Let's break down and solve the problem in steps based on the given image:

Understanding the Problem:

1. Vectors and Geometry:
   The shape is a regular hexagon (ABCDOE) with side length $5 \, \text{cm}$.
   Vectors $\vec{a}$ and $\vec{b}$ are defined as $\vec{OA}$ and $\vec{OB}$, respectively.

2. Sub-questions:
   We need to:

   • Express the vectors $\vec{BA}, \vec{OD}, \vec{BD}$ in terms of $\vec{a}$ and $\vec{b}$.
   • Calculate:
     • The angle of the $\triangle BCD$.
     • The area of $\triangle BCD$.
     • The length of $\vec{BD}$.
     • The area of the regular hexagon.

Step 1: Expressing Vectors

(a) $\vec{BA}$:

$\vec{BA} = \vec{OA} - \vec{OB} = \vec{a} - \vec{b}.$

(b) $\vec{OD}$:

The symmetry of the hexagon shows that $\vec{OD}$ points opposite to $\vec{b}$. Hence: $\vec{OD} = -\vec{b}.$

(c) $\vec{BD}$:

Using the vector addition rule: $\vec{BD} = \vec{OD} - \vec{OB} = (-\vec{b}) - \vec{b} = -2\vec{b}.$

Step 2: Calculating Geometry

(a) Angle of $\triangle BCD$:

In $\triangle BCD$, we know the hexagon’s internal angles are $120^\circ$. Hence, the angle $\angle BCD = 120^\circ$.

(b) Area of $\triangle BCD$:

For $\triangle BCD$, we use the area formula: $\text{Area} = \frac{1}{2} \cdot |\vec{BC} \times \vec{CD}|.$

1. Length of $BC = 5 \, \text{cm}$, and $CD = 5 \, \text{cm}$.
2. $\angle BCD = 120^\circ$.
3. Using the formula $|\vec{u} \times \vec{v}| = |u| |v| \sin\theta$: $\text{Area} = \frac{1}{2} \cdot 5 \cdot 5 \cdot \sin 120^\circ = \frac{1}{2} \cdot 25 \cdot \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{4} \, \text{cm}^2.$

(c) Length of $\vec{BD}$:

From Step 1(c), $\vec{BD} = -2\vec{b}$. The magnitude is: $|\vec{BD}| = 2|\vec{b}| = 2 \cdot 5 = 10 \, \text{cm}.$

(d) Area of the Regular Hexagon:

The area of a regular hexagon with side length $a$ is given by: $\text{Area} = \frac{3\sqrt{3}}{2} \cdot a^2.$ Substitute $a = 5$: $\text{Area} = \frac{3\sqrt{3}}{2} \cdot 25 = \frac{75\sqrt{3}}{2} \, \text{cm}^2.$

Final Answers:

1. $\vec{BA} = \vec{a} - \vec{b}$,
2. $\vec{OD} = -\vec{b}$,
3. $\vec{BD} = -2\vec{b}$,
4. $\angle BCD = 120^\circ$,
5. Area of $\triangle BCD = \frac{25\sqrt{3}}{4} \, \text{cm}^2$,
6. Length of $BD = 10 \, \text{cm}$,
7. Area of hexagon = $\frac{75\sqrt{3}}{2} \, \text{cm}^2$.

Let me know if you'd like further explanations or visualizations!
Here are some related questions to explore:

1. How are vector magnitudes related to hexagon geometry?
2. What is the significance of the cross-product in finding areas?
3. Can you generalize the area of any polygon using vectors?
4. How can vectors simplify regular polygon calculations?
5. How do transformations like rotations relate to vector operations?

Tip: Always use symmetry in regular polygons to simplify calculations!