Suppose a particle is displaced through $4$ $\text{m}$ and then $3$ $\text{m}$ due east. If these two displacement vectors are represented by $\overrightarrow{P}$ and $\overrightarrow{Q}$, then the sum $\overrightarrow{P}+\overrightarrow{Q}$ of the two displacement vectors will be the displacement of $7$ $\text{m}$ due east.

Now, suppose the particle is displaced through $4$ $\text{m}$ due east and then $3$ $\text{m}$ due west, then the sum $\overrightarrow{P}+(-\overrightarrow{Q})$ of the two displacement vectors will be a displacement of $1$ $\text{m}$ due east.

Suppose a particle is first displaced through $4$ $\text{m}$ due east and then through $3$ $\text{m}$ due north. These two displacements have been represented by $\overrightarrow{AB}$ and $\overrightarrow{BC}$ as shown in the figure given below. The measurements tell that the magnitude of displacement $\overrightarrow{AC}$ is $5$ $\text{m}$.

We can also measure the angle $\theta$ which gives the direction of $\overrightarrow{AC}$ due north of east. Therefore, instead of moving a distance of $4$ $\text{m}$ from $A$ to $B$ and then a distance of $3$ $\text{m}$ from $B$ to $C$ (a total distance of $4+3=7$ $\text{m}$), if the particle moves a distance of $5$ $\text{m}$ from $A$ to $C$ at an angle $\theta$ with $AB$, it would reach the same point.

Thus, the displacement of $5$ $\text{m}$ along $AC$ produces the same effect as the displacements of $4$ $\text{m}$ along $AB$ and $3$ $\text{m}$ along $BC$ together do. Hence, we can say that the displacement vector $\overrightarrow{AC}$ is the sum (or resultant) of the displacement vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$.

\[\therefore\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\]

If these vectors $\overrightarrow{AB}$, $\overrightarrow{BC}$ and $\overrightarrow{AC}$ are represented by vectors $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ respectively, then,

\[\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}\]

From the discussion in the three cases, it follows that the vectors are not added algebraically but are added geometrically.

We have seen that when two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ are inclined to each other as shown in the figure given below, their sum is given by \[\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}\]

If we observe it carefully, then we will find out that the vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ have been represented by the two sides $AB$ and $BC$ of a triangle taken in same order and their sum $\overrightarrow{R}$ is given by the side $AC$ taken in opposite side.

In the figure given above, the two vectors act at right angles to each other, but this conclusion is applied to the two vectors inclined at any angle and leads to the various laws of vector addition such as triangle law, parallelogram law and polygon law of vector addition.

**Previous:** Types of Vectors

© 2022 AnkPlanet | All Rights Reserved