Step 1: Understanding the Question:
We need to find the magnitude of a vector triple product $|\vec{a} \times (\vec{b} \times \vec{c})|$ where the vectors are coplanar.
Step 2: Key Formula or Approach:
1. Magnitude of cross product: $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta_{uv}$.
2. Dot product: $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta_{bc}$.
Step 3: Detailed Explanation:
First, find $|\vec{c}|$ using the dot product information:
$\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 45^\circ = 8$
$2 \times |\vec{c}| \times \frac{1}{\sqrt{2}} = 8 \implies |\vec{c}| \sqrt{2} = 8 \implies |\vec{c}| = \frac{8}{\sqrt{2}} = 4\sqrt{2}$.
Now, consider the vector $\vec{v} = \vec{b} \times \vec{c}$.
The magnitude is $|\vec{v}| = |\vec{b}| |\vec{c}| \sin 45^\circ = 2 \times 4\sqrt{2} \times \frac{1}{\sqrt{2}} = 8$.
The vector $\vec{b} \times \vec{c}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$.
Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar, $\vec{a}$ lies in that same plane.
Therefore, $\vec{a}$ is perpendicular to $\vec{b} \times \vec{c}$ (the angle is $90^\circ$).
Then, $|\vec{a} \times (\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b} \times \vec{c}| \sin 90^\circ$.
$|\vec{a} \times (\vec{b} \times \vec{c})| = 1 \times 8 \times 1 = 8$.
Step 4: Final Answer:
The value is 8.