Let G be the centroid of `Delta` ABC , If `vec(AB) = vec a , vec(AC) = vec b,` then the `vec(AG),` in terms of `vec a and vec b, ` is

Options

A. `(2)/(3) ( vec a + vec b)`
B. `(1)/(6) ( vec a + vec b)`
C. `(1)/(3) ( vec a + vec b)`
D. `(1)/(2) ( vec a + vec b)`

Answer

To find the vector \( \vec{AG} \) in terms of \( \vec{a} \) and \( \vec{b} \), we will follow these steps:

### Step 1: Understand the Position Vectors
Let the position vector of point \( A \) be \( \vec{A} = \vec{0} \) (the origin). Then, we can express the position vectors of points \( B \) and \( C \) as:
- \( \vec{B} = \vec{A} + \vec{AB} = \vec{0} + \vec{a} = \vec{a} \)
- \( \vec{C} = \vec{A} + \vec{AC} = \vec{0} + \vec{b} = \vec{b} \)

### Step 2: Find the Centroid \( G \)
The centroid \( G \) of triangle \( ABC \) is given by the formula:
\[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]
Substituting the position vectors we have:
\[ \vec{G} = \frac{\vec{0} + \vec{a} + \vec{b}}{3} = \frac{\vec{a} + \vec{b}}{3} \]

### Step 3: Find the Vector \( \vec{AG} \)
The vector \( \vec{AG} \) can be calculated as:
\[ \vec{AG} = \vec{G} - \vec{A} \]
Since \( \vec{A} = \vec{0} \), we have:
\[ \vec{AG} = \vec{G} - \vec{0} = \vec{G} = \frac{\vec{a} + \vec{b}}{3} \]

### Final Result
Thus, the vector \( \vec{AG} \) in terms of \( \vec{a} \) and \( \vec{b} \) is:
\[ \vec{AG} = \frac{\vec{a} + \vec{b}}{3} \]