JEE Main & Advanced Physics Vectors Resolution of Vector Into Components

Consider a vector \[\overrightarrow{R}\,\] in X-Y plane as shown in fig.  If we draw orthogonal vectors \[{{\overrightarrow{R}}_{x}}\] and \[{{\overrightarrow{R}}_{y}}\] along x and \[y\]axes respectively, by law of vector addition, \[\vec{R}={{\vec{R}}_{x}}+{{\vec{R}}_{y}}\]

Now as for any vector \[\overrightarrow{A}=A\,\hat{n}\] 

so, \[{{\overrightarrow{R}}_{x}}=\hat{i}{{R}_{x}}\] and \[{{\overrightarrow{R}}_{y}}=\hat{j}{{R}_{y}}\]

so \[\overrightarrow{R}=\hat{i}{{R}_{x}}+\hat{j}{{R}_{y}}\]                   ...(i)

But from figure \[{{R}_{x}}=R\cos \theta \]                                           ...(ii)      

and \[{{R}_{y}}=R\sin \theta \]                                                              ...(iii)

Since R and q are usually known, Equation (ii) and (iii) give the magnitude of the components of \[\overrightarrow{R}\] along x and y-axes respectively.

Here it is worthy to note once a vector is resolved into its components, the components themselves can be used to specify the vector as

(1) The magnitude of the vector\[\overrightarrow{R}\] is obtained by squaring and adding equation (ii) and (iii), i.e. \[R=\sqrt{R_{x}^{2}+R_{y}^{2}}\]                                

(2) The direction of the vector \[\overrightarrow{R}\] is obtained by dividing equation (iii) by (ii), i.e.                         

\[\tan \theta =({{R}_{y}}/{{R}_{x}})\] or \[\theta ={{\tan }^{-1}}({{R}_{y}}/{{R}_{x}})\]