Let ABC be a triangle, where the points A,B and C have position vectors a,b and c respectively. Show that the centroid of triangle ABC has position vector (1)/(3)(a+b+c).

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Definition of Centroid: The centroid of a triangle is the point where the three medians of the triangle intersect. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The centroid divides each median in a 2:1 ratio, with the longer segment being closer to the vertex. To find the position vector of the centroid, we need to average the position vectors of the vertices.Calculation of Position Vector: Let G be the centroid of triangle ABC. The position vector of G, denoted as g, is the average of the position vectors of the vertices A, B, and C. Mathematically, this can be expressed as g = (1/3)(a + b + c).Verification through Medians: To verify this, we can consider the medians from each vertex. For example, the median from vertex A to the midpoint of BC would have a position vector that is the average of b and c, which is (1/2)(b + c). Since the centroid divides this median in a 2:1 ratio, the position vector of the centroid must be closer to A by a factor of 2 compared to the midpoint of BC.Position Vector Calculation: Therefore, the position vector of the centroid G is two times the vector a plus one time the vector (1/2)(b + c), all divided by 3. This simplifies to g = (2a + (b + c))/3 = (2a + b + c)/3 = (a + b + c)/3.Confirmation of Centroid Position: This confirms that the position vector of the centroid G is indeed (1/3)(a + b + c), which is the average of the position vectors of the vertices A, B, and C.

Let $A B C$ be a triangle, where the points $A, B$ and $C$ have position vectors $a, b$ and $c$ respectively. Show that the centroid of triangle $A B C$ has position vector $\frac{1}{3}(a+b+c)$ .

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Let ABC A B C ABC be a triangle, where the points A,B A, B A,B and C C C have position vectors a,b a, b a,b and c c c respectively. Show that the centroid of triangle ABC A B C ABC has position vector 13(a+b+c) \frac{1}{3}(a+b+c) 31​(a+b+c).

Let $A B C$ be a triangle, where the points $A, B$ and $C$ have position vectors $a, b$ and $c$ respectively. Show that the centroid of triangle $A B C$ has position vector $\frac{1}{3}(a+b+c)$ .