Explain the resolution of a vector in three dimensions.

(N/A) Let $\alpha, \beta,$ and $\gamma$ be the angles between the vector $\overrightarrow{A}$ and the $x, y,$ and $z$-axes,respectively.
The components of the vector $\overrightarrow{A}$ along the $x, y,$ and $z$-axes are given by:
$A_{x} = A \cos \alpha$
$A_{y} = A \cos \beta$
$A_{z} = A \cos \gamma$
In general,the vector $\overrightarrow{A}$ can be expressed in terms of its components as:
$\overrightarrow{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$
The magnitude of the vector $\overrightarrow{A}$ is given by:
$|\overrightarrow{A}| = A = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}$
Similarly,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 the $x, y,$ and $z$-axes,respectively.
The magnitude of the position vector $\vec{r}$ is:
$|\vec{r}| = \sqrt{x^{2} + y^{2} + z^{2}}$