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Vectors in Cartesian Three-Space Help (page 3)

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Unit Vectors

Any vector a, reduced to standard form so its starting point is at the origin, ends up at some point ( x a , y a , z a ). This vector can be broken down into the sum of three mutually perpendicular vectors, each of which lies along one of the coordinate axes as shown in Fig. 9-9:

a = ( x a , y a , z a )

= ( x a ,0,0) + (0, y a ,0) + (0,0, z a )

= x a (1,0,0) + y a (0,1,0) + z a (0,0,1)

The vectors (1,0,0), (0,1,0), and (0,0,1) are called unit vectors because their length is 1. It is customary to name these vectors i , j , and k , because they come in handy:

(1,0,0) = i

(0,1,0) = j

(0,0,1) = k

Therefore, the vector a shown in Fig. 9-9 breaks down this way:

a = ( x a , y a , z a ) = x a i + y a j + z a k

Vectors and Cartesian Three-Space Vectors in Cartesian Three-Space Unit Vectors

Fig. 9-9 . Any vector in Cartesian three-space can be broken up into a sum of three component vectors, each of which lies on one of the coordinate axes.

PROBLEM 9-6

Break the vector b = (−2, 3, −7) down into a sum of multiples of the unit vectors i , j , and k .

SOLUTION 9-6

This is a simple process—almost trivial—but envisioning it requires a keen “mind’s eye.” If you have any trouble seeing this in your imagination, think of i as “one unit of width going to the right,” j as “one unit of height going up,” and k as “one unit of depth coming towards you.” Here we go:

b = (−2,3, − 7)

= −2 × (1,0,0) + 3 × (0,1,0) + [−7 × (0,0,1)]

= −2 i + 3 j + (−7) k

= −2 i + 3 j − 7 k

Practice problems for these concepts can be found at:  Vectors And Cartesian Three-Space Practice Test.

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