Understanding Position Vectors and Triangle Centroid

This problem involves understanding position vectors and the concept of a triangle's centroid in vector algebra. We are given the position vectors of the vertices of a triangle ABC, denoted by \(\vec a\), \(\vec b\), and \(\vec c\) for vertices A, B, and C respectively. G is the centroid of the triangle. We need to find the vector \(\overrightarrow{AG}\).

What is a Position Vector?

A position vector is a vector that indicates the position of a point in space relative to a fixed origin. If O is the origin, the position vector of a point P is \(\overrightarrow{OP}\).

What is a Triangle Centroid?

The centroid of a triangle is the point where the three medians of the triangle intersect. A median is a line segment from a vertex to the midpoint of the opposite side. The centroid is the center of mass of the triangle.

Finding the Position Vector of the Centroid

If the vertices of a triangle are A, B, and C with position vectors \(\vec a\), \(\vec b\), and \(\vec c\) respectively, the position vector of the centroid G, denoted by \(\vec g\), is given by the formula:

\[ \vec g = \frac{\vec a + \vec b + \vec c}{3} \]

Calculating the Vector \(\overrightarrow{AG}\)

The vector \(\overrightarrow{AG}\) is the vector from point A to point G. In terms of position vectors, this vector is given by the position vector of the terminal point (G) minus the position vector of the initial point (A):

\[ \overrightarrow{AG} = \text{Position vector of G} - \text{Position vector of A} \]

\[ \overrightarrow{AG} = \vec g - \vec a \]

Now, we substitute the formula for \(\vec g\) into this equation:

\[ \overrightarrow{AG} = \left(\frac{\vec a + \vec b + \vec c}{3}\right) - \vec a \]

To simplify this expression, we find a common denominator:

\[ \overrightarrow{AG} = \frac{\vec a + \vec b + \vec c}{3} - \frac{3\vec a}{3} \]

\[ \overrightarrow{AG} = \frac{(\vec a + \vec b + \vec c) - 3\vec a}{3} \]

Combine the terms with \(\vec a\):

\[ \overrightarrow{AG} = \frac{\vec a - 3\vec a + \vec b + \vec c}{3} \]

\[ \overrightarrow{AG} = \frac{-2\vec a + \vec b + \vec c}{3} \]

This can be written as:

\[ \overrightarrow{AG} = \frac{\vec b + \vec c - 2\vec a}{3} \]

Comparing with Options

Let's compare our result with the given options:

1. \(\frac{\vec a + \vec b +\vec c}{3}\)
2. \(\frac{2\vec a - \vec b -\vec c}{3}\)
3. \(\frac{\vec b + \vec c - 2\vec a}{3}\)
4. \(\frac{\vec a - 2\vec b - 2\vec c}{3}\)

Our derived expression \(\frac{\vec b + \vec c - 2\vec a}{3}\) matches Option 3.

Revision Table: Key Vector Concepts

Additional Information on Triangle Centroid and Vectors

The centroid G divides each median in a 2:1 ratio. For example, on the median from A to the midpoint M of BC, the centroid G is located such that AG : GM = 2 : 1. This property can also be used to derive the formula for the centroid's position vector.

Using the section formula, the position vector of a point R dividing a line segment PQ with position vectors \(\vec p\) and \(\vec q\) in the ratio \(m:n\) is given by \(\vec r = \frac{n\vec p + m\vec q}{m+n}\).

Let M be the midpoint of BC. Its position vector is \(\vec m = \frac{\vec b + \vec c}{2}\). The centroid G divides the median AM in the ratio 2:1. So, G divides the segment joining A (position vector \(\vec a\)) and M (position vector \(\vec m\)) in the ratio 2:1. Using the section formula:

\[ \vec g = \frac{1 \cdot \vec a + 2 \cdot \vec m}{2+1} = \frac{\vec a + 2\left(\frac{\vec b + \vec c}{2}\right)}{3} = \frac{\vec a + \vec b + \vec c}{3} \]

This confirms the centroid position vector formula used in the solution. The vector \(\overrightarrow{AG}\) represents the displacement from vertex A to the centroid G, which is often useful in various vector geometry problems involving triangles.