If `G` is the centroid of a triangle `A B C ,` prove that ` vec G A+ vec G B+ vec G C= vec0dot`

Options

A. `vec0`
B. `3 vec(GA)`
C. `3 vec(GB)`
D. `3 vec(GC)`

Explanation

To prove that \( \vec{GA} + \vec{GB} + \vec{GC} = \vec{0} \) where \( G \) is the centroid of triangle \( ABC \), we can follow these steps:

### Step 1: Define the position vectors
Let the position vectors of points \( A \), \( B \), and \( C \) with respect to the origin \( O \) be:
\[ \vec{OA} = \vec{A}, \quad \vec{OB} = \vec{B}, \quad \vec{OC} = \vec{C} \]

### Step 2: Find the position vector of the centroid \( G \)
The centroid \( G \) of triangle \( ABC \) is given by the formula:
\[ \vec{OG} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]

### Step 3: Express the vectors \( \vec{GA} \), \( \vec{GB} \), and \( \vec{GC} \)
Using the definition of vector subtraction, we can express the vectors from \( G \) to \( A \), \( B \), and \( C \):
\[ \vec{GA} = \vec{OA} - \vec{OG} = \vec{A} - \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]
\[ \vec{GB} = \vec{OB} - \vec{OG} = \vec{B} - \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]
\[ \vec{GC} = \vec{OC} - \vec{OG} = \vec{C} - \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]

### Step 4: Substitute and simplify
Now we can add these vectors:
\[ \vec{GA} + \vec{GB} + \vec{GC} = \left( \vec{A} - \frac{\vec{A} + \vec{B} + \vec{C}}{3} \right) + \left( \vec{B} - \frac{\vec{A} + \vec{B} + \vec{C}}{3} \right) + \left( \vec{C} - \frac{\vec{A} + \vec{B} + \vec{C}}{3} \right) \]

Combining the terms:
\[ = \vec{A} + \vec{B} + \vec{C} - 3 \cdot \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]
\[ = \vec{A} + \vec{B} + \vec{C} - (\vec{A} + \vec{B} + \vec{C}) = \vec{0} \]

### Conclusion
Thus, we have shown that:
\[ \vec{GA} + \vec{GB} + \vec{GC} = \vec{0} \]
This completes the proof.