Page 46 - Dynamics of Mechanical Systems

0593_C02_fm Page 27 Monday, May 6, 2002 1:46 PM

Review of Vector Algebra 27

Z
V
n
z

| A | = 4
B
B |
Y | = 3
n
y
30°
X n x A

FIGURE 2.6.4 FIGURE 2.6.5
A vector V and coordinate axes X, Y, and Z. Vectors A and B.

This result will prove to be useful in many ways. It demonstrates the analytical advantage
of vector algebra. For example, if we want to obtain the magnitude of a vector, it is simply:

A = A = A A = A 2 + A 2 + A 2 (2.6.24)
⋅
2
1 2 3
Also, if we want to find the angle θ between two vectors, we have:

⋅
cosθ= AB = A B + AB 2 + AB 3 (2.6.25)
1 1
3
2
AB 2 2 2 1 2 2 + ) 1
2
A
( A 1 + A 2 + ) (B 1 + B 2 B 3 2
2
3
The application and utility of the scalar product may be further illustrated by a few
examples, as in the following paragraphs.
Example 2.6.1: Scalar Product Definition
Consider the vectors A and B, shown in Figure 2.6.5. Let the magnitude of A and B be 4
and 3, respectively. Evaluate the scalar product.
Solution: From Figure 2.6.5, the angle between A and B is 30˚; therefore, from Eq. (2.6.1)
the scalar product is:

⋅
AB = A B cosθ = ()( )( 3 2 ) = 6 3 (2.6.26)
43
Example 2.6.2: Magnitude of a Vector
Let the vector A be expressed in terms of mutually perpendicular unit vectors n , n , and
2
1
n as:
3
A = 5 n − 2 n + 4 n (2.6.27)
1 2 3
Determine the magnitude of A.
Solution: From Eqs. (2.6.4) and (2.6.23) we have:

⋅
2
A = A = A A = () +− ( 2) + () = 45 (2.6.28)
2
2
2
2
4
5