Section 9.2 The Dot Product
The Dot Product of Vectors
EXAMPLES:
(a) If u = h3, −2i and v = h4, 5i then
u · v = (3)(4) + (−2)(5) = 12 − 10 = 2
(b) If u = 2i + j and v = 5i − 6j then
u · v = (2)(5) + (1)(−6) = 10 − 6 = 4
Proof: We prove only the last property. Let u = ha, bi . Then
u · u = ha, bi · ha, bi = a · a + b · b = a2 + b2 =
√
a2 + b 2
EXAMPLE: If u = h3, −2i and v = h4, 5i then
|u|2 = u · u = (3)(3) + (−2)(−2) = 9 + 4 = 13
and
|v|2 = v · v = (4)(4) + (5)(5) = 16 + 25 = 41
therefore
|u| =
√
13 and |v| =
1
√
41
2
= |u|2
EXAMPLE: Find the angle between the vectors u = h2, 5i and v = h4, −3i .
Solution: By the formula for the angle between two vectors, we have
cos θ =
u·v
8 − 15
−7
7
(2)(4) + (5)(−3)
√
√
=√
= √ √ =− √
=√
|u||v|
4 + 25 16 + 9
22 + 52 42 + 32
29 25
5 29
Thus the angle between u and v is
θ = cos
−1
7
− √
≈ 105.1◦
5 29
EXAMPLE: Determine whether the vectors in each pair are perpendicular.
(a) u = h3, 5i and v = h2, −8i
(b) u = h2, 1i and v = h−1, 2i
Solution:
(a) We have
u · v = (3)(2) + (5)(−8) = 6 − 40 = −34 6= 0
so u and v are not perpendicular.
(b) We have
u · v = (2)(−1) + (1)(2) = −2 + 2 = 0
so u and v are perpendicular.
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The Component of u Along v
The component of u along v (or the component of u in the direction of v) is defined
to be
|u| cos θ
where θ is the angle between u and v.
Proof: Since cos θ =
u·v
, we have
|u||v|
component of u along v = |u| cos θ = |u| ·
u·v
u·v
=
|u||v|
|v|
EXAMPLE: Let u = h1, 4i and v = h−2, 1i . Find the component of u along v.
Solution: We have
component of u along v =
u·v
2
(1)(−2) + (4)(1)
√
=√
=
|v|
4+1
5
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The Projection of u onto v
The projection of u onto v, denoted by projv u, is the vector whose direction is the same as v
and whose length is the component of u along v.
To find an expression for projv u, we first find a unit vector in the direction of v and then
multiply it by the component of u along v:
u·v v
u·v
projv u = (component of u along v)(unit vector in direction of v) =
v
=
|v| |v|
|v|2
EXAMPLE: Let u = h−2, 9i and v = h−1, 2i .
(a) Find projv u.
(b) Resolve u into u1 and u2 , where u1 is parallel to v and u2 is orthogonal to v.
Solution:
(a) By the formula for the projection of one vector onto another we have
h−2, 9i · h−1, 2i
(−2)(−1) + (9)(2)
u·v
v=
h−1, 2i =
h−1, 2i
projv u =
|v|2
(−1)2 + 22
1+4
2 + 18
=
h−1, 2i = 4 h−1, 2i = h−4, 8i
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(b) By the formula in the preceding box we have u = u1 + u2 , where
u1 = projv u = h−4, 8i
u2 = u − projv u = h−2, 9i − h−4, 8i = h2, 1i
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