The Central Idea
A vector can be described by how far it moves along each coordinate axis.
In two dimensions,
\[ \boxed{\mathbf{v}=\langle a,b\rangle} \]means:
- \(a\) units of horizontal directed change;
- \(b\) units of vertical directed change.
The pair \(\langle a,b\rangle\) is called the component form of the vector.
Components Are Directed Changes
Consider
\[ \mathbf{v}=\langle 4,3\rangle. \]The first component, \(4\), indicates movement \(4\) units in the positive horizontal direction.
The second component, \(3\), indicates movement \(3\) units in the positive vertical direction.
Together they determine one directed displacement.
Signs Encode Direction
Component signs tell us which way the vector moves relative to the axes.
| Component | Positive | Negative |
|---|---|---|
| Horizontal | Right | Left |
| Vertical | Up | Down |
For example,
\[ \langle -5,2\rangle \]means \(5\) units left and \(2\) units up.
Vectors in Each Quadrant
In the standard coordinate plane, the signs of the components correspond to the familiar quadrants.
| Quadrant | Component Pattern |
|---|---|
| I | \(\langle +,+\rangle\) |
| II | \(\langle -,+\rangle\) |
| III | \(\langle -,-\rangle\) |
| IV | \(\langle +,-\rangle\) |
Horizontal and Vertical Vectors
If the second component is zero, the vector is horizontal:
\[ \langle a,0\rangle. \]If the first component is zero, the vector is vertical:
\[ \langle 0,b\rangle. \]Examples include
\[ \langle 6,0\rangle \]and
\[ \langle 0,-4\rangle. \]The Zero Vector in Component Form
The zero vector in two dimensions is
\[ \boxed{\mathbf{0}=\langle 0,0\rangle}. \]It represents no horizontal change and no vertical change.
Its magnitude is \(0\), and its direction is not defined.
Finding Components from Two Points
If
\[ A=(x_1,y_1) \]and
\[ B=(x_2,y_2), \]then
\[ \boxed{\overrightarrow{AB} =\langle x_2-x_1,\;y_2-y_1\rangle}. \]The rule is simple: terminal minus initial.
Example: Components from Two Points
Let
\[ A=(-2,1),\qquad B=(4,5). \]Then
\[ \overrightarrow{AB} =\langle 4-(-2),\;5-1\rangle. \]Therefore,
\[ \boxed{\overrightarrow{AB}=\langle 6,4\rangle}. \]Magnitude from Geometry
A vector
\[ \mathbf{v}=\langle a,b\rangle \]forms the hypotenuse of a right triangle whose legs have lengths \(|a|\) and \(|b|\).
By the Pythagorean theorem,
\[ \|\mathbf{v}\|^2=a^2+b^2. \]Therefore,
\[ \boxed{\|\mathbf{v}\|=\sqrt{a^2+b^2}}. \]Magnitude Is Always Nonnegative
Since magnitude represents length,
\[ \|\mathbf{v}\|\ge 0. \]Squaring the components removes their signs:
\[ (-a)^2=a^2. \]Thus opposite vectors have equal magnitude.
Example: Magnitude of a Vector
Find the magnitude of
\[ \mathbf{v}=\langle 3,4\rangle. \]Apply the magnitude formula:
\[ \|\mathbf{v}\| =\sqrt{3^2+4^2} =\sqrt{9+16} =\sqrt{25}. \]Therefore,
\[ \boxed{\|\mathbf{v}\|=5}. \]Example with Negative Components
Let
\[ \mathbf{u}=\langle -5,12\rangle. \]Then
\[ \|\mathbf{u}\| =\sqrt{(-5)^2+12^2} =\sqrt{25+144} =\sqrt{169}. \]Therefore,
\[ \boxed{\|\mathbf{u}\|=13}. \]Magnitude of a Vector Between Two Points
Since
\[ \overrightarrow{AB} =\langle x_2-x_1,\;y_2-y_1\rangle, \]its magnitude is
\[ \boxed{ \|\overrightarrow{AB}\| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }. \]This is exactly the distance formula.
