Subtraction of Vectors

July 9th, 2017 by admin Leave a reply »

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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