Coordinate Vectors

Section 4.4 Coordinate Vectors

In Section 4.3, we visualized vectors as directed line segments with magnitudes and direction. Now, we will explore vectors through a coordinate approach, where we express vectors in terms of their components along the coordinate axis.

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Subsection 4.4.1 Vectors in Coordinates

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Definition 4.4.1. Component Form of a Vector.

A vector

v

with initial point

P (x_{1}, y_{1})

and terminal point

Q (x_{2}, y_{2})

can be expressed in various ways. One common representation is as the directed line segment from point

P

to point

Q,

denoted as

\vec{P Q} .

Alternatively, we can describe it using the changes in the

x -

and

y -

coordinates between points

P

and

Q .

These changes are referred to as the horizontal component and vertical component of the vector, denoted as

v_{x}

and

v_{y}

respectively. Mathematically, these representations are all equivalent:

\begin{aligned} v & = \vec{P Q} \\ = ⟨ x_{2} - x_{1}, y_{2} - y_{1} ⟩ \\ = ⟨ v_{x}, v_{y} ⟩ \end{aligned}

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as illustrated in the following figure.

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Remark 4.4.2. Notation.

To avoid confusion with the notation for a point and an interval, we use the symbol

⟨ v_{1}, v_{2} ⟩

for an ordered pair that represents a vector in component form. Lowercase letters are used to represent the components.

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Definition 4.4.3. Position Vector.

A vector,

v

with an initial point at

(0, 0)

and terminal point at

(a, b)

is called the position vector and can be written as

v = ⟨ a - 0, b - 0 ⟩ = ⟨ a, b ⟩ .

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Remark 4.4.4.

The position vector represents the terminal point of any vector

v

when its initial point is at the origin. Essentially, it is

v

translated so that its initial side starts at the origin. This translation does not change the magnitude or direction, making the two vectors equivalent. Thus, we can refer to both vectors as

v .

In other words, the terminal point of a vector starting at the origin is defined by its horizontal and vertical components.

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Example 4.4.5.

Find the position vector of the vector that goes from

P (2, - 1)

Q (5, 3) .

Solution.

From Definition 4.4.1, our original vector can be written as:

v = \vec{P Q} = ⟨ 5 - 2, 3 - (- 1) ⟩ = ⟨ 3, 4 ⟩ .

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Then by Definition 4.4.3, the position vector is:

v = ⟨ 3, 4 ⟩ .

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This vector starts at the origin with 3 as its horizontal component and 4 as its vertical component, as illustrated in the figure below.

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Definition 4.4.6. Zero Vector.

The zero vector is the vector

0 = ⟨ 0, 0 ⟩

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Definition 4.4.7. Magnitude of a Vector.

Given a vector

v = ⟨ v_{x}, v_{y} ⟩,

the magnitude or length is

‖ v ‖ = \sqrt{v_{x}^{2} + v_{y}^{2}}

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Remark 4.4.8.

This formula is derived from the Pythagorean Theorem or the distance formula where the square of the magnitude equals the square of the horizontal component plus the square of the vertical component:

‖ v ‖^{2} = v_{x}^{2} + v_{y}^{2}

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This relationship can be visually understood using the Pythagorean Theorem, as illustrated in the following figure.

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Example 4.4.9.

Find the magnitude of each vector.

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(a)

u = ⟨ 10, - 5 ⟩

Solution.

‖ u ‖ = \sqrt{10^{2} + (- 5)^{2}} = \sqrt{125} = \sqrt{25 \cdot 5} = \sqrt{25} \cdot \sqrt{5} = 5 \sqrt{5}

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(b)

v = ⟨ - 2, 7 ⟩

Solution.

‖ v ‖ = \sqrt{(- 2)^{2} + 7^{2}} = \sqrt{4 + 49} = \sqrt{53}

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(c)

w = ⟨ \frac{12}{13}, \frac{5}{13} ⟩

Solution.

‖ w ‖ = \sqrt{{(\frac{12}{13})}^{2} + {(\frac{5}{13})}^{2}} = \sqrt{\frac{144}{169} + \frac{25}{169}} = \sqrt{\frac{169}{169}} = 1

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Definition 4.4.10. Unit Vector.

A vector

u

with magnitude 1,

‖ u ‖ = 1,

is called a unit vector

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Definition 4.4.11. Algebraic operations of vectors.

u = ⟨ u_{x}, u_{y} ⟩,

v = ⟨ v_{x}, v_{y} ⟩,

and

c

is a scalar, then

\begin{aligned} u + v & = ⟨ u_{x} + v_{x}, u_{y} + v_{y} ⟩ \\ u - v & = ⟨ u_{x} - v_{x}, u_{y} - v_{y} ⟩ \\ c u & = ⟨ c u_{x}, c u_{y} ⟩ \end{aligned}

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Example 4.4.12.