Why the Distance Formula Is a Vector Magnitude Formula
The distance between two points is the length of the displacement vector connecting them.
Thus, the familiar distance formula is not a separate idea. It is the magnitude of
\[ \overrightarrow{AB}. \]Vector language unifies distance and directed change.
Direction in the Plane
A nonzero vector in the plane can be assigned a direction angle.
In standard position, the vector is imagined with its initial point at the origin.
Its direction angle \(\theta\) is measured from the positive \(x\)-axis, usually counterclockwise.
Direction Angle and Components
For
\[ \mathbf{v}=\langle a,b\rangle, \]the right-triangle relationship gives
\[ \tan\theta=\frac{b}{a} \]when \(a\neq 0\).
A first calculation may therefore be
\[ \theta=\tan^{-1}\left(\frac{b}{a}\right). \]However, the quadrant must still be checked.
Why the Quadrant Must Be Checked
The tangent function repeats every \(180^\circ\), so the same tangent value can correspond to two opposite directions.
For example,
\[ \tan 45^\circ=1 \]and
\[ \tan 225^\circ=1. \]Component signs tell us which direction is actually correct.
Example: Direction in Quadrant I
Let
\[ \mathbf{v}=\langle 3,3\rangle. \]Then
\[ \tan\theta=\frac{3}{3}=1. \]Since both components are positive, the vector lies in Quadrant I.
Therefore,
\[ \boxed{\theta=45^\circ}. \]Example: Direction in Quadrant II
Let
\[ \mathbf{v}=\langle -3,3\rangle. \]The reference angle is
\[ 45^\circ. \]But the vector lies in Quadrant II, so the direction angle is
\[ \boxed{\theta=135^\circ}. \]Example: Direction in Quadrant III
Let
\[ \mathbf{v}=\langle -2,-2\rangle. \]The reference angle is \(45^\circ\), but both components are negative.
Therefore, the vector lies in Quadrant III:
\[ \boxed{\theta=225^\circ}. \]Example: Direction in Quadrant IV
Let
\[ \mathbf{v}=\langle 4,-4\rangle. \]The reference angle is \(45^\circ\).
Since the vector lies in Quadrant IV,
\[ \boxed{\theta=315^\circ}. \]Axis-Aligned Direction Angles
Some vectors point directly along an axis.
| Direction | Typical Angle |
|---|---|
| Positive \(x\)-axis | \(0^\circ\) |
| Positive \(y\)-axis | \(90^\circ\) |
| Negative \(x\)-axis | \(180^\circ\) |
| Negative \(y\)-axis | \(270^\circ\) |
Degrees and Radians
Direction angles may be expressed in degrees or radians.
For example,
\[ 45^\circ=\frac{\pi}{4}, \] \[ 90^\circ=\frac{\pi}{2}, \] \[ 180^\circ=\pi. \]The underlying direction is the same regardless of the angular unit used.
Direction from Magnitude and Components
If
\[ \mathbf{v}=\langle a,b\rangle \]has magnitude
\[ r=\|\mathbf{v}\|, \]then
\[ a=r\cos\theta \]and
\[ b=r\sin\theta. \]Therefore,
\[ \boxed{ \mathbf{v} = \langle r\cos\theta,\;r\sin\theta\rangle }. \]Building a Vector from Magnitude and Direction
Suppose a vector has magnitude \(10\) and direction angle \(30^\circ\).
Then
\[ \mathbf{v} = \langle 10\cos 30^\circ,\; 10\sin 30^\circ \rangle. \]Since
\[ \cos 30^\circ=\frac{\sqrt{3}}{2} \]and
\[ \sin 30^\circ=\frac12, \]we obtain
\[ \boxed{ \mathbf{v} = \langle 5\sqrt{3},5\rangle }. \]Direction Determines Component Ratios
Vectors pointing in the same direction have proportional components.
