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Question: If G is the centroid of a triangle ABC, prove that \[\overrightarrow{GA}+\overrightarrow{GB}+\overri...

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

Explanation

Solution

Hint: Use the formula that the vector of the centroid ( G→\overrightarrow{G} ) is given as 3G→=A→+B→+C→3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} . Also, apply the concept that GA→=A→−G→\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 GA→+GB→+GC→=0\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0
Now, we know that vector AB→=B→−A→\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A} . Here both A→\overrightarrow{A} and B→\overrightarrow{B} are the point vectors that represent the vector from the origin. Also, AB→\overrightarrow{AB} is the line vector that gives the difference of the two vectors A→\overrightarrow{A} and B→\overrightarrow{B} in that order.
So we will apply this concept and will write all the vectors GA→, GB→, GC→\overrightarrow{GA},\overrightarrow{\,GB},\,\overrightarrow{GC} as follows:
GA→=A→−G→,       GB→=B→−G→,      GC→=C→−G→\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 A→\overrightarrow{A} , B→\overrightarrow{B} and C→\overrightarrow{C} , then the centroid vector G→\overrightarrow{G} is given by the formula 3G→=A→+B→+C→3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} .
So here we have to apply this concept and we will simplify it as follows:

& \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{aligned}$$ Next, we know that $$\overrightarrow{GA}=\overrightarrow{A}-\overrightarrow{G},\,\,\,\,\,\,\,\overrightarrow{GB}=\overrightarrow{B}-\overrightarrow{G},\,\,\,\,\,\,\overrightarrow{GC}=\overrightarrow{C}-\overrightarrow{G}$$ , so we get: $$\begin{aligned} & \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{aligned}$$ Hence we have proven that if G is the centroid of a triangle ABC then $$\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.