point division and position vectors PointDivisionAndPositionVectors 2026-08-20 04:33:32 2026-08-21 15:13:09 Topic section formula vector midpoint affine dependence Ceva's theorem Menelaus's theorem vector centroid geometry % this is the default PhysicsLibrary preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \section*{Point Division, Affine Combinations, Ceva, and Menelaus} Let $P$ divide the directed segment $AB$ internally in the ratio $m:n$, meaning \[ \frac{AP}{PB}=\frac{m}{n}. \] With position vectors $\mathbf{a},\mathbf{b},\mathbf{p}$, the modern section formula is \[ \boxed{\mathbf{p}=\frac{n\mathbf{a}+m\mathbf{b}}{m+n}}. \tag{1} \] The midpoint is the special case \[ \mathbf m=\frac{\mathbf{a}+\mathbf{b}}{2}. \tag{2} \] \begin{center} \includegraphics[width=.68\textwidth]{brand_fig_7a.png} \textit{Figure 7a, modernized: point division and position vectors.} \end{center} \subsection*{Midpoints of two vectors} If $M$ and $N$ are the midpoints of $AA'$ and $BB'$, then \[ \overrightarrow{MN} =\frac12\left(\overrightarrow{AB}+\overrightarrow{A'B'}\right). \] \begin{center} \includegraphics[width=.68\textwidth]{brand_fig_7b.png} \textit{Figure 7b, modernized: midpoint theorem for two directed segments.} \end{center} \subsection*{Affine dependence} If \[ a\mathbf{a}+b\mathbf{b}+c\mathbf{c}=\mathbf0, \qquad a+b+c=0, \] with nonzero coefficients, then the points $A,B,C$ are collinear. Similarly, if four position vectors satisfy a nontrivial \PMlinkescapetext{relation} whose coefficients sum to zero, the four points are coplanar. \subsection*{Ceva's theorem} For points $L\in BC$, $M\in CA$, and $N\in AB$, the cevians $AL$, $BM$, and $CN$ are concurrent precisely when the directed division ratios satisfy \[ \frac{BL}{LC}\frac{CM}{MA}\frac{AN}{NB}=1. \] Brand derives this efficiently by expressing the intersection point as an affine combination of the vertex position vectors. \begin{center} \includegraphics[width=.68\textwidth]{brand_fig_7c.png} \textit{Figure 7c, modernized: concurrent cevians used in the vector proof of Ceva's theorem.} \end{center} \subsection*{Menelaus's theorem} For a transversal meeting the extended sides of triangle $ABC$ at $L,M,N$, the directed ratios obey \[ \frac{BL}{LC}\frac{CM}{MA}\frac{AN}{NB}=-1. \] \begin{center} \includegraphics[width=.68\textwidth]{brand_fig_7d.png} \textit{Figure 7d, modernized: transversal geometry used in Menelaus's theorem.} \end{center} \subsection*{Centroid geometry} The medians of a triangle are concurrent at the centroid $G$, and \[ \mathbf r_G=\frac{\mathbf{r}_A+\mathbf{r}_B+\mathbf r_C}{3}. \] Each median is divided by $G$ in the ratio $2:1$ measured from the vertex. \begin{center} \includegraphics[width=.68\textwidth]{brand_fig_7e.png} \textit{Figure 7e, modernized: medians and centroid of a triangle.} \end{center} For a tetrahedron, the segments joining the midpoints of opposite edges meet at their common midpoint. \begin{center} \includegraphics[width=.68\textwidth]{brand_fig_7f.png} \textit{Figure 7f, modernized: midpoint geometry in a tetrahedron.} \end{center} \section*{Source problems} \begin{enumerate} \item If $P,Q,R$ are the midpoints of the sides of triangle $ABC$, prove for any origin $O$ that \[ \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} =\overrightarrow{OP}+\overrightarrow{OQ}+\overrightarrow{OR}. \] \item For quadrilateral $ABCD$, with $P,Q$ the midpoints of $AC,BD$ and $M$ the midpoint of $PQ$, prove \[ \overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+ \overrightarrow{CD}=4\overrightarrow{PQ}, \] and \[ \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+ \overrightarrow{OD}=4\overrightarrow{OM}. \] \item Prove the $2:1$ centroid division theorem for a triangle and $\mathbf r_G=(\mathbf{r}_A+\mathbf{r}_B+\mathbf r_C)/3$. \item If $G,G'$ are the centroids of triangles $ABC,A'B'C'$, prove \[ \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'} =3\overrightarrow{GG'}. \] \item If $E,F$ are midpoints of $AB,BC$ in parallelogram $ABCD$, prove the lines $DE,DF$ trisect diagonal $AC$ in the manner described by Brand. \item If $A,B,C,D$ are the midpoints of the successive sides of any space quadrilateral, prove \[ \overrightarrow{AB}=\overrightarrow{DC},\qquad \overrightarrow{AD}=\overrightarrow{BC}. \] \item Prove the midpoint theorem for opposite edges of a tetrahedron shown in Figure 7f. \item If $G$ is the centroid of $A,B,C$ and $M$ the mean center of $A,B,C,D$, prove that $M$ divides $DG$ in the ratio $3:1$. \item Prove Desargues's theorem using affine vector relations: if triangles $ABC$ and $A'B'C'$ are perspective from a point, then the intersections of corresponding sides are collinear. \end{enumerate} \section*{Modern notation references} The notation and terminology in this modernized article follow standard present-day mechanics and vector-analysis usage, particularly: \begin{enumerate} \item J. R. Taylor, \emph{Classical Mechanics}, University Science Books, 2005. \item D. Kleppner and R. Kolenkow, \emph{An Introduction to Mechanics}, 2nd ed., Cambridge University Press, 2014. \item H. Goldstein, C. Poole, and J. Safko, \emph{Classical Mechanics}, 3rd ed., Addison--Wesley, 2002. \end{enumerate} \section*{Source} This article is a modernized restatement of the corresponding Public \PMlinkescapetext{Domain} article in Louis Brand, \emph{Vectorial Mechanics}, John Wiley \& Sons, New York, 1930, Chapter I, ``Vector Algebra.'' The original 1930 edition is the source basis.