Question

Give the derivation for resultant of two vectors and for the angle between resultant and a vector.

Answer 1

Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.

Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.

So, we have

R = P + Q

Now, expand A to C and draw BC perpendicular to OC.

From triangle OCB,

$O{B}^2=C^2+B{C}^2$

or $O{B}^2=(OA+AC{)}^2+B{C}^2$. . . . . . ( i )

Intriangle ABC,

$cos\theta =\frac{AC}{AB}$

or , AC = AB cos$\theta$

or , AC = OD cos$\theta$

= Q cos$\theta$ [ AB = OD = Q ]

Also,

$cos\theta =\frac{BC}{AB}$

or , BC = AB sin$\theta$

or , BC = OD sin$\theta$

= Q sin$\theta$ [ AB = OD = Q }

Magnitude of resultant:

Substituting value of AC and BC in ( i ), we get

$R^2=(P+Qcos\theta {)}^2+(Qsin\theta {)}^2$

or , $R^2=P^2+2PQcos\theta +Q^{2}co{s}^2\theta +Q^{2}si{n}^2\theta$

or , $R^2=P^2+2PQcos\theta +Q^2$

R = $\sqrt{P^2+2PQcos\theta +Q^2}$

Which is the magnitude of resultant.

Direction of resultant :

Let $\varphi$ be the angle made by resultant R with P . Then,

From triangle OBC,

$tan\varphi =\frac{BC}{OC}=\frac{BC}{OA+AC}$

or , $tan\varphi =\frac{Qsin\theta }{P+Qcos\theta }$

$\varphi =ta{n}^{-1}\left(\frac{Qsin\theta }{P+Qcos\theta }\right)$

which is the direction of resultant.