Question

Let $|\overrightarrow{a}|=7,|\overrightarrow{b}|=11,|\overrightarrow{a}+\overrightarrow{b}|=10\sqrt{3}$.Find the angle between $(\overrightarrow{a}+\overrightarrow{b})$ and $(\overrightarrow{a}+\overrightarrow{b}).$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. None of the above

Answer 1

The correct option is D None of the above

$|\overrightarrow{a}+\overrightarrow{b}|=10\sqrt{3}$

$\Rightarrow \sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta }=10\sqrt{3}$

$\Rightarrow \sqrt{7^2+{11}^2+2\times 7\times 11\cos \theta }=10\sqrt{3}$

squaring both sides

$\Rightarrow 170+154\cos \theta =300$

$\Rightarrow \cos \theta =\frac{130}{154}$ ....... $\left(i\right)$

We know that, the angle $\theta$ between two vector $\overrightarrow{a}$ and

$\overrightarrow{b}$ is given by

$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}$

Thus, the angle between $(\overrightarrow{a}+\overrightarrow{b})$ and $(\overrightarrow{a}+\overrightarrow{b})$ is

$\cos \varphi =\frac{(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{a}+\overrightarrow{b}|}$

$\cos \varphi =\frac{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta }{|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{a}+\overrightarrow{b}|}$

$\cos \varphi =\frac{7^2+{11}^2+2\times 7\times 11\cos \theta }{{\left(10\sqrt{3}\right)}^2}$

$\cos \varphi =\frac{300}{300}$ ...... From $\left(i\right)$

$\cos \varphi =1$

$\varphi =\arccos 1$

$\varphi =0^o$