For example,
\[ \langle 2,1\rangle \]and
\[ \langle 6,3\rangle \]have the same direction because
\[ \langle 6,3\rangle =3\langle 2,1\rangle. \]Their magnitudes differ, but their component ratio is the same.
What Is a Unit Vector?
A unit vector is a vector whose magnitude is \(1\).
If \(\mathbf{u}\) is a unit vector, then
\[ \boxed{\|\mathbf{u}\|=1}. \]Unit vectors are useful because they represent direction without carrying an arbitrary size.
Direction Without Size
Suppose
\[ \mathbf{v}=\langle 6,8\rangle. \]Its magnitude is
\[ \|\mathbf{v}\|=10. \]Dividing the vector by \(10\) gives
\[ \left\langle \frac{6}{10},\frac{8}{10}\right\rangle = \left\langle \frac35,\frac45\right\rangle. \]This new vector points in the same direction but has magnitude \(1\).
Normalizing a Vector
To convert a nonzero vector into a unit vector with the same direction, divide by its magnitude.
\[ \boxed{ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} }. \]This process is called normalization.
Why Normalization Works
If
\[ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}, \]then its magnitude is
\[ \|\mathbf{u}\| = \left\| \frac{\mathbf{v}}{\|\mathbf{v}\|} \right\|. \]Scaling a vector by \(1/\|\mathbf{v}\|\) scales its magnitude by the same positive factor:
\[ \|\mathbf{u}\| = \frac{\|\mathbf{v}\|}{\|\mathbf{v}\|} =1. \]Example: Find a Unit Vector
Let
\[ \mathbf{v}=\langle 3,4\rangle. \]Since
\[ \|\mathbf{v}\|=5, \]the unit vector in the same direction is
\[ \mathbf{u} = \frac{1}{5}\langle 3,4\rangle. \]Therefore,
\[ \boxed{ \mathbf{u} = \left\langle\frac35,\frac45\right\rangle }. \]Check the Unit Vector
Verify
\[ \mathbf{u} = \left\langle\frac35,\frac45\right\rangle. \]Its magnitude is
\[ \|\mathbf{u}\| = \sqrt{ \left(\frac35\right)^2+ \left(\frac45\right)^2 }. \]Thus,
\[ \|\mathbf{u}\| = \sqrt{\frac{9}{25}+\frac{16}{25}} = \sqrt{1} =1. \]The normalization is correct.
The Zero Vector Cannot Be Normalized
The normalization formula is
\[ \frac{\mathbf{v}}{\|\mathbf{v}\|}. \]For the zero vector,
\[ \|\mathbf{0}\|=0. \]Normalizing would require division by zero.
Therefore,
\[ \boxed{\text{the zero vector cannot be normalized}}. \]Standard Unit Vectors
In the coordinate plane, two especially important unit vectors are
\[ \boxed{\mathbf{i}=\langle 1,0\rangle} \]and
\[ \boxed{\mathbf{j}=\langle 0,1\rangle}. \]The vector \(\mathbf{i}\) points along the positive \(x\)-axis, while \(\mathbf{j}\) points along the positive \(y\)-axis.
Component Form and \(\mathbf{i},\mathbf{j}\) Form
The vector
\[ \mathbf{v}=\langle a,b\rangle \]can also be written
\[ \boxed{ \mathbf{v}=a\mathbf{i}+b\mathbf{j} }. \]This notation makes the horizontal and vertical contributions explicit.
