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Question: If G is the centroid of a triangle ABC then\(\overset{\to }{\mathop{\text{GA}}}\,+\overset{\to }{\ma...

If G is the centroid of a triangle ABC thenGA→ +GB→ +GC→ \overset{\to }{\mathop{\text{GA}}}\,+\overset{\to }{\mathop{\text{GB}}}\,+\overset{\to }{\mathop{\text{GC}}}\, is equal to :
A . 0→ \overset{\to }{\mathop{0}}\,
B. 3GA→ 3\overset{\to }{\mathop{\text{GA}}}\,
C. 3GB→ 3\overset{\to }{\mathop{\text{GB}}}\,
D. 3GC→ 3\overset{\to }{\mathop{\text{GC}}}\,

Explanation

Solution

Hint: In triangle ABC we can define centroid as G→=A→+B→+C→3\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3} where A→,B→,C→\overrightarrow{A},\overrightarrow{B},\overrightarrow{C} are sides vector of triangle ABC. If have length of sides of triangle then we can use it as in same form.

Complete step by step solution:
If sides vectors of triangle ABC are A→,B→,C→\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}. Then we can write centroid of triangle as
G→=A→+B→+C→3\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}
We can arrange it as
3G→=A→+B→+C→3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}..................................(i)
We can also write
GA→=A→−G→\overrightarrow{GA}=\overrightarrow{A}-\overrightarrow{G}
GB→=B→−G→\overrightarrow{GB}=\overrightarrow{B}-\overrightarrow{G}
GC→=C→−G→\overrightarrow{GC}=\overrightarrow{C}-\overrightarrow{G}
Hence given expression can be written as
⇒GA→+GB→+GC→=A→−G→+B→−G→+C→−G→\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{A}-\overrightarrow{G}+\overrightarrow{B}-\overrightarrow{G}+\overrightarrow{C}-\overrightarrow{G}
⇒GA→+GB→+GC→=A→+B→+C→−3G→\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}-3\overrightarrow{G}
From equation (i) we can write value of A→+B→+C→\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}
⇒GA→+GB→+GC→=3G→−3G→\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=3\overrightarrow{G}-3\overrightarrow{G}
⇒GA→+GB→+GC→=0\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0
Hence option a is correct.

Note: We can use the same formula of centroid in terms of coordinate as well. Also we can write AB→\overrightarrow{AB} as difference of individual vector A and vector B as below:
AB→=B→−A→\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}
But we always write vector B first and then vector A. We need to remember this point.