815 8.4 Algebraically Defined Vectors and the Dot Product For example, to prove the first part of property (d), 1k u2 # v = k1u # v2, we let u = 8a, b9 and v = 8c, d9. 1k u2 # v = Ak8a, b9B # 8c, d9 Substitute. = 8ka, kb9 # 8c, d9 Multiply by scalar k. = kac + kbd Dot product = k1ac + bd2 Distributive property = kA 8a, b9 # 8c, d9B Dot product = k1u # v2 Substitute. The proofs of the remaining properties are similar. The dot product of two vectors can be positive, 0, or negative. A geometric interpretation explains when each of these cases occurs. This interpretation involves the angle between two vectors. Consider the two vectors u = 8a1, a29 and v = 8b1, b29, as shown in Figure 43. The angle U between u and v is defined to be the angle having the two vectors as its sides for which 0° … u … 180°. We can use the law of cosines to develop a formula to find angle u in Figure 43. u - v 2 = u 2 + v 2 - 2 u v cos u Law of cosines applied to Figure 43 A 21a1 - b122 + 1a 2 - b222 B 2 = A 2a1 2 + a 2 2 B 2 + A 2b1 2 + b 2 2 B 2 Magnitude of a vector - 2 u v cos u a1 2 - 2a 1b1 + b1 2 + a 2 2 - 2a 2 b2 + b2 2 Square. = a1 2 + a 2 2 + b 1 2 + b 2 2 - 2 u v cos u -2a1b1 - 2a2 b2 = -2 u v cos u a1b1 + a2 b2 = u v cos u Divide by -2. u # v = u v cos u Definition of dot product cos u = u # v u v x y u v u – v 0 u kb1, b2l ka1, a2l Figure 43 Subtract like terms from each side. Divide by u v and rewrite. Geometric Interpretation of Dot Product If u is the angle between the two nonzero vectors u and v, where 0° … u … 180°, then the following holds true. cos U = u # v ∣ u∣ ∣ v∣
RkJQdWJsaXNoZXIy NjM5ODQ=