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0593_C02_fm Page 31 Monday, May 6, 2002 1:46 PM

Review of Vector Algebra 31

D , B , and C are the projections of D, B, and C along A; hence, let D , B , and C be

⊥

⊥

⊥
defined as:
−
−
D = D D , B = B B , and C = C C || (2.7.12)
−
⊥
⊥
⊥
||
||
D , B , and C are the components of D, B, and C perpendicular to A. Because D is the
⊥
⊥
⊥
resultant of B and C (D = B + C), Eq. (2.7.11) shows that:
D = B + C ⊥ (2.7.13)
⊥
⊥
Hence, let D , B , and C be depicted as in Figure 2.7.3.
⊥
⊥
⊥
Consider the products n × D , n × B , and n × C : because n  is 1.0, and because
⊥
⊥
⊥
A
A
A
A
n is perpendicular to D , B , and C , the definition of Eq. (2.7.1) shows that n × D ,
⊥
⊥
⊥
⊥
A
A
n × B , and n × C are vectors in the plane of D , B , and C with the same magnitudes
⊥
⊥
⊥
⊥
⊥
A
A
ˆ
ˆ
,
as D , B , and C and with directions perpendicular to D , B , and C . Let D ⊥ B ⊥ , and
⊥
⊥
⊥
⊥
⊥
⊥
ˆ
ˆ
ˆ
,
C ⊥ represent n × D , n × B , and n × C . Then, in view of Figure 2.7.3, D ⊥ B ⊥ , and
⊥
⊥
⊥
A
A
A
ˆ
C ⊥ may be represented as in Figure 2.7.4. Hence, we have:
ˆ
ˆ
ˆ
D = B + C ⊥ (2.7.14)
⊥
⊥
or,
n × D = n × B + n × C (2.7.15)
A ⊥ A ⊥ A ⊥
Then, from Eq. (2.7.13) we have:
n ×( B + ) = n × B + n ×
A ⊥ C ⊥ A ⊥ A C ⊥ (2.7.16)
By multiplying by A (a scalar) we then obtain (see Eq. (2.3.6)):
A ×( B + ) = A B + A C ⊥ (2.7.17)
×
×
C
⊥
⊥
⊥
C ˆ ⊥
ˆ
D ⊥
D
⊥
C
ˆ
A ⊥ B
⊥
B ⊥
FIGURE 2.7.3 FIGURE 2.7.4
A representation of vectors perpendicular to a A representation of D ˆ ⊥ (= n A × D ⊥ ), ˆ B ⊥ (= n A × B ⊥ ),
vector A. and C ˆ ⊥ (= n A × C ⊥ ).