Question

$\mathrm{If}\mathrm{two}\mathrm{vectors}\overrightarrow{A}\mathrm{and}\overrightarrow{B}\mathrm{are}\mathrm{such}\mathrm{that}\left(\overrightarrow{A}+\overrightarrow{B}\right|>\left(\overrightarrow{A}-\overrightarrow{B}\right|\mathrm{then}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{A}\mathrm{and}\overrightarrow{B}\mathrm{is}$

• A. ${90}^o$
• B. Less than ${90}^o$
• C. More than ${90}^o$

Answer 1

The correct option is B Less than ${90}^o$

We know that,
$\left(\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{A^2+B^2+2\mathrm{ABcos\theta }}S\mathrm{imilary},\left(\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt{A^2+{(-B)}^2+2A(-B)\mathrm{cos\theta }}\because \left(\overrightarrow{A}+\overrightarrow{B}\right|>\left(\overrightarrow{A}-\overrightarrow{B}\right|\Rightarrow {\left(\overrightarrow{A}+\overrightarrow{B}\right|}^2>{\left(\overrightarrow{A}-\overrightarrow{B}\right|}^2\Rightarrow A^2+B^2+2\mathrm{AB}\cos \theta >A^2+B^2-2\mathrm{AB}c\mathrm{os}\theta On\mathrm{solving}\mathrm{we}\mathrm{get}\cos \theta >0\therefore \theta <90°$