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If $Q$ is the point of intersection of medians of a triangle $\triangle ABC$ then show that $\vec{QA} + \vec{QB} + \vec{QC} = 0$

I am taking help from this link.

If Q is the point of concurrence of the?

But two problems first it is not well written and second its not involved use of vectors.

3 Answers 3

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$$\vec{QA} + \vec{QB} + \vec{QC} = 0 \Leftrightarrow (\vec{A}-\vec{Q})+(\vec{B}-\vec{Q})+(\vec{C}-\vec{Q})=0 \Leftrightarrow \vec{Q}=\frac{\vec{A}+\vec{B}+\vec{C}}{3}$$

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Let the triangle be $\triangle ABC$. Its medians are $\vec{AA_1} = \frac 12 (\vec{AB}+\vec{AC}) $, $\vec{BB_1} = \frac 12 (\vec{BA}+\vec{BC}) $, and $\vec{CC_1} = \frac 12 (\vec{CB}+\vec{CA}) $.

Now from basic goemetry we know that the common point of all medians $Q$ splits the medians in relation $2:1$, on other words, $\vec{AQ} = \frac 23 \vec{AA_1}$, $\vec{CQ} = \frac 23 \vec{CC_1}$, and $\vec{BQ} = \frac 23 \vec{BB_1}$.

Finally, $$\vec{QA} + \vec{QB} + \vec{QC} = - \frac23 (\vec{AA_1} + \vec{BB_1} + \vec{CC_1}) = -\frac 13(\vec{AB}+\vec{AC} + \vec{BA}+\vec{BC}+\vec{CB}+\vec{CA})=\vec 0. $$

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    Can you explain first step. – 2017-01-12
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    @JohnSr Parallelogramm rule of vector addition plus a basic fact that the point of intersection of parallelogramm diagonals splits them in half. – 2017-01-12
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    How you finded medians? – 2017-01-12
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    Can you send me the link of this property please. – 2017-01-12
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    If you have a parallelogram $XYZT$ with diagonals intersecting at a point $Q$, then the triangles $\triangle XQY$ and $\triangle ZQT$ are equal by a criterion "a side and two adjacent angles are the same for both triangles". This gives you immediately $XQ=QZ$ and $YQ=QT$. – 2017-01-12
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    You are confusing me question is here has a triangle you are taking the example of parallelogram. Your solution is clear except how you find medians. – 2017-01-12
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Write $\vec A$ for the position vector $\vec{OA}$. The median line $m_A$ through $A$ is then given by the parametrization $$m_A:\quad t\mapsto (1-t)\>\vec A+t\>{\vec B+\vec C\over2}\qquad(-\infty

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    Sir not understanding anything except last step. – 2017-01-12