# Triangle Law of Vectors

It states that: If two vector quantities acting simultaneously on a body are represented in both magnitude and direction by two sides of triangle taken in order, the resultant of the vector in both magnitude and direction is given by the third side taken in the opposite order.

To find the magnitude and direction of the resultant of P and Q, a line OA is produced. From the point B, perpendicular is drawn on OA and produced to C. Let ${\theta}$ be the angle between ${\vec{R}}$ and  ${\vec{P}}$ . Let ${\phi}$ be the angle made by the resultant ${\vec{R}}$ with  ${\vec{P}}$.

Using the basic laws of vector addition,

In triangle OBC,

OB2 = OC2 + CB2

or, R2 = (OA + AC)2 + CB2

or, R2 = OA2 + 2OA ·AC + AC2 + CB2……… (i)

In triangle BAC,

$$cos \theta = \frac{AC}{BA} = \frac{AC}{Q}$$
$$\therefore AC = Qcos \theta$$

Also, $$sin \theta = \frac{CB}{BA} = \frac{CB}{Q}$$
$$\therefore CB = Qcos \theta$$

Substituting the value of OA, AC and CB in equation (i), we get:

R2 = P2 + 2PQCos  + (QCosθ)2 + (QSinθ)2

or, R2 = P2 + 2PQCos  + Q2 (Cos2θ + Sin2θ)

Since, Cos2 θ + Sin2θ = 1,

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

The above equation gives the resultant of P and Q.

To find the direction of resultant R, suppose ${\beta}$ be the angle made by the resultant with direction of P.

In triangle OBC,

$$tan \phi = \frac{BC}{OC}$$
$$or, tan \phi = \frac{BC}{OA + AC}$$
$$tan \phi = \frac{Qsin \theta }{P +Qcos \theta }$$
$$tan \therefore \phi = tan^{-1} (\frac{Qsin \theta }{P + Q cos \theta })……(ii)$$

The above equation gives the direction of resultant vector.

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