Let

u = ⟨ 5, 1 ⟩

and

v = ⟨ 4, 6 ⟩ .

Compute the following:

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(a)

u + v

Solution.

u + v = ⟨ 5, 1 ⟩ + ⟨ 4, 6 ⟩ = ⟨ 5 + 4, 1 + 6 ⟩ = ⟨ 9, 7 ⟩

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(b)

u - v

Solution.

u - v = ⟨ 5, 1 ⟩ - ⟨ 4, 6 ⟩ = ⟨ 5 - 4, 1 - 6 ⟩ = ⟨ 1, - 5 ⟩

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(c)

3 u

Solution.

3 u = 3 ⟨ 5, 1 ⟩ = ⟨ 3 \cdot 5, 3 \cdot 1 ⟩ = ⟨ 15, 3 ⟩

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Subsection 4.4.2 Properties of Addition, Length, and Scalar Multiplication

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Definition 4.4.13. Properties of Vectors.

u,

v,

and

w

are vectors,

0

is the zero vector, and

c

and

d

are scalars, then

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Vector Addition

$u + v = v + u$
$u + (v + w) = (u + v) + w$
$u + 0 = u$
$u + (- u) = 0$

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Length of a Vector

$‖ c u ‖ = | c | ‖ u ‖$

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Scalar Multiplication

$c (u + v) = c u + c v$
$(c + d) u = c u + d u$
$(c d) u = c (d u) = d (c u)$
$1 u = u$
$0 u = 0$
$c 0 = 0$

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Subsection 4.4.3 Finding a Unit Vector

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Sometimes it is useful to look at problems involving vectors in terms of a vector in the same direction, but with a magnitude of one. Recall that a unit vector

u

is a vector whose length is one,

‖ u ‖ = 1 .

To find a unit vector in the same direction, you will need to divide that vector by its length or magnitude.

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Definition 4.4.14. Unit Vector That Has the Same Direction as $v$ .

Let

v

be any vector. The vector

u = \frac{v}{‖ v ‖}

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is a unit vector that has the same direction as

v .

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Example 4.4.15.

Find the unit vector that has the same direction as

v = ⟨ 10, - 5 ⟩

Solution.

From Example 4.4.9, we know

‖ v ‖ = 5 \sqrt{5} .

Then

\begin{aligned} u & = \frac{v}{‖ v ‖} \\ = \frac{⟨ 10, - 5 ⟩}{5 \sqrt{5}} \\ = ⟨ \frac{10}{5 \sqrt{5}}, - \frac{5}{5 \sqrt{5}} ⟩ \\ = ⟨ \frac{2}{\sqrt{5}}, - \frac{1}{\sqrt{5}} ⟩ \\ = ⟨ \frac{2 \sqrt{5}}{5}, - \frac{\sqrt{5}}{5} ⟩ \end{aligned}

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We can verify that this is in fact a unit vector by computing the magnitude:

\begin{aligned} ‖ u ‖ & = ‖ ⟨ \frac{2 \sqrt{5}}{5}, - \frac{\sqrt{5}}{5} ⟩ ‖ \\ = \sqrt{{(\frac{2 \sqrt{5}}{5})}^{2} + {(- \frac{\sqrt{5}}{5})}^{2}} \\ = \sqrt{\frac{4 \cdot 5}{25} + \frac{5}{25}} \\ = \sqrt{\frac{25}{25}} \\ = 1 \end{aligned}

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Two useful unit vectors are

i

and

j .

The vector

i

represents the unit vector whose direction is along the positive

x

-axis and the vector

j

represents the unit vector whose direction is along the positive

y

-axis.

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Definition 4.4.16. $i$ and $j$ .

The unit vectors

i

and

j

are defined as

i = ⟨ 1, 0 ⟩, j = ⟨ 0, 1 ⟩ .

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We can now represent any vector using the unit vectors

i

and

j .

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Definition 4.4.17. Horizontal and Vertical Components of a Vector.

Let

v

be any vector. Then we can express

v

in terms of its horizontal and vertical components

v = ⟨ v_{x}, v_{y} ⟩ = v_{x} i + v_{y} j

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Subsection 4.4.4 Vector Components and Direction

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Recall that a vector is composed of a direction and magnitude. Earlier in this section we learned how to calculate the magnitude. Now we will discuss how to calculate the direction.

