Q.The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to vector $\vec{A}$, and its magnitude is equal to half the magnitude of vector $\vec{B}$. Then the angle between $\vec{A}$ and $\vec{B}$ is

The triangle law says that placing two vectors head-to-tail along two sides of a triangle gives their resultant as the closing third side, with magnitude R = square root of (A^2+B^2+2AB*cos(theta)).

Statement: If two vectors acting on a particle can be represented, both in magnitude and direction, by the two sides of a triangle taken in order (i.e., the tail of the second vector coincides with the head of the first), then the resultant of these two vectors is represented, in magnitude and direction, by the third side of the triangle taken in the reverse order (from the tail of the first vector to the head of the second).

Derivation of the magnitude and direction:

Let vector A = OP be represented by one side, and vector B = PQ be represented by the second side (drawn from the head of A), making an angle theta with A (the angle between A and B). By the triangle law, the resultant R = OQ is the third side, closing the triangle.

Drop a perpendicular QN from Q to the extension of OP. In right triangle ONQ:

QN = Bsin(theta) (the perpendicular height) ON = A + Bcos(theta) (the extended base)

Magnitude of resultant: Using the Pythagorean theorem in triangle ONQ,

OQ^2 = ON^2 + QN^2

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

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

R^2 = A^2 + B^2 + 2ABcos(theta) (since cos^2(theta) + sin^2(theta) = 1) So R = square root of (A^2 + B^2 + 2ABcos(theta))

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