If \(\vec a, \vec b, \vec c\) , are the position vectors of the vertices A, B, C respectively of a triangle ABC and G is the centroid of the triangle, then what is \(\overrightarrow{AG}\) equal to ? .
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:
- \(\frac{\vec a + \vec b +\vec c}{3}\)
- \(\frac{2\vec a - \vec b -\vec c}{3}\)
- \(\frac{\vec b + \vec c - 2\vec a}{3}\)
- \(\frac{\vec a - 2\vec b - 2\vec c}{3}\)
Our derived expression \(\frac{\vec b + \vec c - 2\vec a}{3}\) matches Option 3.
Revision Table: Key Vector Concepts
| Concept | Description | Vector Formula (relative to origin O) |
|---|---|---|
| Position Vector of Point P | Vector from origin O to point P | \(\overrightarrow{OP} = \vec p\) |
| Vector from Point A to B | Vector \(\overrightarrow{AB}\) | \(\overrightarrow{OB} - \overrightarrow{OA} = \vec b - \vec a\) |
| Midpoint M of AB | Point halfway between A and B | \(\vec m = \frac{\vec a + \vec b}{2}\) |
| Centroid G of Triangle ABC | Intersection of medians | \(\vec g = \frac{\vec a + \vec b + \vec c}{3}\) |
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.
Important Questions from Centroid
If the centroid of a triangle formed by (7, x), (y, -6) and (9, 10) is (6, 3), then the values of x and y are respectively
What is the area of the triangle formed by these lines?
The centroid of the triangle is at which one of the following points?
The centroid of the triangle with vertices A(2, -3, 3), B(5, -3, -4) and C(2, -3, -2) is the point
If a vertex of a triangle is (1, 1) and the midpoints of two sides of the triangle through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is
