point division and position vectors

Point Division, Affine Combinations, Ceva, and Menelaus

Let P divide the directed segment AB internally in the ratio m : n, meaning

AP m ----= --. P B n

With position vectors a,b,p, the modern section formula is

|--------------| | na + mb | p = --------- . (1) ------m--+-n---

The midpoint is the special case

a + b m = ------. (2) 2

Figure 7a, modernized: point division and position vectors.

Midpoints of two vectors

If M and N are the midpoints of AA′ and BB′, then

( −−→ ) −M−→N = 1- −AB→ + A′B ′ . 2

Figure 7b, modernized: midpoint theorem for two directed segments.

Affine dependence

aa + bb + cc = 0, a + b + c = 0,

with nonzero coefficients, then the points A,B,C are collinear. Similarly, if four position vectors satisfy a nontrivial relation whose coefficients sum to zero, the four points are coplanar.

Ceva’s theorem

For points L ∈ BC, M ∈ CA, and N ∈ AB, the cevians AL, BM, and CN are concurrent precisely when the directed division ratios satisfy

BL CM AN -------------= 1. LC M A N B

Brand derives this efficiently by expressing the intersection point as an affine combination of the vertex position vectors.

Figure 7c, modernized: concurrent cevians used in the vector proof of Ceva’s theorem.

Menelaus’s theorem

For a transversal meeting the extended sides of triangle ABC at L,M,N, the directed ratios obey

BL--CM---AN-- LC M A N B = − 1.

Figure 7d, modernized: transversal geometry used in Menelaus’s theorem.

Centroid geometry

The medians of a triangle are concurrent at the centroid G, and

rA + rB + rC rG = -------------. 3

Each median is divided by G in the ratio 2 : 1 measured from the vertex.

Figure 7e, modernized: medians and centroid of a triangle.

For a tetrahedron, the segments joining the midpoints of opposite edges meet at their common midpoint.

Figure 7f, modernized: midpoint geometry in a tetrahedron.

Source problems

If P,Q,R are the midpoints of the sides of triangle ABC, prove for any origin O that $−→ −−→ − → −→ −→ − → OA + OB + OC = OP + OQ + OR.$
For quadrilateral ABCD, with P,Q the midpoints of AC,BD and M the midpoint of PQ, prove $−→ −−→ −−→ −−→ −→ AB + AD + CB + CD = 4P Q,$
and
$−→ −−→ − → −−→ −−→ OA + OB + OC + OD = 4 OM .$
Prove the 2 : 1 centroid division theorem for a triangle and r_G = (r_A + r_B + r_C)∕3.
If G,G′ are the centroids of triangles ABC,A′B′C′, prove $−−→ −−→ −−→ −−→ AA ′ + BB ′ + CC ′ = 3GG ′.$
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.
If A,B,C,D are the midpoints of the successive sides of any space quadrilateral, prove $− → −−→ −−→ −−→ AB = DC, AD = BC.$
Prove the midpoint theorem for opposite edges of a tetrahedron shown in Figure 7f.
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.
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.

Modern notation references

The notation and terminology in this modernized article follow standard present-day mechanics and vector-analysis usage, particularly:

J. R. Taylor, Classical Mechanics, University Science Books, 2005.
D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.
H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison–Wesley, 2002.

Source

This article is a modernized restatement of the corresponding Public Domain article in Louis Brand, Vectorial Mechanics, John Wiley & Sons, New York, 1930, Chapter I, “Vector Algebra.” The original 1930 edition is the source basis.

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