Question

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $\left|\overrightarrow{a}\right|=\sqrt{37},\left|\overrightarrow{b}\right|=3,\left|\overrightarrow{c}\right|=4$, then the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is C ${60}^0$

Given $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $\left|\overrightarrow{a}\right|=\sqrt{37},\left|\overrightarrow{b}\right|=3$ and $\left|\overrightarrow{c}\right|=4$

Now, $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\Rightarrow \overrightarrow{a}=-(\overrightarrow{b}+\overrightarrow{c})$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2={|-(\overrightarrow{b}+\overrightarrow{c})|}^2$

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

$\Rightarrow 9+16+24\cos \theta =37$

$\Rightarrow 37=25+24\cos \theta$

$\Rightarrow 24\cos \theta =12\Rightarrow \theta ={60}^0$