Q.State Triangle law of Vector addition. Give its analytical treatment to find the magnitude and direction of a Resultant vector.

Triangle law: place two vectors head-to-tail; the resultant is the vector from the tail of the first to the head of the second. Using geometry (dropping a perpendicular), R = sqrt(A^2 + B^2 + 2ABcos(theta)), with direction beta = tan^-1[Bsin(theta)/(A + B*cos(theta))].

Statement of the Triangle Law of Vector Addition:

If two vectors can be represented, in both magnitude and direction, by the two sides of a triangle taken in the same order (i.e. the tail of the second vector coincides with the head/tip of the first vector), then their resultant vector is represented completely, in both magnitude and direction, by the third side of the triangle taken in the OPPOSITE order (from the tail of the first vector to the head of the second vector).

Analytical (geometric) treatment:

Consider two vectors A (represented by OP) and B (represented by PQ), placed head-to-tail, with theta being the angle between A and B (i.e., the angle between OP extended and PQ). Their resultant R = A + B is given by the third side OQ of the triangle OPQ.

To find R analytically, drop a perpendicular from Q to the line OP extended, meeting it at point S. Then:

• In right triangle PQS: PS = Bcos(theta) and QS = Bsin(theta) (since angle QPS = theta, the angle between A and B, taken as the exterior angle at P)
• OS = OP + PS = A + B*cos(theta)

Magnitude of the resultant:

In right triangle OQS (right-angled at S), by the Pythagorean theorem:

OQ^2 = OS^2 + QS^2

R^2 = (A + Bcos(theta))^2 + (Bsin(theta))^2

R^2 = A^2 + 2ABcos(theta) + B^2cos^2(theta) + B^2*sin^2(theta)

R^2 = A^2 + 2ABcos(theta) + B^2(cos^2(theta) + sin^2(theta))

R^2 = A^2 + B^2 + 2AB*cos(theta) [since cos^2(theta) + sin^2(theta) = 1]

R = sqrt(A^2 + B^2 + 2AB*cos(theta))

Direction of the resultant:

Let beta be the angle that the resultant R makes with vector A (i.e., angle QOS). In right triangle OQS:

tan(beta) = QS / OS = (Bsin(theta)) / (A + Bcos(theta))

beta = tan^-1 [ (Bsin(theta)) / (A + Bcos(theta)) ]

This fully determines both the magnitude R and the direction beta of the resultant of two vectors A and B added by the triangle law.

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