Resolution of Vector Into Components
Category : JEE Main & Advanced
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 \[\theta ={{\tan }^{-1}}({{R}_{y}}/{{R}_{x}})\]
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