Explain in detail the triangle law of addition.

Explanation

Let us consider two vectors `vecA` and ` vecB` as shown in figure. To find the resultant of the two vectors we apply the triangular law of addition as follows:
Represent the vectors `vecA` and `vecB` by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector `vecA` is connected to the tail of the second vector `vec`B. Let `theta` be the angle between `vecA` and `vecB` . Then `vecR` is the resultant vector connecting the tail of the first vector `vecA` to the head of the second vector `vecB`. The magnitude of `vecR` (resultant) is given geometrically by the length of `vecR (OQ)` and the direction of the resultant vector is the angle between `vecR` and `vecA` . Thus we write `vecR = vecA + vecB`
`vec(OQ) = vec(OP) + vec(PQ)`
1. Magnitude of resultant vector: The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows. From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure ,let R is the magnitude of the resultant of `vecA` and `vecB`
`cos theta = (AN)/(B) therefore AN = B cos theta " and " sin theta = (BN)/(B) therefore BN = B sin theta `
For `Delta OBN` , we have `OB^2 = ON^2 + BN^2`
` rArr R^2 = (A + B cos theta)^2 + (B sin theta)^2`
` rArr R^2 =A^2 + B^2 cos^2 theta + 2 AB cos theta + B^2 sin^2 theta`
` rArr R^2 = A^2 + B^2 (cos^2 theta + sin^2 theta) + 2AB cos theta `
` rArr R^2 = sqrt(A^2 + B^2 + 2AB cos theta)`
2. Direction of resultant vectors : if ` theta` is the angle between `vecA` and `vecB` then
`|vecA + vecB | = sqrt(A^2 + B^2 + 2AB cos theta)`
If `vecR` makes an angle `alpha ` with `vecA` , then in `Delta OBN`
` tan alpha = (BN)/(ON) = (BN)/(OA + AN)`
` tan alpha = (B sin theta)/(A + B cos theta) rArr alpha tan^(-1) ( (B sin theta)/(A + Bcos theta) )`