

 Let initial position is (x0 ,y0) then  y (t)  , y as a function of time will be given as

 y(t)= y_0+u_y t+ 1/2 gt^2    ; u_y= u sinθ ,    t=((x-xo))/(u cosθ)

When (x0 , yo)= (0,0)   when object starts from origin

 Then equation of path of projectile motion will be given as

 y=x tanθ-(g x^2)/(2 u^2 cos^2 θ)

This is an equation of a parabola and hence we get a parabolic path for an object moving in a plane under gravity.

Discussion over oblique projectiles.

  • Range of the projectile is maximum for θ=45.

R=u^2 sin2θ/g  R is max when sin 2θ=1; 2θ=90; θ=45

  • Projectile for different initial speeds for a given angle.

  •  The velocity with which the object strikes the ground is same as with it was thrown.

Horizontal projectile

In a horizontal projectile, an object is thrown from a height ‘ h’  with a horizontal velocity ‘vo’. Here the only force acting on the object is gravity.

There is no force acting in the horizontal direction. This is exactly like an oblique case.

Fx=0 , Fy=mg

Also ax=0 , ay=g  (downward)

We can find the time of flight ‘T’, horizontal distance ‘s’ and final velocity ‘v’ by using the equation of motion in x and y directions.

To find the time of flight we will use the equation of motion in the y-direction

Here Y=h, uy=0  as initially there was only horizontal velocity ‘vo’

y=u_y t+1/2 gt^2   ;  h= 0+ 1/2 gt^(2   )  ;      t^2=2h/g ;  t=√(2h/g)

Therefore the time of flight T=√(2h/g)

Now to find the horizontal distance traveled we will use motion along the x-direction. As motion along the x-direction is uniform motion

So  X distance= speed in x-direction* time of flight

s= vo √(2h/g)

Therefore horizontal distance traveled is  s= vo √(2h/g)

To find the final velocity of the object we need to use both motions along x and y-direction.  We will find the final velocity along the x and y direction and then find its magnitude.

v_y=u_y+ a_y t    ;    v_y=0 +gt=g √(2h/g)

 Thus final velocity along y   v_y=√2gh

Final velocity along x  v_x=vo, as there is no change in velocity along the x-direction.

The magnitude of final velocity v=√(vx^2+ vy^2 )= √( vo^2+2gh)

As you can see in the illustration given below the horizontal remains the same but the vertical velocity was initially zero but keeps on increasing as the object is falling.

Example of horizontal projectile

Consider a horizontal projectile with an initial speed vo= 10 m/s. It is given that it takes 5 seconds to reach the ground level. We have to find the height, final vertical velocity and the horizontal distance traveled.

Uniform circular motion

Motion of an object in a circular path with uniform speed is called uniform circular motion.

In a uniform circular motion, the speed of the object remains the same but the velocity changes throughout the motion as the direction of velocity changes. As you can see from the picture given below that velocity is always tangent to the path and its direction changes continuously.

Force needed to keep the object moving in the circular path is called centripetal force which is always directed towards the center. So there is an acceleration acting on the particle which is toward the center and called centripetal acceleration.

We have centripetal force Fc= -m v^2/r  and hence centripetal acceleration   a_c= -v^2/r, where r is the radius of the circle.

Time period of uniform circular motion T= circumference/ velocity

T=2Πr/v, this would be the time taken by the particle in one complete revolution.

Frequency of the uniform circular motion f=v/2Πr

Angular frequency ‘w’ is the rate of change of angular displacement θ


If the distance traveled in Δt is Δs then the velocity will be given as

v=(Δs )/Δt    ;  here  s=rθ    ; v=r Δθ/Δt=r w ;    v=rw         

We can express  fc, ac, T, and f in terms of  w

centripetal force   fc=-m v^2/r=-mw^2 r 

centripetal acceleration  ac= -w^2 r 

Time period  T= 2Π/w

frequency  f= w/2Π