# Subtraction of Vectors

Subtraction of vectors

Since, $\displaystyle \overrightarrow{A}-\overrightarrow{B}=\overrightarrow{A}+(-\overrightarrow{B})$ and

$\displaystyle |\overrightarrow{A}+\overrightarrow{B}|\,=\,\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$

$\displaystyle \Rightarrow$ $\displaystyle |\overrightarrow{A}-\overrightarrow{B}|\,=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \,({{180}^{o}}-\theta )}$

Since, $\displaystyle \cos \,(180-\theta )=-\cos \theta$

$\displaystyle \Rightarrow$ $\displaystyle |\overrightarrow{A}-\overrightarrow{B}|\,=\,\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta }$

$\displaystyle \tan {{\alpha }_{1}}=\frac{B\sin \theta }{A+B\cos \theta }$

and $\displaystyle \tan {{\alpha }_{2}}=\frac{B\sin \,(180-\theta )}{A+B\cos \,(180-\theta )}$

But $\displaystyle \sin (180-\theta )=\sin \theta$ and $\displaystyle \cos (180-\theta )=-\cos \theta$

$\displaystyle \Rightarrow$$\displaystyle \tan {{\alpha }_{2}}=\frac{B\sin \theta }{A-B\cos \theta }$

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