Proof using vectors.

Question

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.

Answer 1

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

Answer 2

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

Answer 3

Write $\vec A$ for the position vector $\vec{OA}$. The median line $m_A$ through $A$ is then given by the parametrization $$m_A:\quad t\mapsto (1-t)\>\vec A+t\>{\vec B+\vec C\over2}\qquad(-\infty<t<\infty)\ .$$ Putting $t:={2\over3}$ here produces the point $$\vec Q:={1\over3}\bigl(\vec A+\vec B+\vec C\bigr)\in m_A\ .$$ Due to symmetry this point will lie on all three medians, hence is the point of intersection of the medians. It follows that $$\vec{QA}+\vec{QB}+\vec{QC}=(\vec A-\vec Q)+(\vec B-\vec Q)+(\vec C-\vec Q)=3\left({\vec A+\vec B+\vec C\over 3}-\vec Q\right)=\vec0\ .$$