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Multiplication of Vectors

−→
a
φ
−→
. . . . b

Figure 5.1 – Dot product

permanent force on the straight line segment of the way equals to the dot product of a vector of
force on a moving vector — this is the physical essence of the dot product.

Formula (5.1) could be written in another way. A multiple b · cos φ is the projection of
−→

−→ −→ − →
− → −→
vector b on the axis, defined by vector a — the projection on vector a : b · cos φ=pr−→ b .
a
So,
−→ −→
−→ −→
a · b = | a | · pr−→ b .
a

−→
− →
− → −→
In an analogy, a · b = b · pr −→ a . Projections of a vector on another vector can be found
b
by formulas:
−→ −→
−→ −→
− → a · b − → a · b
pr−→ b = −→ , pr −→ a = (5.2)
a
| a | b −→
b
Based on formula (5.1), a formula for calculating cosine of angle φ can be written as:
−→
−→
a · b
cos φ = (5.3)
− → −→
| a | · b
Let’s consider main properties of a dot product.
−→ −→ − →
−→
1. A dot product of two vectors is commutative: a · b = b · a . Proof can
be performed by using the definition of a dot product.
2. A dot product of two vectors has an associative property related to a
−→ −→ −→
−→
−→
− →
scalar multiplier, that is: λ·( a · b ) = (λ· a )· b = a ·(λ· b ). When proving,
let’s consider a case, when λ > 0(for λ < 0 in the same way). In this case
−→
−→
angle φ between vectors a and b equals to the angle between vectors

−→ −→ − → −→ −→
−→
−→ −→
−→ −→
λ· a and b . So, λ·( a · b ) = λ·| a |· b ·cos φ, (λ· a )· b = |λ · a |· b ·cos φ =
−→ −→
−→
−→
−→
−→
λ · | a | · b · cos φ. Evidently, λ · ( a · b ) = (λ · a ) · b .
− → −→
− →
−→
Formula λ · ( a · b ) = a · (λ · b ) can be proved in the same way.
−→
−→
−→
3. A dot product of two vectors has a distributive property: ( a + b ) · c =
−→
−→ −→ −→
a · c + b · c .
− →
−→
−→
−→
−→
−→
− →
−→
−→
−→
−→
Indeed, ( a + b )· c = | c |·pr ˙ ( a + b ) = | c |·(pr ˙ a +pr ˙ b ) = | c |·pr ˙ a +
A
A
A
A
−→
−→
−→
− →
−→ −→
−→
−→ −→
− →
| c | · pr ˙ b = c · a + c · b = a · c + · b · c .
A
4. If a dot product equals to zero, then equals to zero either one of vectors
or a cosine of an angle between these vectors, meaning vectors are
−→ − →
− →
−→
perpendicular: a ⊥ b . It is true in a contrary: if a ⊥ b , then cos φ = 0, so
−→
−→
a · b =0. Therefore, for two non-zero vectors to be perpendicular it is
necessary and sufficient their dot product equals to zero.
5. Let’s consider the dot product of any vector of itself. Such product is
−→ 2
−→ −→
−→ −→
−→ −→
called a scalar square of a vector: a · a = | a |·| a |·cos 0 = | a |·| a | = | a | .
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