Addition of Vectors: Class 11 Physics Chapter 4 CBSE

Triangular Law of Vector Addition: Following the rule of parallel shifting, draw one vector and then draw another vector such that the head of one vector is joined to the tail of another vector, thereafter joining the tail of the former vector to the head of the later vector, we get the resultant vector of both vectors which is the addition of the vectors.

As an example, a displacement of 3 m east and a displacement of 4 m north, then find the addition of the vector

Let the vector = 3 m east and = 4 m north then their addition is

Hence the resultant of both vectors is 5 m northwest,so there are specific rule for vector addition,we can’t add them like the scaler quantity means the resultant can’t be 3 +4 =7

What is the resultant of two vectors when the angles between them is other than 90°

Let there be two vectors and then their addition is represented as follows

The resultant R of both vectors is given as

The vector sum of both vectors is not a simple addition,the method mentioned above is head to tail method,apart from that there are other methods of solving vector addition.

The resultant of head to tail method is applicable when both vectors are in a same line or perpendicular to each other,so when both vectors have a certain angle θ between them then their resultant vecor is evaluated by parallelogram rule of vector addition.

Parallelogram Law of Vectors addition: In this method, two vectors are drawn tail to tail as they are adjacent sides of a parallelogram and then geometrically completing the parallelogram,then the diagonal obtained by joining the vertex of both vectors to the opposite vertex of the parallelogram shows their resultant.

Draw two vectors P and Q such that they join each other tail to tail

Complete the parallelogram ABCD, then draw a DE⊥CE

The sides AD =Q=AB,DC =P=AB shows magnitudes of the vectors

BD will show the resultant of the vectors

Let both vectors have angle θ between them and the resultant forms an angle of α between them and the magnitude of the resultant vector is R

Considering the right ΔBDE

∠ABC = ∠DCE=θ

BD² = BE² + DE²

R² =(BC+CE)² +DE²

R² = BC² +CE² +2BC.CE + DE²

R² = Q² + CE²+ 2Q.CE + DE²…….(i)

Let us find the value of CE and DE

Considering the triangle ΔDCE

cos θ =CE/DC and sinθ = DE/DC

CE = DCcos θ =Pcos θ …..(ii) and DE =DCsin θ= Psin θ …..(iii)

Putting the value of CE and DE in equation (i)

R² = Q² + (Pcos θ)²+ 2Q.Pcos θ + (Psin θ )²

R² = Q² + P²cos² θ+ 2Q.Pcos θ + P²sin ²θ

R² = Q² + P²(cos² θ+sin ²θ)+2Q.Pcos θ

R² = Q² + P²+2PQcos θ

The above equation is used for determining the magnitude of the resultant of two vectors

Now

tanα =DE/BE

BE =BC +CE =Q+ Pcosθ

From equation (ii) and (iii)

tanα = Psinθ/(Q+ Pcosθ)

This equation is used to determine the direction of resultant

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Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
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