SECTION 9.6 Vectors in Space 661 THEOREM Distance Formula in Space If P x y z , , 1 1 1 1 ( ) = and P x y z , , 2 2 2 2 ( ) = are two points in space, the distance d from P1 to P2 is d x x y y z z 2 1 2 2 1 2 2 1 2 ( ) ( ) ( ) = − + − + − (1) Using the Distance Formula Find the distance from P 1, 3, 2 1 ( ) = − to P 4, 2, 5 . 2 ( ) = − Solution EXAMPLE 1 ( ) [ ] [ ] [ ] = −− +−−+− = ++= d 4 1 2 3 5 2 25 25 9 59 2 2 2 The proof, which we omit, utilizes a double application of the Pythagorean Theorem. Now Work PROBLEM 15 2 Find Position Vectors in Space To represent vectors in space, we introduce the unit vectors i , j , and k whose directions are along the positive x -axis, the positive y -axis, and the positive z -axis, respectively. If v is a vector with initial point at the origin O and terminal point at P a b c , , , ( ) = then we can represent v in terms of the vectors i , j , and k as a b c v i j k = + + See Figure 82. The scalars a , b , and c are called the components of the vector a b c v i j k, = + + with a being the component in the direction i , b the component in the direction j , and c the component in the direction k. A vector whose initial point is at the origin is called a position vector. The next result states that any vector whose initial point is not at the origin is equal to a unique position vector. THEOREM Suppose that v is a vector with initial point P x y z , , , 1 1 1 1 ( ) = not necessarily the origin, and terminal point P x y z , , . 2 2 2 2 ( ) = If PP v , 1 2 = then v is equal to the position vector x x y y z z v i j k 2 1 2 1 2 1 ( ) ( ) ( ) = − + − + − (2) Figure 83 illustrates this result. Figure 82 y x z v 5 ai 1 bj 1 ck k j i P 5 (a, b, c) O Figure 83 y x z v 5 P1P2 5 (x2 2 x1)i 1 (y2 2 y1)j 1 (z2 2 z1)k P2 5 (x2, y2, z2) P1 5 (x1, y1, z1) O NOTE Most handheld graphing calculators cannot graph in 3 dimensions. However, a quick Internet search will find online tools available that can graph in 3-space. j
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