Question

State parallelogram law of vectors. Derive an expression for the magnitude and direction of the

resultant vector.


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Answer 1

Answer:

Parallelogran law is the law used for the addition of vectors which starts from the Same point or whose tails are connected.

Explanation:

The resultant will be given by diagonal which starts from the starting point of 2 vectors.

Magnitude of a resultant vector=

R=

$\sqrt{ {a}^{2} } + \sqrt{ {b}^{2} } + \sqrt{ 2ab \: cos \: theata }$

Answer 2

Parallelogram Law

To find:

Derivation of   expression for the magnitude and direction of the  resultant vector.

Explanation:

Parallelogram law:

If two vectors are represented by the two adjacent sides of a parallelogram then the diagonal passing through the intersection of given vectors represents their resultant (both in direction and magnitude ). It is also known as parallelogram of vector addition.

For derivation ,

I have attached the required figure.

In the figure   given, X and Y are two vectors having magnitudes equal to length OA and OB respectively . Angle θ  is between them.

Construction:

Complete the parallelogram, OACB,

Diagonal OC is joined, which makes angle α with vector  X.

 To find the magnitude of resultant vector , a perpendicular CD is produced to meet OA  which is further produced to D.

From △ OCD,

$OC^{2}+ OD^{2} =CD^{2}$

and  

$\vec{CD}=\vec{AC}sin\theta=\vec{Y}sin\theta$

$\vec{AD}=\vec{AC}cos\theta=\vec{Y}cos\theta$

$\vec{OD}=\vec{OA}+\vec{AD}=X+\vec{Y}cos\theta$

Put the values and represent the resultant vector OC by   R , magnitude of the resultant is given by,

$R^2=(\vec{X} +\vec{Y }cos\theta)^2+(\vec{Y}sin\theta)^2$

$\implies R^2=\vec{X}^2+\vec{Y}^2+2\vec{X}\vec{Y}cos\theta$

In ΔOCD,

$tan\ \alpha=\frac{CD}{OD} =\frac{\vec{Y}sin\theta}{\vec{X}+\vec{Y}cos\theta}$

Resultant acts in the direction making an angle α

$tan^{-1}(\frac{\vec{Y}sin\theta}{\vec{X}+\vec{Y}cos\theta})$  with direction of vector X.