Mathematics - Vector Question with Solution | TestHub

Given,

$\left|\overrightarrow{a}\right|=\sqrt{14}, \left|\overrightarrow{b}\right|=\sqrt{6} \& \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{48}$

Now we know that,

${\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\left(\overrightarrow{a}·\overrightarrow{b}\right)\right|}^2={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2{\sin }^2\theta +{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2{\cos }^2\theta$

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

$\left\{\text{as} {\cos }^2\theta +{\sin }^2\theta =1\right\}$

$\Rightarrow 48+{\left|\overrightarrow{a}·\overrightarrow{b}\right|}^2=14\times 6$

$\Rightarrow {\left|\overrightarrow{a}·\overrightarrow{b}\right|}^2=84-48=36$.