Question

If $|\overrightarrow{a}+\overrightarrow{b}|<|\overrightarrow{b}-\overrightarrow{a}|$ then angle between $\overrightarrow{a}$and $\overrightarrow{b}$ is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. acute
• D. obtuse

Answer 1

The correct option is D obtuse

$|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\overrightarrow{a}.\overrightarrow{b}}$

$|\overrightarrow{b}-\overrightarrow{a}|=\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\overrightarrow{a}.\overrightarrow{b}}$

$|\overrightarrow{a}+\overrightarrow{b}|<|\overrightarrow{b}-\overrightarrow{a}|$

$⟹\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}}<\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\overrightarrow{a}.\overrightarrow{b}}$

$⟹{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}<{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\overrightarrow{a}.\overrightarrow{b}$

$⟹4\overrightarrow{a}.\overrightarrow{b}<0$

$⟹\overrightarrow{a}.\overrightarrow{b}<0$

$⟹\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|\cos \theta <0$

$⟹\cos \theta <0$

$\therefore \theta isobtuse.$

$\therefore Opt:\left(D\right)$ is the Answer