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The Dot Product If u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ), then the dot product of u and v is u · v = u1 v1 + u2 v2 + u3 v3 . For instance, the dot product of u = i − 2 j − 3 k and v = 2 j − k is u · v = 1 · 0 + (−2) · 2 + (−3)(−1) = −1. Properties of the Dot Product. Let u, v, and w be three vectors and let c be a real number. Then u · v = v · u, (u + v) · w = u · w + v · w, (cu) · v = c(u · v). Further, u · u = |u|2 . Thus, if u = 0 is the zero vector, then u · u = 0, and otherwise u · u > 0. 1 Orthogonality Two vectors u and v are said to be orthogonal (perpendicular), if the angle between them is 90◦ . Theorem. Two vectors u and v are orthogonal if and only if u · v = 0. −−→ Consider the following triangle ABC, where AB = v −−→ −→ and BC = u. Then AC = u + v. Extend AB −−→ −−→ −−→ to D such that BD = AB. Then BD = v and −−→ DC = −v + u = u − v. C u-v u+ v u A v B 2 v D We observe that |u + v|2 = (u + v) · u + (u + v) · v =u·u+v·u+u·v+v·v = |u|2 + 2u · v + |v|2 . Similarly, |u − v|2 = |u|2 − 2u · v + |v|2 . Thus, |u + v|2 = |u − v|2 if and only if u · v = 0. Suppose that u and v are orthogonal. Then 6 ABC = 6 DBC = 90◦ . By Pythagoras’ theorem, |AC|2 = |AB|2 +|BC|2 and |DC|2 = |DB|2 +|BC|2 . It follows that |u + v|2 = |u − v|2 . Hence, u · v = 0. Conversely, if u·v = 0, then |u + v|2 = |u − v|2 . Thus, |AC| = |DC|. Further, |AB| = |DB| and BC is the common side of 4ABC and 4DBC. Hence, 4ABC ∼ = 4DBC. It follows that 6 ABC = 6 DBC. But 6 ABC + 6 DBC = 180◦ . So 6 ABC = 90◦ . 3 Orthogonal Projections Let u and v be two vectors with u 6= 0. In −−→ the following figure, u is represented by AB and v −→ is represented by AC. Let L be the line passing through C and perpendicular to AB. We denote by D the intersection point of line L with line AB. The −−→ vector p = AD is called the vector projection of v onto u. C v θ C v-p A p D B D 4 v θ p A u B Let w = u/|u| be the unit vector in the direction of u. Since the vector p and w lie on the same line, there exists a real number c such that p = cw. −−→ −→ −−→ Then DC = AC − AD = v − p. Since CD ⊥ AB, we have (v − p) · u = 0. It follows that p · u = u · v. But p = cw = cu/|u|. Hence, u c ·u=u·v |u| or c|u| = u · v. Consequently, c=v· u . |u| The number c is called the scalar projection of v onto u. Thus the vector projection of v onto u is p = cw = c v·u u u = . |u| |u| |u| Clearly, |p| = |cw| = |c||w| = |c|. 5 Example. Find the scalar projection and vector projection of v = (1, 1, 2) onto u = (−2, 3, 1). p √ 2 2 2 Solution. We have |u| = (−2) + 3 + 1 = 14 and 1 u = √ (−2, 3, 1). |u| 14 The scalar projection of v onto u is u 1 3 c=v· = √ (−2 + 3 + 2) = √ . |u| 14 14 The vector projection of v onto u is 3 9 3 u 3 p=c = (−2, 3, 1) = − , , . |u| 14 7 14 14 6 The Angle between Two Vectors Let u and v be two nonzero vectors, and let w = u/|u|. Recall that the scalar projection of v onto u is c = v · w. Let θ be the angle between u and v. We claim c = |v| cos θ. Clearly, this is true if θ = 0, π/2, or π. Moreover, if 0 < θ < π/2, then c > 0 and 6 CAD = θ. Hence, c = |AD| = |AC| cos 6 CAD = |v| cos θ. If π/2 < θ < π, then c < 0 and 6 CAD = π − θ. Consequently, c = −|AD| = −|AC| cos 6 CAD = −|v| cos(π − θ) = |v| cos θ. 7 Now we have c = u · v/|u| and c = |v| cos θ. Hence, u · v/|u| = |v| cos θ. It follows that u · v = |u||v| cos θ and cos θ = u·v . |u||v| Since | cos θ| ≤ 1, we obtain the Schwarz inequality: |u · v| ≤ |u||v|. Example. Find the angle between the two vectors u = (1, −1, 1) and v = (−2, 2, 2). Solution. We have p √ 2 2 2 |u| = 1 + (−1) + 1 = 3, p √ 2 2 2 |v| = (−2) + 2 + 2 = 12, u · v = 1(−2) + (−1)2 + 1(2) = −2. Thus, we obtain cos θ = u·v −2 1 = √ √ =− . |u||v| 3 3 12 Consequently, θ ≈ 1.91 (radian) or θ ≈ 109.47◦ . 8