Example: Convert to \(\mathbf{i},\mathbf{j}\) Form
Let
\[ \mathbf{v}=\langle -4,7\rangle. \]Then
\[ \boxed{ \mathbf{v} = -4\mathbf{i}+7\mathbf{j} }. \]Example: Convert Back to Component Form
Suppose
\[ \mathbf{u} = 5\mathbf{i}-2\mathbf{j}. \]Since
\[ \mathbf{i}=\langle 1,0\rangle \]and
\[ \mathbf{j}=\langle 0,1\rangle, \]we have
\[ \boxed{ \mathbf{u}=\langle 5,-2\rangle }. \]Unit Vectors from Direction Angles
A vector of magnitude \(1\) with direction angle \(\theta\) has component form
\[ \boxed{ \mathbf{u} = \langle \cos\theta,\sin\theta\rangle }. \]This works because
\[ \cos^2\theta+\sin^2\theta=1. \]Checking the Trigonometric Unit Vector
Let
\[ \mathbf{u} = \langle \cos\theta,\sin\theta\rangle. \]Then
\[ \|\mathbf{u}\| = \sqrt{ \cos^2\theta+\sin^2\theta }. \]By the Pythagorean identity,
\[ \boxed{\|\mathbf{u}\|=1}. \]Building Any Vector from a Unit Direction
If \(\mathbf{u}\) is a unit vector and \(r>0\), then
\[ \mathbf{v}=r\mathbf{u} \]has magnitude \(r\) and the same direction as \(\mathbf{u}\).
Thus a vector can be thought of as
\[ \boxed{ \text{magnitude}\times\text{unit direction} }. \]Magnitude-Direction Form
Combining the previous ideas gives
\[ \mathbf{v} = \|\mathbf{v}\| \left\langle \cos\theta,\sin\theta \right\rangle. \]Therefore,
\[ \boxed{ \mathbf{v} = \langle \|\mathbf{v}\|\cos\theta,\; \|\mathbf{v}\|\sin\theta \rangle }. \]This formula separates size from direction.
Example: From Unit Direction to Vector
Suppose
\[ \mathbf{u} = \left\langle \frac{12}{13}, \frac{5}{13} \right\rangle \]is a unit vector.
A vector of magnitude \(26\) in the same direction is
\[ \mathbf{v}=26\mathbf{u}. \]Therefore,
\[ \mathbf{v} = \left\langle 24,10 \right\rangle. \]Hence,
\[ \boxed{\mathbf{v}=\langle 24,10\rangle}. \]Three-Dimensional Components
In three dimensions,
\[ \mathbf{v} = \langle a,b,c\rangle. \]The components record directed change along the \(x\)-, \(y\)-, and \(z\)-axes.
Magnitude in Three Dimensions
For
\[ \mathbf{v}=\langle a,b,c\rangle, \]the magnitude is
\[ \boxed{ \|\mathbf{v}\| = \sqrt{a^2+b^2+c^2} }. \]This is the three-dimensional extension of the Pythagorean theorem.
Example: Three-Dimensional Magnitude
Let
\[ \mathbf{v}=\langle 2,-3,6\rangle. \]Then
\[ \|\mathbf{v}\| = \sqrt{2^2+(-3)^2+6^2}. \]Therefore,
\[ \|\mathbf{v}\| = \sqrt{4+9+36} = \sqrt{49} = \boxed{7}. \]The Standard Unit Vector \(\mathbf{k}\)
In three dimensions, a third standard unit vector is used:
\[ \boxed{\mathbf{k}=\langle 0,0,1\rangle}. \]Thus,
\[ \mathbf{i}=\langle 1,0,0\rangle, \] \[ \mathbf{j}=\langle 0,1,0\rangle, \] \[ \mathbf{k}=\langle 0,0,1\rangle. \]Three-Dimensional \(\mathbf{i},\mathbf{j},\mathbf{k}\) Form
The vector
\[ \mathbf{v}=\langle a,b,c\rangle \]can be written
\[ \boxed{ \mathbf{v} = a\mathbf{i}+b\mathbf{j}+c\mathbf{k} }. \]Normalizing in Three Dimensions
The same normalization rule applies:
\[ \boxed{ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} }. \]If
\[ \mathbf{v}=\langle 2,-3,6\rangle \]and \(\|\mathbf{v}\|=7\), then
\[ \boxed{ \mathbf{u} = \left\langle \frac27,-\frac37,\frac67 \right\rangle }. \]Higher-Dimensional Magnitude
The same pattern extends to
\[ \mathbf{v} = \langle x_1,x_2,\ldots,x_n\rangle. \]Its standard Euclidean magnitude is
\[ \boxed{ \|\mathbf{v}\| = \sqrt{x_1^2+x_2^2+\cdots+x_n^2} }. \]Although higher-dimensional vectors are difficult to visualize directly, the algebra remains systematic.
