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0593_C02_fm Page 53 Monday, May 6, 2002 1:46 PM

Review of Vector Algebra 53

Using these concepts, find the distance d between the lines L and L shown in Figure
2
1
P2.8.3, where the coordinates are expressed in feet. (Observe that n may be obtained from
the vector product (P Q × P Q )/P Q × P Q .)
2
2
2
1
1
1
2
1
Z
n z
P (2,-2,4)
1
n
Q (0,6,3)
2
P (6,-1,4)
2
d n y
O
Y
Q (1,5,0)
1
FIGURE P2.8.3
Distance d between non-parallel, n x
non-intersecting lines. X
P2.8.4: The triple vector product is useful in determining second-moment vectors. Let P
be a particle with mass m located at a point P with coordinates (x, y, z) relative to a Cartesian
coordinate frame X, Y, Z, as in Figure P2.8.4. (If a particle is small, it may be identified by
a point.) Let position vector p locate P relative to the origin O. Let n be an arbitrarily
a
directed unit vector and let n , n , and n be unit vectors parallel to axes X, Y, and Z as
z
y
x
shown. The second moment vector I of P for O for the direction n is then defined as:
a
a
I = m p ×( n × p)
a a
Z
P(x,y,z)
n z
n
p a
O
Y
n y
FIGURE P2.8.4 n x
A particle P in a Cartesian reference frame. X

a. Show that in this expression the parentheses are unnecessary; that is, unlike the
general triple vector products in Eqs. (2.8.12) and (2.8.13), we have here:

mp ×( n × p) = (
m p × ) ×
a n a p
b. Observing that p may be expressed as:

p = x n + y n + z n
x y z
find I , I , and I , the second moment vectors for the directions n , n , and n .
y
z
x
z
x
y