Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}| = 1, |\vec{b}| = 4$$ and $$\vec{a} \cdot \vec{b} = 2$$. If $$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$ and the angle between $$\vec{b}$$ and $$\vec{c}$$ is $$\alpha$$, then $$192\sin^2\alpha$$ is equal to _________

We need to find $$192\sin^2\alpha$$ where $$\alpha$$ is the angle between $$\vec{b}$$ and $$\vec{c}$$.

Since $$|\vec{a}| = 1$$, $$|\vec{b}| = 4$$, $$\vec{a} \cdot \vec{b} = 2$$ and $$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$, the first task is to determine $$|\vec{c}|^2$$.

$$|\vec{c}|^2 = |2(\vec{a}\times\vec{b}) - 3\vec{b}|^2 = 4|\vec{a}\times\vec{b}|^2 - 12(\vec{a}\times\vec{b})\cdot\vec{b} + 9|\vec{b}|^2$$

Since $$\vec{a}\times\vec{b}$$ is perpendicular to $$\vec{b}$$, $$(\vec{a}\times\vec{b})\cdot\vec{b} = 0$$. Next, $$|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 1 \times 16 - 4 = 12$$, which leads to $$|\vec{c}|^2 = 4(12) - 0 + 9(16) = 48 + 144 = 192$$.

In a similar way, the dot product $$\vec{b}\cdot\vec{c}$$ is given by

$$\vec{b} \cdot \vec{c} = \vec{b} \cdot [2(\vec{a}\times\vec{b}) - 3\vec{b}] = 2\vec{b}\cdot(\vec{a}\times\vec{b}) - 3|\vec{b}|^2$$

Again, $$\vec{b}\cdot(\vec{a}\times\vec{b}) = 0$$, so $$\vec{b} \cdot \vec{c} = 0 - 3(16) = -48$$.

Using these results, the cosine of the angle is

$$\cos\alpha = \frac{\vec{b}\cdot\vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4\sqrt{192}} = \frac{-48}{4 \times 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = \frac{-\sqrt{3}}{2}$$

Hence $$\cos^2\alpha = \frac{3}{4}$$ and $$\sin^2\alpha = 1 - \cos^2\alpha = 1 - \frac{3}{4} = \frac{1}{4}$$.

Finally, $$192\sin^2\alpha = 192 \times \frac{1}{4} = 48$$.

The answer is 48.