Rectangular Components of 3-D Vector
Category : JEE Main & Advanced
\[\overrightarrow{R}={{\overrightarrow{R}}_{x}}+{{\overrightarrow{R}}_{y}}+{{\overrightarrow{R}}_{z}}q\] or \[\overrightarrow{R}={{R}_{x}}\hat{i}+{{R}_{y}}\hat{j}+{{R}_{z}}\hat{k}\]
If \[\overrightarrow{R}\] makes an angle a with x axis, b with y axis and \[\gamma \]with z axis, then
\[\Rightarrow q\] \[\cos \alpha =\frac{{{R}_{x}}}{R}=\frac{{{R}_{x}}}{\sqrt{R_{x}^{2}+R_{y}^{2}+R_{z}^{2}}}=l\]
\[\Rightarrow \] \[\cos \beta =\frac{{{R}_{y}}}{R}=\frac{{{R}_{y}}}{\sqrt{R_{x}^{2}+R_{y}^{2}+R_{z}^{2}}}=m\]
\[\Rightarrow \] \[\cos \gamma =\frac{{{R}_{z}}}{R}=\frac{{{R}_{z}}}{\sqrt{R_{x}^{2}+R_{y}^{2}+R_{z}^{2}}}=n\]
Where l, m, n are called Direction Cosines of the vector \[\overrightarrow{R}\] and \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=\]\[{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =\frac{R_{x}^{2}+R_{y}^{2}+R_{z}^{2}}{R_{x}^{2}+R_{y}^{2}+R_{z}^{2}}=1\]
Note :
\[\overset{\to }{\mathop{r}}\,=({{x}_{2}}-{{x}_{1}})\,\hat{i}+({{y}_{2}}-{{y}_{1}})\hat{j}+({{z}_{2}}-{{z}_{1}})\hat{k}\]
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