VECTOR ALGEBRA
O
A
B
C
D
E
G F
a
b
c
Figure 7.6 The centroid of a triangle. The triangle is defined by the pointsA,
BandCthat have position vectorsa,bandc. The broken linesCD,BE,AF
connect the vertices of the triangle to the mid-points of the opposite sides;
these lines intersect at the centroidGof the triangle.
Result (7.6) is a version of theratio theoremand we may use it in solving more
complicated problems.
The vertices of triangleABChave position vectorsa,bandcrelative to some originO
(see figure 7.6). Find the position vector of the centroidGof the triangle.
From figure 7.6, the pointsDandEbisect the linesABandACrespectively. Thus from
the ratio theorem (7.6), withλ=μ=1/2, the position vectors ofDandErelative to the
origin are
d=^12 a+^12 b,
e=^12 a+^12 c.
Using the ratio theorem again, we may writethe position vector of a general point on the
lineCDthat divides the line in the ratioλ:(1−λ)as
r=(1−λ)c+λd,
=(1−λ)c+^12 λ(a+b), (7.7)
where we have expresseddin terms ofaandb. Similarly, the position vector of a general
point on the lineBEcan be expressed as
r=(1−μ)b+μe,
=(1−μ)b+^12 μ(a+c). (7.8)
Thus, at the intersection of the linesCDandBEwe require, from (7.7), (7.8),
(1−λ)c+^12 λ(a+b)=(1−μ)b+^12 μ(a+c).
By equating the coefficents of the vectorsa,b,cwe find
λ=μ,^12 λ=1−μ, 1 −λ=^12 μ.