Mathematics - Vector Question with Solution | TestHub

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$

$\because \frac{\overrightarrow{a}.\left(\overrightarrow{c}-\overrightarrow{b}\right)}{\overrightarrow{c}.\left(\overrightarrow{a}-\overrightarrow{b}\right)}=\frac{\left|\overrightarrow{a}\right|}{\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|}$

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

$\Rightarrow \frac{{\left|\overrightarrow{c}\right|}^2-{\left|\overrightarrow{b}\right|}^2}{{\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2}=\frac{3}{3+4}=\frac{3}{7}$

$\Rightarrow {\left|\overrightarrow{c}\right|}^2=13$

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

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

$\Rightarrow 9+16+2\left(\overrightarrow{a}.\overrightarrow{b}\right)=13$

$\Rightarrow \overrightarrow{a}·\overrightarrow{b}=-6$

Now, ${\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left(\overrightarrow{a}.\overrightarrow{b}\right)}^2={\left|\overrightarrow{a}\right|}^2.{\left|\overrightarrow{b}\right|}^2$

$\Rightarrow {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+36=\left(9\right)\left(16\right)\Rightarrow {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2=108$