SECTION 9.6 Vectors in Space 665 Figure 84 Direction angles y x z 0 # a # p, 0 # b # p, 0 # g # p a b g C 5 (0, 0, c) v A 5 (a, 0, 0) P 5 (a, b, c) B 5 (0, b, 0) Our first goal is to find expressions for , , α β and γ in terms of the components of a vector. Let a b c v i j k = + + denote a nonzero vector. The angle α between v and i , the positive x -axis, obeys a v i v i v cos α = ⋅ = Similarly, b c v v cos and cos β γ = = Since a b c v , 2 2 2 = + + the following result is obtained. THEOREM Direction Angles If a b c v i j k = + + is a nonzero vector in space, the direction angles , , α β and γ obey • a a b c a v cos 2 2 2 α = + + = • b a b c b v cos 2 2 2 β = + + = • c a b c c v cos 2 2 2 γ = + + = (5) The numbers α β cos , cos , and cosγ are called the direction cosines of the vector v . Finding the Direction Angles of a Vector Find the direction angles of v i j k 3 2 6 . = − + − Solution EXAMPLE 8 v 3 2 6 49 7 2 2 2 ( ) ( ) = − + + − = = Using the formulas in equation (5), we have cos 3 7 115.4 cos 2 7 73.4 cos 6 7 149.0 α α β β γ γ = − ≈ ° = ≈ ° = − ≈ ° THEOREM Property of the Direction Cosines If , , α β and γ are the direction angles of a nonzero vector v in space, then cos cos cos 1 2 2 2 α β γ + + = (6) The proof is a direct consequence of the equations in (5).
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