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Definition 4.4.18.

Let

v = ⟨ v_{x}, v_{y} ⟩

be a vector. The angle

θ

represents the direction of

v

and is the smallest positive angle in standard position formed by the positive

x

-axis and

v

(

0^{\circ} \leq θ \leq 360^{\circ}

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Remark 4.4.19. Finding Direction of a Vector.

Recall that

\tan θ = \frac{v_{y}}{v_{x}},

θ = \tan^{- 1} (\frac{v_{y}}{v_{x}})

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where

- 90^{\circ} < θ < 90^{\circ} .

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To find the direction (

θ

), first draw the vector. If

v_{x} = 0

v_{y} = 0,

then use the drawing to find

θ .

v_{x} \neq 0

and

v_{y} \neq 0,

then note the quadrant where

θ

lies in.

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If $θ$ is in Quadrant I, then $θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) .$
If $θ$ is in Quadrant II or III, then $θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) + 180^{\circ} .$
If $θ$ is in Quadrant IV, then $θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) + 360^{\circ} .$

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We now have all the information needed to write a vector.

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Definition 4.4.20. Finding Horizontal and Vertical Components of Vectors from Magnitude and Direction.

For vector

v

with magnitude

‖ v ‖

and direction

θ,

we can use Right Triangle Trigonometry to solve for the horizontal and vertical components, denoted as

v_{x}

and

v_{y},

respectively:

v_{x} = ‖ v ‖ \cos θ, v_{y} = ‖ v ‖ \sin θ

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This allows us to express

v

v = ‖ v ‖ \cos θ i + ‖ v ‖ \sin θ j = ⟨ ‖ v ‖ \cos θ, ‖ v ‖ \sin θ ⟩

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Subsection 4.4.5 Velocity

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Vectors can be used to represent the velocity of a moving canoe. These velocity vectors have a direction and magnitude.

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Example 4.4.21.

Wind is blowing from the house Nā Leo Hoʻolua (N

{22.5}^{\circ}

W) at 10 knots. Write the wind velocity as a vector

v .

Solution.

First, we note that since the velocity is 10 knots, we have

‖ v ‖ = 10 .

Next, we need to find the direction. Since

θ

is measured from the positive

x

-axis, by Definition 4.4.20 we see that

θ = 90^{\circ} + {22.5}^{\circ} = {112.5}^{\circ}

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To find the vector, will need the horizontal and vertical components:

\begin{aligned} v_{x} = & ‖ v ‖ \cos θ = 10 \cos {112.5}^{\circ} \approx - 3.8 \\ v_{y} = & ‖ v ‖ \sin θ = 10 \sin {112.5}^{\circ} \approx 9.2 \end{aligned}

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Thus

v \approx - 3.8 i + 9.2 j = ⟨ - 3.8, 9.2 ⟩

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Example 4.4.22. Calculating Canoe Velocity with Current Drift.

The vaka Marumaru Atua sets sail on a northward voyage from Rarotonga to Hawaiʻi, maintaining a steady speed of 5 knots through the water. However, a 1-knot current flows in the direction of Lā Kona (a heading of 260 degrees). This current’s influence resembles walking on a moving treadmill, where forward progress is limited despite moving legs. Similarly, Marumaru Atua navigates while the water, in the form of a current, also moves. In sailing, the direction and speed at which the current is pushing the vaka are referred to as set and drift, respectively.

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The actual velocity of Marumaru Atua is the resultant of the canoe’s velocity and the current’s velocity. This resultant velocity vector denotes the canoe’s speed and direction relative to fixed objects on Earth, influenced by the current’s impact on Marumaru. The magnitude and direction of this velocity vector are represented by the speed over ground (SOG) and course over ground (COG), respectively. SOG represents the speed of the canoe relative to fixed objects, accounting for both the canoe’s speed through the water and the current’s speed and direction. COG indicates the direction of the canoe’s motion over the Earth’s surface.

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Understanding the difference between the velocity over water and the velocity relative to fixed objects illustrates how set and drift show how the current influences the vaka’s course and speed over ground. Voyagers need to account for set and drift when navigating to ensure they reach their intended destination accurately and safely.

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(a)

Express Marumaru Atua’s velocity vector, represented as

v_{m},

and the velocity vector of the current, denoted as

v_{c},

in terms of their horizontal and vertical components. Round your answer to two decimal places.

Solution.

First, we draw the velocity vectors for Marumaru Atua and the current:

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Since Marumaru Atua is sailing north at 5 knots, the velocity is:

v_{m} = ⟨ 0, 5 ⟩ .

