Let → a and → b be two vector such that | → a | = 14 , | → b | = 6...

Given,

$\left|\overset{\to}{a}\right| = \sqrt{14}, \left|\overset{\to}{b}\right| = \sqrt{6} \text{ and }$ \left|\overset{\to}{a} \times \overset{\to}{b}\right| = \sqrt{48}$</p><p>Now we know that,</p><p>${\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} + {\left|\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)\right|}^{2} = {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}\sin^{2}\theta + {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}\cos^{2}\theta$</p><p>$⇒ {\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} + {\left|\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)\right|}^{2} = {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}$</p><p>\left\{\text{as}\cos^{2}\theta + \sin^{2}\theta = 1\right\}$

$⇒ 48 + {\left|\overset{\to}{a} \cdot \overset{\to}{b}\right|}^{2} = 14 \times 6$

$⇒ {\left|\overset{\to}{a} \cdot \overset{\to}{b}\right|}^{2} = 84 - 48 = 36$.