# What is the particle's displacement from r1 to r2?

## The position vector of a particle is initially r1=(-3.0m) $\hat{i}$ + (4.0m) $\hat{j}$ and then later is r2=(9.0m) $\hat{i}$ + (-3.5m) $\hat{j}$. Please have me an explanation.

Jul 18, 2018

#### Explanation:

The position vector of a particle is initially r1=(-3.0m) $\hat{i}$ + (4.0m) $\hat{j}$ and then later is r2=(9.0m) $\hat{i}$ + (-3.5m) $\hat{j}$.

Let $O$ be the orgin in X-Y plane A and B represent the initial and final positions of the particle.

$\vec{O A} = {\vec{r}}_{1} = \left(- 3.0 m\right) \hat{i} + \left(4.0 m\right) \hat{j}$

$\vec{O B} = {\vec{r}}_{2} = \left(9.0 m\right) \hat{i} + \left(- 3.5 m\right) \hat{j}$

By triangle law of vector addition we have
Displacement vector
$\vec{A B} = \vec{O B} - \vec{O A}$

$= \left(9 - \left(- 3\right)\right) m \hat{i} + \left(\left(- 3.5\right) - 4\right) m \hat{j}$

$= \left(12\right) m \hat{i} - \left(7.5\right) m \hat{j}$