Explain resolution of vector in two dimension. Explain resolution of vector in its perpendicular components.

$O$-$XY$ two dimensional Cartesian coordinate system is represented in figure.

Position vector of point $\mathrm{P}$ is $\overrightarrow{\mathrm{A}}$.

By drawing projection from $P$ to $X$-axis, $OM$ is obtained $\overrightarrow{O M}=\vec{A}_{x}=\mathrm{A}_{x} \hat{i}=\mathrm{X}$-component of $\overrightarrow{\mathrm{A}}$

By drawing projection from $P$ to $Y$-axis, $ON$ is obtained $\overrightarrow{O N}=\overrightarrow{A_{y}}=A_{x} \hat{j}=Y$-component of $\vec{A}$. where, $\mathrm{A}_{x}$ and $\mathrm{A}_{y}$ are real numbers.

From figure,

$\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{A}_{x}}+\overrightarrow{\mathrm{A}_{y}}$

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

Suppose, $\vec{A}$ makes angle ' $\theta$ ' with $X$-axis.

For $\Delta \mathrm{OMP}$,

$\cos \theta=\frac{\mathrm{A}_{x}}{\mathrm{~A}}$

$\therefore \mathrm{A}_{x}=\mathrm{A} \cos \theta$

$\sin \theta=\frac{\mathrm{A}_{y}}{\mathrm{~A}}$

$\therefore \mathrm{A}_{y}=\mathrm{A} \sin \theta$

From equation $(3)$ and $(4)$, it can be said that components can be positive, negative or zero depending upon $\theta$.

Vectors can be represented in a plane by two ways:

$(i)$ By its magnitude and direction.

$(ii)$ By its components ( $x$ and $y$ components)

$O$-$XY$ two dimensional Cartesian coordinate system is represented in figure.

Position vector of point $\mathrm{P}$ is $\overrightarrow{\mathrm{A}}$.

By drawing projection from $P$ to $X$-axis, $OM $ is obtained $\overrightarrow{O M}=\vec{A}_{x}=\mathrm{A}_{x} \hat{i}=\mathrm{X}$-component of $\overrightarrow{\mathrm{A}}$

By drawing projection from $P$ to $Y$-axis, $ON$ is obtained $\overrightarrow{O N}=\vec{A}_{y}=\mathrm{A}_{x} \hat{j}=\mathrm{Y}$-component of $\vec{A}$

where, $\mathrm{A}_{x}$ and $\mathrm{A}_{y}$ are real numbers.

From figure,

$\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{A}_{x}}+\overrightarrow{\mathrm{A}_{y}} $

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

Suppose, $\overrightarrow{\mathrm{A}}$ makes angle ' $\theta$ ' with $X$-axis.

For $\Delta \mathrm{OMP}$,

$\cos \theta=\frac{\mathrm{A}_{x}}{\mathrm{~A}}$

$\therefore \mathrm{A}_{x}=\mathrm{A} \cos \theta$

$\sin \theta=\frac{\mathrm{A}_{y}}{\mathrm{~A}}$

$\therefore \mathrm{A}_{y}=\mathrm{A} \sin \theta$

From equation $( 3 )$ and $(4)$, it can be said that components can be positive, negative or zero depending upon $\theta$.

Vectors can be represented in a plane by two ways :

$(i)$ By its magnitude and direction.

$(ii)$ By its components ( $x$ and $y$ components)

.