College Trigonometry, 2011a

1020 Applications of Trigonometry
Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude
and direction of the vector. We can convert the point (v 1 ,v 2 ) in rectangular coordinates to a pair
(r, θ) in polar coordinates where r ≥ 0. The magnitude of ⃗v, which we said earlier was length
of the directed line segment, is r = √ v 2 1
+ v 2 2
and is denoted by ‖⃗v‖. From Section 11.4, we
know v 1 = r cos(θ) =‖⃗v‖ cos(θ) andv 2 = r sin(θ) =‖⃗v‖ sin(θ). From the definition of scalar
multiplication and vector equality, we get
⃗v = 〈v 1 ,v 2 〉
= 〈‖⃗v‖ cos(θ), ‖⃗v‖ sin(θ)〉
= ‖⃗v‖〈cos(θ), sin(θ)〉
This motivates the following definition.
Definition 11.8. Suppose ⃗v is a vector with component form ⃗v = 〈v 1 ,v 2 〉.Let(r, θ) beapolar
representation of the point with rectangular coordinates (v 1 ,v 2 )withr ≥ 0.
ˆ The magnitude of ⃗v, denoted ‖⃗v‖, isgivenby‖⃗v‖ = r = √ v 2 1
+ v 2 2
ˆ If ⃗v ≠ ⃗0, the (vector) direction of ⃗v, denoted ˆv is given by ˆv = 〈cos(θ), sin(θ)〉
Taken together, we get ⃗v = 〈‖⃗v‖ cos(θ), ‖⃗v‖ sin(θ)〉.
A few remarks are in order. First, we note that if ⃗v ≠ 0 then even though there are infinitely
many angles θ which satisfy Definition 11.8, the stipulation r>0 means that all of the angles are
coterminal. Hence, if θ and θ ′ both satisfy the conditions of Definition 11.8, then cos(θ) = cos(θ ′ )
and sin(θ) =sin(θ ′ ), and as such, 〈cos(θ), sin(θ)〉 = 〈cos(θ ′ ), sin(θ ′ )〉 making ˆv is well-defined. 10 If
⃗v = ⃗0, then ⃗v = 〈0, 0〉, and we know from Section 11.4 that (0,θ) is a polar representation for
the origin for any angle θ. For this reason, ˆ0 is undefined. The following theorem summarizes the
important facts about the magnitude and direction of a vector.
Theorem 11.20. Properties of Magnitude and Direction: Suppose ⃗v is a vector.
ˆ ‖⃗v‖ ≥0and‖⃗v‖ = 0 if and only if ⃗v = ⃗0
ˆ For all scalars k, ‖k⃗v‖ = |k|‖⃗v‖.
( )
ˆ If ⃗v ≠ ⃗0 then⃗v = ‖⃗v‖ˆv, sothatˆv = 1
‖⃗v‖
⃗v.
The proof of the first property in Theorem 11.20 is a direct consequence of the definition of ‖⃗v‖.
If ⃗v = 〈v 1 ,v 2 〉,then‖⃗v‖ = √ √ v 2 1
+ v 2 2
which is by definition greater than or equal to 0. Moreover,
v 2 1
+ v 2 2
= 0 if and only of v 2 1
+ v 2 2
= 0 if and only if v 1 = v 2 = 0. Hence, ‖⃗v‖ = 0 if and only if
⃗v = 〈0, 0〉 = ⃗0, as required.
The second property is a result of the definition of magnitude and scalar multiplication along with
a propery of radicals. If ⃗v = 〈v 1 ,v 2 〉 and k is a scalar then
10 If this all looks familiar, it should. The interested reader is invited to compare Definition 11.8 to Definition 11.2
in Section 11.7.