The Dot Product If U = (U 1,U2,U3) and V = (V 1,V2,V3)

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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 = u1v1 + u2v2 + u3v3.

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 u+v D v B v A

2 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 C v v-p v θ u B B p A θ D A p D

4 Let w = u/|u| be the unit vector in the direc- tion 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 or c|u| = u · v. |u|

Consequently, u c = v · . |u|

The number c is called the scalar projection of v onto u. Thus the vector projection of v onto u is

u v · u u p = cw = c = . |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). √ Solution. We have |u| = p(−2)2 + 32 + 12 = 14 and u 1 = √ (−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

u 3 3 9 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 = |u||v| cos θ and cos θ = . |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 √ |u| = p12 + (−1)2 + 12 = 3, √ |v| = p(−2)2 + 22 + 22 = 12, u · v = 1(−2) + (−1)2 + 1(2) = −2. Thus, we obtain

u · v −2 1 cos θ = = √ √ = − . |u||v| 3 12 3

Consequently, θ ≈ 1.91 (radian) or θ ≈ 109.47◦.