Laws of Vector Addition

Addition or Composition of Vectors

Vectors are added geometrically as they do not follow the ordinary laws of algebra because of direction it possess. We need to find the resultant of the vector by adding two or more vector. The resultant of the vector is called composition of a vector. There are two laws of vector addition for adding two vectors. They are:

1. Triangle law of vector addition
2. Parallelogram law of vector addition

Triangle Law of Vector Addition

Statement: If two sides of a triangle completely represent two vectors both in magnitude and direction taken in same order, then the third side taken in opposite order represents the resultant of the two vectors both in magnitude and direction.

Let there be two vectors and acting on a particle simultaneously represented both in magnitudes and direction by the sides OP and PQ of a triangle OPQ. Let be the angle between two vectors and. According to the law of vectors, the side OQ represents their resultant which makes an angle with one of the vector. So, the resultant vector is $\vec R$

Drop a perpendicular line CD to the extended LINE OP. So, that a right-angled triangle OQN is formed.

$$\text{In} \Delta\text {OQN,}$$

$$OQ^2= ON^2 + NQ^2$$

$$= (OP + PN)^2+ NQ^2$$

$$= OP^2 + 2(OP × PN) + PN^2 + NQ^2$$

$$\text{or,} R^2 = A^2 +2(A × PN) + PN^2 + NQ^2$$

$$\text{In} \Delta \text {PNQ,}$$

$$\cos\theta = \frac {PN}{PQ} = \frac {PN}{B}$$

$$\text{or,} PN = B\cos\theta$$

$$\sin\theta = \frac {QN}{PQ} = \frac {QN}{B}$$

$$\text{or,} QN = B\sin\theta$$

$$\text{from equation (i)}$$

$$R^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta$$

$$\text{or,} R^2 = A^2 + 2AB\cos\theta + B^2$$

$$\boxed {R =\sqrt {( A^2 + 2AB\cos\theta + B^2)}}$$

Direction of resultant $\vec R$ : As resultant $\vec R$ makes an angle $\phi$ with , then in $\Delta\text {OQN,}$

$$\tan\phi= \frac{QN}{ON} = \frac{QN}{OP + PN}$$

$$=\frac{B\sin\theta}{A + B\cos\theta}$$

$$\boxed {\theta=\tan^{-1}\frac{B\sin\theta}{A + B\cos\theta}}$$

Parallelogram Law of Vector Addition

If two vectors acting simultaneously at a point are represented both in magnitude and direction by two adjacent sides of parallelogram drawn from the point, then the diagonal of parallelogram through that point represents the resultant both in magnitude and direction.

As shown in the figure vector n in the figure vector$\vec Aand \vec B$are represented by the sides of a parallelogram OPQS and diagonal is represented by the diagonal OQ such that $\vec R= \vec A+ \vec B$Magnitude of: To calculate the magnitude of the resultant vector, let us drop a perpendicular at N from Q when OS is produced. Let the angle between vectors and be $\theta$.

$$OQ^2 = ON^2 + NQ^2$$

$$\text{or,} OQ^2= (OS + SN)^2+ NQ^2$$

$$\sin\theta = \frac {NQ}{QS} = \frac {NQ}{B}$$

$$\text{or,}NQ = B\sin\theta$$

$$\cos\theta = \frac {SN}{QS} = \frac {SN}{B}$$

$$\text{or,} SN = B\cos\theta$$

$$\text {From Equation} (i)$$

$$OQ^2 = (OS + B\cos\theta)^2 + B^2\sin^2\theta$$

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

$$R^2 = A^2 + 2AB\cos\theta + B^2$$

$$\boxed {R =\sqrt {( A^2 + 2AB\cos\theta + B^2)}}$$

Direction of $\vec R$ : Let the angle made by the resultant $\vec R$with the vector $\vec A$ be $\phi$ . In $\Delta$ONQ,

$$\tan\phi= \frac {QN}{ON} = \frac{QN}{OS + SN}$$

$$\text{or,} =\frac {B\sin\theta}{A + B\cos\theta}$$

$$\boxed {\therefore \phi =\tan^{-1}\frac{B \sin\theta}{A + B \cos\theta}}$$

Special cases: (i) When vectors $\vec Aand \vec B$ and act in the same direction, $\theta$= 0^o and then,

$$R =\sqrt {( A^2 + 2AB\cos\theta + B^2)}$$

$$R^2 =(\sqrt {( A^2 + 2AB + B^2))^2}$$

$$R^2 = A + B$$

$$\boxed {\tan\phi= \frac{B\sin\theta}{A + B\cos\theta} = \frac{B \sin0^o}{A + B \cos0^o}}$$

$$\text{or,} \phi = 0$$

Special cases: (ii) When vectors $\vec Aand \vec B$ and act in the opposite direction, $\theta$= 180^o and then,

$$R =\sqrt {( A^2 + 2AB\cos 180^o + B^2)}$$

$$R^2 =(\sqrt {( A^2 + 2AB -B^2)})^2$$

$$R^2 = A -B \text{(minimum value of R)}$$

$$\tan\phi= \frac {B\sin\theta}{A + B\cos\theta} = \frac {B\sin 180^o}{A + B\cos180^o} = 0$$

$$\boxed {\text{or, } \phi = 0}$$

Thus, the resultant will take the direction of greater value.

Special cases: (iii) When vectors $\vec A and \vec B$ and act in the opposite direction, $\theta$ = 90^o and then,

$$R =\sqrt {( A^2 + 2AB\cos 190^o + B^2)}$$

$$=(\sqrt {( A^2 B^2)})$$

$$\tan\phi= \frac{B\sin\theta}{A + B\cos\theta} = \frac{B\sin 90^o}{A + B\cos90^o} = 0$$

$$\boxed {\text{or, } \phi =\tan^{-1} (\frac{B}{A})}$$

Polygon Laws of Vectors

If a number of vectors be represented both in magnitude and direction by the sides of a polygon taken in same order then the resultant is represented completely in magnitude and direction by the closing side of the polygon taken in the opposite order.

Suppose vectors $\vec A, \vec B, \vec C and \vec D$, and are represented by the four sides OP, PQ, QS and ST of a polygon taken in order as shown in Fig. Then the closing side OT taken in opposite order represents the resultant $\vec R$

$$\vec R =\vec A +\vec B + \vec C + \vec D$$

Properties of Vector Addition

• Vector addition follows a commutative law. If, are three vectors, then

$$\vec A + \vec B + \vec C = \vec C + \vec A+ \vec B = \vec B+ \vec C+ \vec A$$

• Vector addition follows a distributive law.

$$\alpha ( \vec A + \vec B+ \vec C) = \alpha \vec A +\alpha \vec B+ \alpha \vec C$$

Where is a scalar

• Vector addition follows an associative law.

$$(\vec A + \vec B) + \vec C = \vec C + (\vec A+ \vec B)$$