Explain the resolution of vector in three dimension.

$\alpha, \beta$ and $\gamma$ are the angles between $\overrightarrow{\mathrm{A}}$ and the $x, y$ and $z$-axes, respectively. So that,

$\mathrm{A}_{x}=\mathrm{A} \cos \alpha$

$\mathrm{A}_{y}=\mathrm{A} \sin \beta$

$\mathrm{A}_{z}=\mathrm{A} \cos \gamma$

In general, we have

$\overrightarrow{\mathrm{A}}=\mathrm{A}_{x} \hat{i}+\mathrm{A}_{y} \hat{j}+\mathrm{A}_{z} \hat{k}$

The magnitude of vector $\overrightarrow{\mathrm{A}}$ is

$|\overrightarrow{\mathrm{A}}|=\mathrm{A}=\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}+\mathrm{A}_{z}^{2}} \quad \ldots$

A position vector $\vec{r}$ can be expressed as,

$\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$

where, $x, y$ and $z$ are the components of $\vec{r}$ along $x, y, z$-axes, respectively.

Magnitude of $\vec{r}=|\vec{r}|=\sqrt{x^{2}+y^{2}+z^{2}} \quad \ldots$