Application of Trig: Vectors
Name:_____________________________________
In the real world, measurements are not simply about value or magnitude, but direction as well. For example, force,
velocity, and acceleration have components of magnitude and direction, so they must be measured in terms of
both.
To do this, we use vectors which are directed line segments.
A vector runs from its initial point P(x1 , y1 ) where the vector begins to its terminal point Q(x2 , y2 ) where the vector
⃗⃗⃗⃗⃗ .
ends. We denote this vector PQ
Component Form of a vector from initial point P(x1 , y1 ) to terminal point Q(x2 , y2 ) is found by
𝐯 = ⟨x2 − x1 , y2 − y1 ⟩ = ⟨v1 , v2 ⟩
To find the magnitude (or length) of a vector, we use distance formula between its initial and terminal points:
⃗⃗⃗⃗⃗ | = √(x2 − x1 )2 + (y2 − y1 )2 = √v1 2 + v2 2
|PQ
For two vectors to be equivalent, they must have matching magnitudes and component forms.
Vector Operations: given vectors 𝐮 = ⟨u1 , u2 ⟩ and 𝐯 = ⟨v1 , v2 ⟩,
Vector Addition: 𝐮 + 𝐯 = ⟨u1 + v1 , u2 + v2 ⟩
Vector Subtraction: 𝐮 − 𝐯 = ⟨u1 − v1 , u2 − v2 ⟩
Scalar Multiplication: 𝐤𝐮 = k⟨u1 , u2 ⟩ = ⟨ku1 , ku2 ⟩
Direction Angles angle  that represents direction of vector in the plane from the positive x-axis to the vector.
y
y
v2
tan θ = to find θ you will use the inverse θ = tan−1 ( ) or θ = tan−1 ( )
x
x
v1
A unit vector is a vector with magnitude of one. If a vector is not originally a unit vectors, it can be represented by a
unit vector by dividing the vector by its magnitude:
𝐮=
𝐯
v1 v2
=⟨ , ⟩
|v|
|v| |v|
Standard unit vectors are used to express vectors as a linear combination. The two separate unit vectors are i = <1,
0> and j = <0, 1>. Any vector v can be written as a linear combination of these standard unit vectors:
v = <a, b>
v = <a,0> + <0, b>
v = a<1, 0> + b<0, 1>
v = ai +bj
If given a magnitude of a vector and the angle of the vector, you can use these to write the component form or
linear combination of the vector <a, b>:
𝐚 = |𝐯| cos θ
𝐛 = |𝐯| sin θ
Dot Product of two vectors u = < u1 , u2 > and 𝐯 = < v1 , v2 > is found by: u
⃗ ∗v
⃗ = (u1 ∗ v1 ) + (u2 ∗ v2 ).
Note: the dot product is a scalar NOT a vector!
Properties of the Dot Product:
1. ⃗u ∗ v
⃗ =v
⃗ ∗u
⃗
2. 0 ∗ v
⃗ =0
3. u
⃗ ∗ (v
⃗ +w
⃗⃗⃗ ) = (u
⃗ ∗v
⃗ ) + (u
⃗ ∗w
⃗⃗⃗ )
2
4. ⃗v ∗ v
⃗ = (||v
⃗ ||)
5. c(u
⃗ ∗v
⃗ ) = cu
⃗ ∗v
⃗ =u
⃗ ∗ cv
⃗
The vectors u and v are orthogonal (or perpendicular) if u
⃗ ∗v
⃗ = 0.
Angle between vectors can be found by: cos θ
=
⃗ ∗v
⃗
u
⃗ ||∗||v
⃗ ||
||u
Example A ship is heading due north at 12 mph. The current is flowing southwest at 4 mph. Find the actual bearing
and speed of the ship.
Ship’s heading: < 12cos 90⁰, 12sin 90⁰> = < 0, 12>
Current’s heading: < 4cos 225⁰, 4sin 225⁰> = < -2.83, -2.83>
Actual velocity vector: < 0,12> + < -2.83, -2.83> = < -2.83, 9.17>
Actual speed: √(−2.83)2 + (9.17)2 ≈ 9.60 mph
a
2.83
Actual direction angle: cos −1 (|v|) = cos −1 (− 9.60) ≈ 107.15
Since the actual direction angle is 107.15⁰, the actual bearing is 342.15⁰.
Practice Problems:
1. Find the component form, magnitude, and directional angle of a vector that has initial point ( 13, 6) and
terminal point ( -27, 42).
2. Determine the magnitude and direction angle for each vector:
a. < 3, 2>
b. < -2, -5>
3. Let u = < -2, -5> and v = < 3, 2>. Determine:
⃗ + 𝐯⃗
a. 𝐮
⃗
c. 4𝐮
⃗ − 𝐯⃗
b. 𝐮
⃗ + 2𝐯⃗
d. 3𝐮
4. Find the dot product u*v if:
a. u = < 3, 4> and v = < 2, -3>
b. u = < 2, 2> and v = < 1, -1>
5. Find the angle between the two given vectors:
a. u = < 3, 0> and v = < 1, 6>
c. u = < 0, 4> and v = < 3, -2>
b. u = < -6, -3> and v = < -8, 4>
6. A ship leaving port sails for 75 miles in a direction of 35⁰ north of due east. Find the magnitude of the
vertical and horizontal components.