The task involves vector analysis in a triangle. Let me analyze and solve the problem by extracting the relevant data.

Observations from the image:

1. Setup:

   • The triangle $\triangle ABC$ is provided with specific vectors:
     • $\overrightarrow{AC} = \vec{a}$
     • $\overrightarrow{BC} = \vec{b}$
   • $O$ is the intersection point of the triangle's medians.

2. Question:

   • Express $\overrightarrow{AO}$ in terms of $\vec{a}$ and $\vec{b}$.

3. Solution Method:

   • In a triangle, the centroid (intersection of medians) divides each median in a $2:1$ ratio, counting from the vertex to the midpoint of the opposite side.
   • Using this property, we calculate $\overrightarrow{AO}$.

Let me now calculate $\overrightarrow{AO}$ and check which option (A, B, C, D, E) is correct.We are tasked with expressing $\overrightarrow{AO}$ in terms of $\vec{a}$ and $\vec{b}$, based on the geometry and vector setup.

Step-by-Step Solution:

1. Understanding the Geometry:

• $O$ is the centroid of $\triangle ABC$, the intersection of the medians.
• The centroid divides each median in the ratio $2:1$ (vertex to centroid is twice the distance from the centroid to the midpoint of the opposite side).

2. Representing the Median:

• From the diagram:
  • $A$ is a vertex, and $C$ is another vertex of the triangle.
  • The median from $A$ connects to the midpoint of $\overrightarrow{BC}$.

3. Midpoint of $BC$:

• The midpoint of $\overrightarrow{BC}$, denoted $M$, is: $M = \frac{\vec{B} + \vec{C}}{2}$ Since $\vec{C} = \vec{0}$ (assuming the coordinate system places $C$ at the origin): $M = \frac{\vec{B} + \vec{C}}{2} = \frac{\vec{b}}{2}.$

4. Vector $\overrightarrow{AM}$:

• The vector $\overrightarrow{AM}$ (from $A$ to $M$) is: $\overrightarrow{AM} = \overrightarrow{AC} + \overrightarrow{CM}.$ Substituting: $\overrightarrow{AM} = \vec{a} + \frac{\vec{b}}{2}.$

5. $\overrightarrow{AO}$:

• Since $O$ divides $\overrightarrow{AM}$ in the ratio $2:1$, we use the section formula: $\overrightarrow{AO} = \frac{2}{3} \overrightarrow{AM}.$ Substituting $\overrightarrow{AM} = \vec{a} + \frac{\vec{b}}{2}$: $\overrightarrow{AO} = \frac{2}{3} \left( \vec{a} + \frac{\vec{b}}{2} \right).$ Simplify: $\overrightarrow{AO} = \frac{2}{3} \vec{a} + \frac{1}{3} \vec{b}.$

Final Answer:

$\overrightarrow{AO} = \frac{2}{3} \vec{a} + \frac{1}{3} \vec{b}.$

This matches option E in the given problem.

Do you want details or have any questions?

Here are 5 related questions:

1. How is the centroid's position determined geometrically in any triangle?
2. What is the section formula, and how is it applied here?
3. Can $\overrightarrow{AO}$ be calculated differently (e.g., using weighted averages)?
4. What happens if $O$ is outside the triangle (in obtuse triangles)?
5. How does the $2:1$ ratio simplify vector operations in centroid-related problems?

Tip: The centroid always splits the medians of a triangle in the same $2:1$ ratio, regardless of the type of triangle!