Higher Engineering Mathematics, Sixth Edition

276 Higher Engineering Mathematics

26.2 The scalar product of two vectors

When vectoroais multipliedby a scalar quantity,sayk,

the magnitude of the resultant vector will bektimes the

magnitude ofoaand its direction will remain the same.

Thus 2×(5N at 20◦)results in a vector of magnitude

10N at 20◦.

One of the products of twovector quantitiesis called the

scalarordot productof two vectors and is defined as

the product of their magnitudes multipliedby the cosine

of the angle between them. The scalar product ofoaand

obis shown asoa•ob. For vectorsoa=oaatθ 1 ,and

ob=obatθ 2 whereθ 2 >θ 1 ,thescalarproductis:

oa•ob=oa obcos(θ 2 −θ 1 )

For vectorsv 1 andv 2 shown in Fig. 26.4, the scalar

product is:

v 1 • v 2 =v 1 v 2 cosθ



v 2

v 1

Figure 26.4

The commutative law of algebra,a×b=b×aapplies

to scalar products. This is demonstrated in Fig. 26.5. Let

oarepresent vectorv 1 andobrepresent vectorv 2. Then:

oa•ob=v 1 v 2 cosθ(by definition of

a scalar product)

O

v 1

v 2

b



a

Figure 26.5

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

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

(b)

(a)

a

c

b

O

v^2





v 2 cos 

v^1

cos



v^2

v 1

v 1

Figure 26.6

The projection ofobonoais shown in Fig. 26.6(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 )

Similarly the projection of oa onob is shown in

Fig. 26.6(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,

then cosθ=

a•b

ab

(1)