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To find the vector for the current, we need to know

t,

the angle of the velocity vector in standard position (see Definition 1.2.8). We begin by drawing the heading angle of

260^{\circ}

and its reference angle,

t^{'} .

$Illustration of the heading angle and its reference angle$

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We can see from the figure that

260^{\circ} + t^{'} = 270^{\circ},

thus we get our reference angle as

t^{'} = 10^{\circ} .

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We now draw our angle,

t,

in standard position and the reference angle,

t^{'} :

$Illustration of the angle t in standard position and the reference angle t’$

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From Remark 1.5.9, we have

t = 180^{\circ} + 10^{\circ} = 190^{\circ} .

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Thus, by Definition 4.4.20 the velocity for the current is

v_{c} = ⟨ 1 \cdot \cos 190^{\circ}, 1 \cdot \sin 190^{\circ} ⟩ \approx ⟨ - 0.98, - 0.17 ⟩ .

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(b)

Find

v_{g},

the velocity of Marumaru Atua relative to fixed objects on the ground.

Solution.

The velocity relative to fixed objects on the ground is the sum of the two vectors:

\begin{aligned} v_{g} = v_{m} + v_{c} & = ⟨ 0, 5 ⟩ + ⟨ - 0.98, - 0.17 ⟩ \\ = ⟨ - 0.98, 5 - 0.17 ⟩ \\ = ⟨ - 0.98, 4.83 ⟩ \end{aligned}

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This can visually demonstrated in the figure below

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(c)

Find the speed over ground (SOG) and the course over ground (COG) of Marumaru Atua, rounded to one decimal place.

Solution.

The actual speed of Marumaru Atua is simply the magnitude of the velocity over ground:

‖ v_{g} ‖ = ‖ ⟨ - 0.98, 4.83 ⟩ ‖ = \sqrt{(- 0.98)^{2} + ({4.83}^{2})} \approx 4.9 knots .

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Thus, although the vaka sails through the water at 5 knots, its forward progress is slightly slower with a speed over ground at 4.9 knots.

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To find the course over ground, we will use Remark 4.4.19 to find direction of the velocity:

θ = \tan^{- 1} \frac{v_{y}}{v_{x}} = \tan^{- 1} \frac{4.83}{- 0.98} \approx - {78.5}^{\circ} .

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Since our velocity vector is in Quadrant II, our course over ground is

- {78.5}^{\circ} + 180^{\circ} = {101.5}^{\circ} .

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Exercises 4.4.6 Exercises

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Exercise Group.

Draw the vector,

v .

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1.

3 i - 2 j

Section 4.4 Coordinate Vectors

Subsection 4.4.1 Vectors in Coordinates

Definition 4.4.1. Component Form of a Vector.

Remark 4.4.2. Notation.

Definition 4.4.3. Position Vector.

Remark 4.4.4.

Example 4.4.5.

Solution.

Definition 4.4.6. Zero Vector.

Definition 4.4.7. Magnitude of a Vector.

Remark 4.4.8.

Example 4.4.9.

(a)

Solution.

(b)

Solution.

(c)

Solution.

Definition 4.4.10. Unit Vector.

Definition 4.4.11. Algebraic operations of vectors.

Example 4.4.12.

(a)

Solution.

(b)

Solution.

(c)

Solution.

Subsection 4.4.2 Properties of Addition, Length, and Scalar Multiplication

Definition 4.4.13. Properties of Vectors.

Subsection 4.4.3 Finding a Unit Vector

Definition 4.4.14. Unit Vector That Has the Same Direction as v.

Example 4.4.15.

Solution.

Definition 4.4.16. i and j.

Definition 4.4.17. Horizontal and Vertical Components of a Vector.

Subsection 4.4.4 Vector Components and Direction

Definition 4.4.18.

Remark 4.4.19. Finding Direction of a Vector.

Definition 4.4.20. Finding Horizontal and Vertical Components of Vectors from Magnitude and Direction.

Subsection 4.4.5 Velocity

Example 4.4.21.

Solution.

Example 4.4.22. Calculating Canoe Velocity with Current Drift.

(a)

Solution.

(b)

Solution.

(c)

Solution.

Exercises 4.4.6 Exercises

Exercise Group.

1.

2.

3.

4.

5.

6.

7.

8.

Exercise Group.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Exercise Group.

21.

22.

23.

24.

Exercise Group.

25.

26.

Definition 4.4.14. Unit Vector That Has the Same Direction as $v$ .

Definition 4.4.16. $i$ and $j$ .