Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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).

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

Full solution

Q. Let ABC A B C be a triangle, where the points A,B A, B and C C have position vectors a,b a, b and c c respectively. Show that the centroid of triangle ABC A B C has position vector 13(a+b+c) \frac{1}{3}(a+b+c) .
  1. 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 2:12:1 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.
  2. Calculation of Position Vector: Let GG be the centroid of triangle ABCABC. The position vector of GG, denoted as gg, is the average of the position vectors of the vertices AA, BB, and CC. Mathematically, this can be expressed as g=13(a+b+c)g = \frac{1}{3}(a + b + c).
  3. Verification through Medians: To verify this, we can consider the medians from each vertex. For example, the median from vertex AA to the midpoint of BCBC would have a position vector that is the average of bb and cc, which is (12)(b+c)(\frac{1}{2})(b + c). Since the centroid divides this median in a 2:12:1 ratio, the position vector of the centroid must be closer to AA by a factor of 22 compared to the midpoint of BCBC.
  4. Position Vector Calculation: Therefore, the position vector of the centroid GG is 22 times the vector a\mathbf{a} plus one time the vector (12)(b+c)(\frac{1}{2})(\mathbf{b} + \mathbf{c}), all divided by 33. This simplifies to g=2a+(b+c)3=2a+b+c3=a+b+c3\mathbf{g} = \frac{2\mathbf{a} + (\mathbf{b} + \mathbf{c})}{3} = \frac{2\mathbf{a} + \mathbf{b} + \mathbf{c}}{3} = \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}.
  5. Confirmation of Centroid Position: This confirms that the position vector of the centroid GG is indeed (13)(a+b+c)(\frac{1}{3})(a + b + c), which is the average of the position vectors of the vertices AA, BB, and CC.

More problems from Write a quadratic function from its x-intercepts and another point