Let → a = a _ i ^ i + a _ 2 ^ j + a _ 3 ^ k and → b = b _ 1 ^ i + b...

Given $| \overset{\to}{a} | = 1, | \overset{\to}{b} | = 4, \overset{\to}{a} \cdot \overset{\to}{b} = 2$

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

Applying dot product with $\overset{\to}{a}$ on both sides of equation \left(i\right)$</p><p>$⇒ \overset{\to}{c} \cdot \overset{\to}{a} = - 6 . . . \left(ii\right)$</p><p>\left\{\text{as} ( \overset{\to}{a} \times \overset{\to}{b} ) \cdot \overset{\to}{a} = ( \overset{\to}{a} \times \overset{\to}{b} ) \cdot \overset{\to}{b} = 0\right\}$

Again, applying dot product with $\overset{\to}{b}$ on both sides of equation \left(i\right)$</p><p>$⇒ \overset{\to}{b} \cdot \overset{\to}{c} = - 48 . . . \left(iii\right)$</p><p>Now, using equation \left(i\right)$

$⇒ {\left|\overset{\to}{c}\right|}^{2} = 4 | \overset{\to}{a} \times \overset{\to}{b}{|}^{2} + 9 | \overset{\to}{b}{|}^{2} - 12\left(\overset{\to}{a} \times \overset{\to}{b}\right) \cdot \overset{\to}{b}$

We know that, ${\left(\overset{\to}{a} \times \overset{\to}{b}\right)}^{2} + {\left(\overset{\to}{a} . \overset{\to}{b}\right)}^{2} = {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}$

$⇒ | \overset{\to}{c}{|}^{2} = 4\left[| \overset{\to}{a}{|}^{2} | \overset{\to}{b}{|}^{2} - ( \overset{\to}{a} \cdot \overset{\to}{b}{)}^{2}\right] + 9 | \overset{\to}{b}{|}^{2}$

$⇒ | \overset{\to}{c}{|}^{2} = 4\left[( 1 ) ( 4{)}^{2} - ( 4 )\right] + 9 ( 16 )$

$⇒ | \overset{\to}{c}{|}^{2} = 4 \times 12 + 144$

$⇒ | \overset{\to}{c}{|}^{2} = 48 + 144$

$⇒ | \overset{\to}{c}{|}^{2} = 192$

Now, $cos\theta = \dfrac{\overset{\to}{b} \cdot \overset{\to}{c}}{| \overset{\to}{b} | | \overset{\to}{c} |}$

$⇒ cos\theta = \dfrac{- 48}{\sqrt{192} \times 4}$

$⇒ cos\theta = \dfrac{- 48}{8\sqrt{3} . 4}$

$⇒ cos\theta = \dfrac{- 3}{2\sqrt{3}}$

$⇒ cos\theta = \dfrac{- \sqrt{3}}{2}$

$⇒ \theta = \cos^{- 1}\left(\dfrac{- \sqrt{3}}{2}\right)$