Let ABC be a triangle having its centroid its centroid at G. If S is any point in the plane of the triangle, then `vec(SA) + vec(SB)+vec(SC)=`

Options

A. ` vec(SG)`
B. `2 vec(SG)`
C. `3 vec(SG)`
D. `vec(0)`

Explanation

To solve the problem, we need to find the expression for the sum of the vectors from a point \( S \) to the vertices of triangle \( ABC \). Let's denote the position vectors of points \( A \), \( B \), and \( C \) as \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) respectively. The centroid \( G \) of triangle \( ABC \) is given by the formula:

\[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]

Now, we want to find \( \vec{SA} + \vec{SB} + \vec{SC} \).

### Step 1: Write the vectors from point \( S \) to points \( A \), \( B \), and \( C \)
The vectors from point \( S \) to points \( A \), \( B \), and \( C \) can be expressed as:

\[ \vec{SA} = \vec{A} - \vec{S} \]
\[ \vec{SB} = \vec{B} - \vec{S} \]
\[ \vec{SC} = \vec{C} - \vec{S} \]

### Step 2: Sum the vectors
Now, we can sum these vectors:

\[ \vec{SA} + \vec{SB} + \vec{SC} = (\vec{A} - \vec{S}) + (\vec{B} - \vec{S}) + (\vec{C} - \vec{S}) \]

### Step 3: Simplify the expression
Combining like terms, we have:

\[ \vec{SA} + \vec{SB} + \vec{SC} = \vec{A} + \vec{B} + \vec{C} - 3\vec{S} \]

### Step 4: Substitute the centroid formula
We know that the centroid \( G \) is given by:

\[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]

Thus, we can express \( \vec{A} + \vec{B} + \vec{C} \) as:

\[ \vec{A} + \vec{B} + \vec{C} = 3\vec{G} \]

### Step 5: Final expression
Substituting this back into our equation gives:

\[ \vec{SA} + \vec{SB} + \vec{SC} = 3\vec{G} - 3\vec{S} = 3(\vec{G} - \vec{S}) \]

Thus, we conclude that:

\[ \vec{SA} + \vec{SB} + \vec{SC} = 3(\vec{G} - \vec{S}) \]

### Final Answer
The final expression is:

\[ \vec{SA} + \vec{SB} + \vec{SC} = 3(\vec{G} - \vec{S}) \]

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