Magnitude as a Measure of Size
The magnitude formula gives a single scalar that summarizes the size of a vector.
This makes it possible to compare vector sizes even when the vectors point in different directions.
For example, if
\[ \|\mathbf{u}\|=5 \]and
\[ \|\mathbf{v}\|=9, \]then \(\mathbf{v}\) has the greater magnitude regardless of direction.
Direction Is Undefined for the Zero Vector
Direction formulas assume a nonzero vector.
For
\[ \mathbf{0}=\langle 0,0\rangle, \]both components vanish.
There is no unique direction angle, and normalization would require division by zero.
Common Error: Adding Components Before Squaring
For
\[ \mathbf{v}=\langle 3,4\rangle, \]the magnitude is not
\[ 3+4=7. \]Instead,
\[ \boxed{ \|\mathbf{v}\| = \sqrt{3^2+4^2} =5 }. \]Common Error: Forgetting the Square Root
The expression
\[ a^2+b^2 \]is the square of the magnitude.
The magnitude itself is
\[ \boxed{ \sqrt{a^2+b^2} }. \]Common Error: Letting Negative Components Produce Negative Magnitude
A component may be negative, but magnitude is a length.
For
\[ \langle -3,-4\rangle, \]we have
\[ \sqrt{(-3)^2+(-4)^2} = \sqrt{9+16} = 5. \]Magnitude is never negative.
Common Error: Using Inverse Tangent Without Checking the Quadrant
Suppose
\[ \mathbf{v}=\langle -1,1\rangle. \]Computing
\[ \tan^{-1}(-1) \]may return \(-45^\circ\), but the vector lies in Quadrant II.
The standard direction angle is
\[ \boxed{135^\circ}. \]Common Error: Confusing a Reference Angle with a Direction Angle
A reference angle is the acute angle made with the \(x\)-axis.
A direction angle identifies the actual orientation of the vector in the plane.
In Quadrants II, III, and IV, these are generally not the same number.
Common Error: Dividing by Components Instead of Magnitude
To normalize
\[ \mathbf{v}=\langle a,b\rangle, \]divide both components by the single scalar
\[ \|\mathbf{v}\|. \]Do not divide \(a\) by itself and \(b\) by itself.
Common Error: Normalizing the Zero Vector
Since
\[ \|\mathbf{0}\|=0, \]the expression
\[ \frac{\mathbf{0}}{\|\mathbf{0}\|} \]would require division by zero.
Therefore, normalization applies only to nonzero vectors.
Common Error: Confusing \(\mathbf{i}\) with the Number \(1\)
The symbol
\[ \mathbf{i} \]here represents the unit vector
\[ \langle 1,0\rangle, \]not the scalar number \(1\).
Context matters because the same letter \(i\) is also used elsewhere in mathematics for the imaginary unit.
A Reliable Procedure for Finding Magnitude
- Identify the vector components.
- Square each component.
- Add the squares.
- Take the nonnegative square root.
- Include units when the vector represents a measured quantity.
In two dimensions:
\[ \boxed{ \|\langle a,b\rangle\| = \sqrt{a^2+b^2} }. \]A Reliable Procedure for Finding a Direction Angle
- Identify the signs of the components.
- Determine the vector's quadrant or axis.
- Use a trigonometric ratio to find a reference angle when needed.
- Adjust the angle to match the actual quadrant.
- State the angle in the requested unit.
- Check that the final angle points in the same direction as the components.
A Reliable Procedure for Normalizing a Vector
- Verify that the vector is nonzero.
- Compute its magnitude.
- Divide every component by that magnitude.
- Write the resulting unit vector.
- Check that its magnitude equals \(1\).
