Higher Engineering Mathematics

238 VECTOR GEOMETRY

22.2 The scalar product of two vectors

When vectoroais multiplied by a scalar quantity,
sayk, the magnitude of the resultant vector will be
ktimes the magnitude ofoaand its direction will
remain the same. Thus 2×(5 N at 20◦) results in a
vector of magnitude 10 N at 20◦.
One of the products of two vector quantities is
called thescalarordot productof two vectors
and is defined as the product of their magnitudes
multiplied by the cosine of the angle between them.
The scalar product ofoaandobis shown asoa•ob.
For vectorsoa=oaatθ 1 , andob=obatθ 2 where
θ 2 >θ 1 ,the scalar product is:

oa•ob=oa ob cos(θ 2 −θ 1 )

For vectorsv 1 andv 2 shown in Fig. 22.4, the scalar

product is:

v 1 • v 2 =v 1 v 2 cosθ

v 2

v 1

θ

Figure 22.4

The commutative law of algebra, a×b=b×a
applies to scalar products. This is demonstrated in
Fig. 22.5. Letoarepresent vectorv 1 andobrepresent
vectorv 2. Then:

oa•ob=v 1 v 2 cosθ(by definition of

a scalar product)

b

a

v 1

v 2

O θ

Figure 22.5

Similarly,ob•oa=v 2 v 1 cosθ=v 1 v 2 cosθby the

commutative law of algebra. Thusoa•ob=ob•oa.

The projection ofobonoais shown in Fig. 22.6(a)

and by the geometry of triangleobc, it can be seen

that the projection isv 2 cosθ. Since, by definition

oa•ob=v 1 (v 2 cosθ),

it follows that

oa•ob=v 1 (the projection ofv 2 onv 1 )

b

a

c

v 1

v 2 cos

O

v^2

v^2

v 1

(b)

(a)

θ

θ

θ

θ

v^1 cos

Figure 22.6

Similarly the projection ofoaonobis shown in

Fig. 22.6(b) and isv 1 cosθ. Since by definition

ob•oa=v 2 (v 1 cosθ),

it follows that

ob•oa=v 2 (the projection ofv 1 onv 2 )

This shows that the scalar product of two vectors

is the product of the magnitude of one vector and

the magnitude of the projection of the other vector

on it.

Theangle between two vectorscan be expressed in

terms of the vector constants as follows:

Becausea•b=abcosθ,