Mathematical Methods for Physics and Engineering - Matematica.NET

VECTOR ALGEBRACEAGFDacBbOFigure 7.6 The centroid of a triangle. The triangle is defined by the points A,B and C that have position vectors a, b and c. The broken lines CD, BE, AFconnect the vertices of the triangle to the mid-points of the opposite sides;these lines intersect at the centroid G of the triangle.Result (7.6) is a version of the ratio theorem and we may use it in solving morecomplicated problems.◮The vertices of triangle ABC have position vectors a, b and c relative to some origin O(see figure 7.6). Find the position vector of the centroid G of the triangle.From figure 7.6, the points D and E bisect the lines AB and AC respectively. Thus fromthe ratio theorem (7.6), with λ = µ =1/2, the position vectors of D and E relative to theorigin ared = 1 a + 1 b,2 2e = 1 a + 1 c.2 2Using the ratio theorem again, we may write the position vector of a general point on theline CD that divides the line in the ratio λ :(1− λ) asr =(1− λ)c + λd,=(1− λ)c + 1 λ(a + b), (7.7)2where we have expressed d in terms of a and b. Similarly, the position vector of a generalpoint on the line BE can be expressed asr =(1− µ)b + µe,=(1− µ)b + 1 µ(a + c). (7.8)2Thus, at the intersection of the lines CD and BE we require, from (7.7), (7.8),(1 − λ)c + 1 λ(a + b) =(1− µ)b + 1 µ(a + c).2 2By equating the coefficents of the vectors a, b, c we find1λ = µ, λ =1− µ, 1 − λ = 1 µ.2 2216