Example

Question:

Find the magnitude and direction of the resultant of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) in terms of their magnitudes and the angle \(\theta\) between them.

Solution:

Let \(OP\) and \(OQ\) represent vectors \(\mathbf{A}\) and \(\mathbf{B}\) making an angle \(\theta\) (see Fig. 3.10). By the parallelogram method, the resultant vector is: \[ \mathbf{R} = \mathbf{A} + \mathbf{B} \] From the geometry,
\[ OS^2 = ON^2 + SN^2 \] But \(ON = A + B\cos\theta\),
\(SN = B\sin\theta\), so \[ OS^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2 \] \[ R^2 = A^2 + B^2 + 2AB\cos\theta \] \[ R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \] For direction, by the law of sines: \[ \frac{R}{\sin\theta} = \frac{B}{\sin\alpha} = \frac{A}{\sin\beta} \] And, \[ \sin\alpha = \frac{B\sin\theta}{R} \] \[ \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \] Equation above gives the direction of the resultant. These are known as the law of cosines (for magnitude) and law of sines (for direction).