Question

Find the magnitude of resultant of two vectors A & B in terms of their magnitudes

& angle β between them.

​

Answer 1

Explanation:

Let OP and OQ represent the two vectors A and B making an angle

. Then, using the parallelogram method of vector addition, OS represents the resultant vector R. <br> R= A+B <br> SN is normal to OP and OP and PM is normal to OS. <br> From the geometry of the figure. <br>

<br> but ON = OP + PN = A + Bcos

<br> SN = B

<br>

<br> or,

.............(4.24a) <br> In

OSN, SN= OS

, and in

PSN, SN= PS

<br> Therefore,

<br> or,

................(4.24b) <br> Similarly, <br> PM= A

<br> or

............(4.24c)<br> Combining Eqs. we get, <br>

................(4.24d) <br> Using Eqs, we get <br>

...............(4.24 e)<br> Where R is given by Eq. <br> or

...............(4.24f) <br> Eqs. (4.24a) givestjhe magnitude of the resultant and Eqs. (4.24e) and (4.24f) its direction. Equation (4.24a) is known as the law of cosines and Eq. (4.24d) as the law of sines

Answer 2

OP and OQ represent the two vectors A and B making an angle θ. Then, using the parallelogram method of vector addition, OS represents the resultant vector R.

R= A+B

SN is normal to OP and OP and PM is normal to OS.

From the geometry of the figure.

(OS)2=(ON)2+(SN)2

but ON = OP + PN = A + Bcosθ

SN = Bsinθ

(OS)2=(A+Bcosθ)2+(Bsinθ)2

or, R2=A2+B2+2ABcosθ−−−−−−−−−−−−−−−−−√.............(4.24a)

In Δ OSN, SN= OSsinα=Rsinα, and in ΔPSN, SN= PS sinθ

Therefore, Rsinα=Bsinθ

or, Rsinθ=Bsinα................(4.24b)

Similarly,

PM= Asinα=Bsinβ

or Asinβ=Bsinα ............(4.24c)

Combining Eqs. we get,

Rsinθ=Asinβ=Bsinα................(4.24d)Using Eqs, we get

sinalph=BRsinθ ...............(4.24 e)

Where R is given by Eq.

or tanα=SNOP+PN=BsinθA+Bcosθ...............(4.24f)

Eqs. (4.24a) givestjhe magnitude of the resultant and Eqs. (4.24e) and (4.24f) its direction. Equation (4.24a) is known as the law of cosines and Eq. (4.24d) as the law of sines.