Physics · Ch 2 — Mathematical Methods

Law of parallelogram of vectors

2.3.4

Law of parallelogram of vectors

Another geometric method for adding two vectors is the parallelogram law of vector addition, stated as follows: if two vectors of the same type, with their tails at the same point, are represented in magnitude and direction by two adjacent sides of a parallelogram, then their resultant vector is given, in magnitude and direction, by the diagonal of the parallelogram drawn from that same point.

Let $\vec{OA} = \vec P$ and $\vec{OB} = \vec Q$ be two vectors originating from a common point O, inclined to each other at angle $\theta$ . Completing the parallelogram OACB, the diagonal $\vec{OC} = \vec R$ represents the resultant.

Magnitude of the resultant. Dropping a perpendicular from C to the extension of OA at D, and applying the Pythagoras theorem in the right-angled triangles ODC and ADC, one obtains

R^2 = P^2+Q^2+2PQ\cos\theta \qquad \Rightarrow \qquad R = \sqrt{P^2+Q^2+2PQ\cos\theta} \qquad \text{--- (2.7), (2.8)}

Direction of the resultant. The angle $\alpha$ that $\vec R$ makes with $\vec P$ is obtained from the same geometric construction:

\tan\alpha = \dfrac{Q\sin\theta}{P+Q\cos\theta} \qquad \text{--- (2.10)}

A similarly derived formula gives the angle $\beta$ that $\vec R$ makes with $\vec Q$ . …

Figure 2.7Parallelogram law of vector addition

What this figure shows. A parallelogram OACB is drawn with O at one vertex. Side $\vec{OA}$ represents vector $\vec P$ and side $\vec{OB}$ represents vector $\vec Q$ , both originating from O and inclined to each other at angle $\theta$ (marked at O). The diagonal $\vec{OC}$ , drawn from O to the opposite vertex C of the completed parallelogram, represents the resultant vector $\vec R=\vec P+\vec Q$ . A perpendicular is dropped from C down to the line OA extended, meeting it at point D, forming the right-angled triangle ODC (right angle at D) — this construction is used to derive the magnitude and direction formulas for $\vec R$ . Points O, A, B, C, D and the angle $\theta$ at O …

Misc Ex.2.3River-crossing boat velocity problem

Worked out. A river flows east at 5 km/hr relative to the shore. A boat, whose speed relative to still water is 20 km/hr, is steered heading due North across the stream. The problem asks for the boat's actual resultant velocity (magnitude and direction) as seen from the shore. The method adds the boat's own 20 km/hr northward velocity vector and the water's 5 km/hr eastward velocity vector using the triangle/parallelogram construction (a right-angled triangle OAB, since the two velocities are perpendicular to each other), giving the resultant by the Pythagoras theorem and its directio …

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