Unformatted text preview: In three dimensions, the vector a OP
point P a1, a2, a3 . (See Figure 13.) Let’s consider any other representation AB of a, where
the initial point is A x 1, y1, z1 and the terminal point is B x 2 , y2 , z2 . Then we must have
x 1 a 1 x 2, y1 a 2 y2, and z1 a 3 z2 and so a 1 x 2 x 1, a 2 y2 y1, and
a 3 z2 z1. Thus, we have the following result.
1 Given the points A x 1, y1, z1 and B x 2 , y2 , z2 , the vector a with represenl
tation AB is
x 2 x 1, y2 y1, z2 z1 EXAMPLE 3 Find the vector represented by the directed line segment with initial point A 2, 3, 4) and terminal point B 2, 1, 1 .
l SOLUTION By (1), the vector corresponding to AB is a 2 2, 1 3 ,1 4 4, 4, 3 The magnitude or length of the vector v is the length of any of its representations and
is denoted by the symbol v or v . By using the distance formula to compute the length
of a segment OP, we obtain the following formulas.
The length of the two-dimensional vector a a 1, a 2 is sa 2
1 a a2
2 The length of the three-dimensional vector a
y a (a¡+b¡, a™+b™) a+b b¡
0 a™ a™ a¡ a2
3 a 1, a 2 and
How do we add vectors algebraically? Figure 14 shows that if a
b 1, b 2 , then the sum is a b
a1 b1, a2 b2 , at least for the case where the
components are positive. In other words, to add algebraic vectors we add their components. Similarly, to subtract vectors we subtract components. From the similar triangles in
Figure 15 we see that the components of c a are ca1 and ca2. So to multiply a vector by a
scalar we multiply each component by that scalar. b™ b sa 2
1 a 1, a 2 , a 3 is x b¡ If a FIGURE 14 a 1, a 2 and b
a b a1 b1, b2 , then
b1, a 2 b2 a
ca b a1 b1, a 2 ca1, ca2 Similarly, for three-dimensional vectors,
ca ca™ a™
a¡ FIGURE 15 ca¡ a 1, a 2 , a 3 b1, b2 , b3 a1 b1, a 2 b2 , a 3 b3 a 1, a 2 , a 3 a b1, b2 , b3 a1 b1, a 2 b2 , a 3 b3 c a 1, a 2 , a 3 ca1, ca2 , ca3 b2 5E-13(pp 838-847) 838 ❙❙❙❙ 1/18/06 11:13 AM Page 838 CHAPTER 13 VECTORS AND THE GEOMETRY OF SPACE E XAMPLE 4 If a 3 b, and 2 a 4, 0, 3 and b 2, 1, 5 , ﬁnd a and the vectors a b, a b, 5 b. SOLUTION s...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.
- Fall '09