Question

If G is the centroid of a triangle ABC, prove that \[\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\]

Answer 1

Hint: Use the formula that the vector of the centroid ( \[\overrightarrow{G}\] ) is given as \[3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}\] . Also, apply the concept that \[\overrightarrow{GA}=\overrightarrow{A}-\overrightarrow{G}\] .
Complete step by step solution:
In the question, we have to prove that if G is the centroid of a triangle ABC, then \[\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\]
Now, we know that vector \[\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}\] . Here both \[\overrightarrow{A}\] and \[\overrightarrow{B}\] are the point vectors that represent the vector from the origin. Also, \[\overrightarrow{AB}\] is the line vector that gives the difference of the two vectors \[\overrightarrow{A}\] and \[\overrightarrow{B}\] in that order.
So we will apply this concept and will write all the vectors \[\overrightarrow{GA},\overrightarrow{\,GB},\,\overrightarrow{GC}\] as follows:
 \[\overrightarrow{GA}=\overrightarrow{A}-\overrightarrow{G},\,\,\,\,\,\,\,\overrightarrow{GB}=\overrightarrow{B}-\overrightarrow{G},\,\,\,\,\,\,\overrightarrow{GC}=\overrightarrow{C}-\overrightarrow{G}\]
Now, we also know that the If the vertices of the triangle ABD are represented by the vectors \[\overrightarrow{A}\] , \[\overrightarrow{B}\] and \[\overrightarrow{C}\] , then the centroid vector \[\overrightarrow{G}\] is given by the formula \[3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}\] .
So here we have to apply this concept and we will simplify it as follows:
 \[\begin{align}
  & \Rightarrow 3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} \\
 & \Rightarrow \overrightarrow{G}+\overrightarrow{G}+\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} \\
 & \Rightarrow 0=\left( \overrightarrow{A}-\overrightarrow{G} \right)+\left( \overrightarrow{B}-\overrightarrow{G} \right)+\left( \overrightarrow{C}-\overrightarrow{G} \right) \\
 & \Rightarrow \left( \overrightarrow{A}-\overrightarrow{G} \right)+\left( \overrightarrow{B}-\overrightarrow{G} \right)+\left( \overrightarrow{C}-\overrightarrow{G} \right)=0 \\
\end{align}\]
Next, we know that \[\overrightarrow{GA}=\overrightarrow{A}-\overrightarrow{G},\,\,\,\,\,\,\,\overrightarrow{GB}=\overrightarrow{B}-\overrightarrow{G},\,\,\,\,\,\,\overrightarrow{GC}=\overrightarrow{C}-\overrightarrow{G}\] , so we get:
 \[\begin{align}
  & \Rightarrow \left( \overrightarrow{A}-\overrightarrow{G} \right)+\left( \overrightarrow{B}-\overrightarrow{G} \right)+\left( \overrightarrow{C}-\overrightarrow{G} \right)=0 \\
 & \Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0 \\
\end{align}\]
Hence we have proven that if G is the centroid of a triangle ABC then \[\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\] .

Note: Here, we have to very careful that \[\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}\] and not \[\overrightarrow{AB}=\overrightarrow{A}-\overrightarrow{B}\] . So this is one of the common mistakes that needs to be avoided while solving the given question.