Step 1: Understanding the Concept:
We need to find the angle $\theta$ between vectors $\bar{b}$ and $\bar{c}$.
The cosine of the angle is given by $\cos \theta = \frac{\bar{b} \cdot \bar{c}}{|\bar{b}| |\bar{c}|}$.
We need to compute the dot product $\bar{b} \cdot \bar{c}$ and the magnitude $|\bar{c}|$.
Step 2: Key Formula or Approach:
Dot product definition: $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$.
Properties of cross product: $\vec{v} \cdot (\vec{u} \times \vec{v}) = 0$ because $\vec{u} \times \vec{v}$ is perpendicular to $\vec{v}$.
Lagrange's Identity: $|\vec{u} \times \vec{v}|^2 = |\vec{u}|^2 |\vec{v}|^2 - (\vec{u} \cdot \vec{v})^2$.
Step 3: Detailed Explanation:
Given: $|\bar{a}| = 1, |\bar{b}| = 4, \bar{a} \cdot \bar{b} = 2$, and $\bar{c} = 2(\bar{a} \times \bar{b}) - 3\bar{b}$.
First, let's find the dot product $\bar{b} \cdot \bar{c}$:
\[ \bar{b} \cdot \bar{c} = \bar{b} \cdot \left( 2(\bar{a} \times \bar{b}) - 3\bar{b} \right) \]
\[ \bar{b} \cdot \bar{c} = 2 \left( \bar{b} \cdot (\bar{a} \times \bar{b}) \right) - 3(\bar{b} \cdot \bar{b}) \]
Since $\bar{a} \times \bar{b}$ is orthogonal to both $\bar{a}$ and $\bar{b}$, the dot product $\bar{b} \cdot (\bar{a} \times \bar{b}) = 0$.
Also, $\bar{b} \cdot \bar{b} = |\bar{b}|^2$.
\[ \bar{b} \cdot \bar{c} = 2(0) - 3|\bar{b}|^2 = -3(4^2) = -3(16) = -48 \]
Next, we need the magnitude of $\bar{c}$. Let's calculate $|\bar{c}|^2$:
\[ |\bar{c}|^2 = \bar{c} \cdot \bar{c} = \left( 2(\bar{a} \times \bar{b}) - 3\bar{b} \right) \cdot \left( 2(\bar{a} \times \bar{b}) - 3\bar{b} \right) \]
Using the property that $(\vec{u} \times \vec{v}) \cdot \vec{v} = 0$:
\[ |\bar{c}|^2 = 4|\bar{a} \times \bar{b}|^2 - 12(\bar{a} \times \bar{b}) \cdot \bar{b} + 9|\bar{b}|^2 \]
\[ |\bar{c}|^2 = 4|\bar{a} \times \bar{b}|^2 - 0 + 9(16) = 4|\bar{a} \times \bar{b}|^2 + 144 \]
We need to find $|\bar{a} \times \bar{b}|^2$. Using Lagrange's identity:
\[ |\bar{a} \times \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2 \]
\[ |\bar{a} \times \bar{b}|^2 = (1^2)(4^2) - (2)^2 = 16 - 4 = 12 \]
Substitute this back into the equation for $|\bar{c}|^2$:
\[ |\bar{c}|^2 = 4(12) + 144 = 48 + 144 = 192 \]
\[ |\bar{c}| = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3} \]
Now, calculate the angle $\theta$:
\[ \cos \theta = \frac{\bar{b} \cdot \bar{c}}{|\bar{b}| |\bar{c}|} = \frac{-48}{4 \cdot 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} \]
Rationalizing the denominator:
\[ \cos \theta = \frac{-3\sqrt{3}}{2 \cdot 3} = -\frac{\sqrt{3}}{2} \]
The principal value for which $\cos \theta = -\frac{\sqrt{3}}{2}$ is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Step 4: Final Answer:
The angle is $\frac{5\pi}{6}$.