I have the points $O, A, B$ and $C$.

Relative to $O$, the position vectors of $A$, $B$ and $C$ are $(1,4, 2 )$, $ (3, 3, 3) $, $( 2, -1, 1)$

Want to show that the lines $OB$ and $AC$ bisect each other.

Is it sufficient to show that $\frac{1}{2} \vec{OB} = \vec{OA} + \frac{1}{2} \vec{AC}$?

Are there other ways using vectors?

  • $\begingroup$ If the diagonals of a quadrilateral bisect each other, it is a parallelogram. So, B=A+C. $\endgroup$ – Anubhab Ghosal Apr 12 '19 at 15:37

Yes, your method is correct.

You may also show that $$ \vec{OA} = \vec{CB} $$ and $$ \vec{OC} = \vec{AB} $$ which make the quadrilateral OABC into a parallelogram where the diagonals bisect each other.

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