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.
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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◦ .
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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|.
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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
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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.
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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◦ .
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