Multiplication by a scalar

VECTOR ALGEBRA
b
a
c
b+c
b
a
c
b+c
a + (b + c)
b
c
a+b
a
a+b
(a + b) + c
Figure 7.2 Addition of three vectors showing the associativity relation.
−b
a
a−b
a
b
Figure 7.3 Subtraction of two vectors.
The subtraction of two equal vectors yields the zero vector, 0, which has zero
magnitude and no associated direction.
7.3 Multiplication by a scalar
Multiplication of a vector by a scalar (not to be confused with the ‘scalar
product’, to be discussed in subsection 7.6.1) gives a vector in the same direction
as the original but of a proportional magnitude. This can be seen in figure 7.4.
The scalar may be positive, negative or zero. It can also be complex in some
applications. Clearly, when the scalar is negative we obtain a vector pointing
in the opposite direction to the original vector. Multiplication by a scalar is
associative, commutative and distributive over addition. These properties may be
summarised for arbitrary vectors a and b and arbitrary scalars λ and µ by
(λµ)a = λ(µa) = µ(λa),
(7.3)
λ(a + b) = λa + λb,
(7.4)
(λ + µ)a = λa + µa.
(7.5)
214
7.3 MULTIPLICATION BY A SCALAR
λa
a
Figure 7.4 Scalar multiplication of a vector (for λ > 1).
B
µ
b
P
λ
p
A
a
O
Figure 7.5 An illustration of the ratio theorem. The point P divides the line
segment AB in the ratio λ : µ.
Having defined the operations of addition, subtraction and multiplication by a
scalar, we can now use vectors to solve simple problems in geometry.
A point P divides a line segment AB in the ratio λ : µ (see figure 7.5). If the position
vectors of the points A and B are a and b, respectively, find the position vector of the
point P .
As is conventional for vector geometry problems, we denote the vector from the point A
to the point B by AB. If the position vectors of the points A and B, relative to some origin
O, are a and b, it should be clear that AB = b − a.
Now, from figure 7.5 we see that one possible way of reaching the point P from O is
first to go from O to A and to go along the line AB for a distance equal to the the fraction
λ/(λ + µ) of its total length. We may express this in terms of vectors as
λ
AB
λ+µ
λ
=a+
(b − a)
λ+µ
λ
λ
a+
= 1−
b
λ+µ
λ+µ
λ
µ
a+
b,
=
λ+µ
λ+µ
OP = p = a +
(7.6)
which expresses the position vector of the point P in terms of those of A and B. We would,
of course, obtain the same result by considering the path from O to B and then to P . 215
VECTOR ALGEBRA
C
E
G
A
F
D
a
c
B
b
O
Figure 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, AF
connect 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 more
complicated 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 from
the ratio theorem (7.6), with λ = µ = 1/2, the position vectors of D and E relative to the
origin are
d = 12 a + 12 b,
e = 12 a + 12 c.
Using the ratio theorem again, we may write the position vector of a general point on the
line CD that divides the line in the ratio λ : (1 − λ) as
r = (1 − λ)c + λd,
= (1 − λ)c + 12 λ(a + b),
(7.7)
where we have expressed d in terms of a and b. Similarly, the position vector of a general
point on the line BE can be expressed as
r = (1 − µ)b + µe,
= (1 − µ)b + 12 µ(a + c).
Thus, at the intersection of the lines CD and BE we 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 vectors a, b, c we find
λ = µ,
1
λ
2
= 1 − µ,
216
1 − λ = 12 µ.
(7.8)