Let → a and → b be two vectors such that | → a | = 1, | → b | = 4...

Given,

$\overset{\to}{a} and$\overset{\to}{b}$be two vectors such that \left|\overset{\to}{a}\right| = 1, \left|\overset{\to}{b}\right| = 4$ and $\overset{\to}{a} ⋅ \overset{\to}{b} = 2$.

Also, given

$\overset{\to}{c} = \left(2\overset{\to}{a} \times \overset{\to}{b}\right) - 3\overset{\to}{b}$

Taking dot product with $\overset{\to}{b}$ both side we get,

$⇒ \overset{\to}{b} ⋅ \overset{\to}{c} = \left(2\overset{\to}{a} \times \overset{\to}{b}\right) ⋅ \overset{\to}{b} - 3{\left|b\right|}^{2}$

$⇒ \left|\overset{\to}{b}\right|\left|\overset{\to}{c}\right|cos\alpha = 0 - 3{\left|\overset{\to}{b}\right|}^{2}$

$⇒ \left|\overset{\to}{c}\right|cos\alpha = - 12$ {as \left|\overset{\to}{b}\right| = 4}</p><p>Now, solving\overset{\to}{a} ⋅ \overset{\to}{b} = 2$</p><p>$⇒ \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|cos\theta = 2$</p><p>$⇒ cos\theta = \dfrac{1}{2} ⇒ \theta = \dfrac{\pi}{3}$</p><p>Now, solving</p><p>${\left|\overset{\to}{c}\right|}^{2} = {\left|\left(2\overset{\to}{a} \times \overset{\to}{b}\right) - 3\overset{\to}{b}\right|}^{2}$</p><p>$⇒ {\left|\overset{\to}{c}\right|}^{2} = 64 \times \dfrac{3}{4} + 144 = 192$</p><p>Now, squaring both side of \left|\overset{\to}{c}\right|cos\alpha = - 12$ we get,

${\left|\overset{\to}{c}\right|}^{2}\cos^{2}\alpha = 144$

$⇒ 192\cos^{2}\alpha = 144$

$⇒ 192\left(1 - \sin^{2}\alpha\right) = 144$

$⇒ 192\sin^{2}\alpha = 48$