Let a and b include the angle θ. By definition a⋅b = |a||b|\mathop{cos}\nolimits θ. Thus \mathop{cos}\nolimits θ = a⋅b∕|a||b|, or \mathop{cos}\nolimits θ = a∕|a|⋅b∕|b|, or \mathop{cos}\nolimits θ =\hat{ a}⋅\hat{b}. Thus \mathop{cos}\nolimits θ is the dot product of the unit vectors \hat{a} and \hat{b}.
Example 3.8:
Consider the vectors u = (2,−1,1) and v = (1,1,2). Find u⋅v and determine the angle between u and v.
Solution:
First Calculate
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u⋅v = {u}_{1}{v}_{1} + {u}_{2}{v}_{2} + {u}_{3}{v}_{3} = (2)(1) + (−1)(1) + (1)(2) = 3.
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Also |u| = \sqrt{6} and |v| = \sqrt{6}, so
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\mathop{cos}\nolimits θ = {u⋅v\over
|u||v|} = {3\over
\sqrt{6}\sqrt{6}} = {3\over
6} = {1\over
2}.
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Hence θ = {π\over 3} (or 60∘).