Question

Parallelogram law of Victor addition . plz give correct answer, don't spam..... Answer in long 7 mark​

Answer 1

Answer:

Hello Mate♥️♥️♥️

Explanation:

Here is ur answer

It states that if two vectors acting on a particle sat the same time be represented in magnitude and direction by the two adjaoent sides of a parallelogram drawn from a point, than their resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from the same point.   

Let the two vectors of represented in magnitude and direction by two adjecent sides OP and OS of parllelogram OPQS. drawn from a point O.

According to the parllelogram law of resultent vectors R will be represented by diagonal OQ of the parllelogram

Mangnitude of R

Draw QN perpendicular OP produced

From the figure,OP = A,OS = PQ = B, OQ =R ∠SOP=∠QPN=θ

In triangle QPN,PN=PQ cosθ=Bcosθ

QN=PQ sinθ=Bsinθ

In right angle triangle QNP ,we have

$OQ ^{2} \: = ON ^{2} + NQ ^{2} \\ \: \: \: \: \: \: \: \: \: \: = ({OP + PN})^{2} + {NQ}^{2} \\ or,R ^{2} = (A + Bcosθ )^{2} + (Bsinθ )^{2}) \\ or,R ^{2} = A ^{2} + 2AB \cosθ+ B ^{2} ( \cosθ ^{2} + sinθ ^{2} ) \\ or,R ^{2} = A ^{2} + 2ABcosθ + B ^{2} \\ or,R^{2} = \sqrt{ {A}^{2} + 2AB\cosθ + {B}^{2} }$

Hope it help uh...☺☺☺

thank me if u like my answer,♥️♥️♥️