Example: Full Component-Magnitude-Direction Workflow
Let
\[ A=(1,2),\qquad B=(5,5). \]First find the displacement vector:
\[ \overrightarrow{AB} = \langle 5-1,\;5-2\rangle = \langle 4,3\rangle. \]Its magnitude is
\[ \|\overrightarrow{AB}\| = \sqrt{4^2+3^2} = 5. \]Its direction angle satisfies
\[ \tan\theta=\frac34. \]Since the vector lies in Quadrant I,
\[ \theta = \tan^{-1}\left(\frac34\right) \approx 36.87^\circ. \]Example: Continue by Normalizing
For
\[ \mathbf{v}=\langle 4,3\rangle \]with magnitude \(5\), the unit vector is
\[ \mathbf{u} = \frac{1}{5}\langle 4,3\rangle. \]Therefore,
\[ \boxed{ \mathbf{u} = \left\langle \frac45,\frac35 \right\rangle }. \]The unit vector keeps the direction while removing the original size.
Practice 1: Read the Components
Describe the horizontal and vertical directed changes represented by
\[ \langle -7,4\rangle. \]Practice 2: Find Components from Two Points
Let
\[ A=(-1,3),\qquad B=(4,-2). \]Find
\[ \overrightarrow{AB}. \]Practice 3: Find Magnitude
Find the magnitude of
\[ \mathbf{v}=\langle 8,15\rangle. \]Practice 4: Find Magnitude with Negative Components
Find
\[ \|\langle -6,8\rangle\|. \]Practice 5: Find a Direction Angle
Find the standard direction angle of
\[ \mathbf{v}=\langle 1,1\rangle. \]Practice 6: Quadrant Check
Find the standard direction angle of
\[ \mathbf{v}=\langle -1,-1\rangle. \]Practice 7: Normalize a Vector
Find a unit vector in the same direction as
\[ \mathbf{v}=\langle 5,12\rangle. \]Practice 8: Convert to \(\mathbf{i},\mathbf{j}\) Form
Rewrite
\[ \langle -3,9\rangle \]using \(\mathbf{i}\) and \(\mathbf{j}\).
Practice 9: Build a Vector from Magnitude and Direction
Find the component form of a vector with magnitude \(8\) and direction angle \(60^\circ\).
Practice 10: Three-Dimensional Magnitude
Find the magnitude of
\[ \mathbf{v}=\langle 1,2,2\rangle. \]Practice Answers
Practice 1
The vector represents \(7\) units left and \(4\) units up.
Practice 2
\[ \overrightarrow{AB} = \langle 4-(-1),\;-2-3\rangle = \boxed{\langle 5,-5\rangle}. \]Practice 3
\[ \|\mathbf{v}\| = \sqrt{8^2+15^2} = \sqrt{64+225} = \sqrt{289} = \boxed{17}. \]Practice 4
\[ \|\langle -6,8\rangle\| = \sqrt{(-6)^2+8^2} = \sqrt{36+64} = \boxed{10}. \]Practice 5
\[ \tan\theta=\frac11=1. \]The vector lies in Quadrant I, so
\[ \boxed{\theta=45^\circ}. \]Practice 6
The reference angle is \(45^\circ\), but the vector lies in Quadrant III.
\[ \boxed{\theta=225^\circ}. \]Practice 7
First compute
\[ \|\mathbf{v}\| = \sqrt{5^2+12^2} = 13. \]Therefore,
\[ \boxed{ \mathbf{u} = \left\langle \frac{5}{13}, \frac{12}{13} \right\rangle }. \]Practice 8
\[ \boxed{ -3\mathbf{i}+9\mathbf{j} }. \]Practice 9
\[ \mathbf{v} = \langle 8\cos60^\circ,\; 8\sin60^\circ \rangle. \]Therefore,
\[ \boxed{ \mathbf{v} = \langle 4,4\sqrt{3}\rangle }. \]Practice 10
\[ \|\mathbf{v}\| = \sqrt{1^2+2^2+2^2} = \sqrt{1+4+4} = \boxed{3}. \]What to Remember
- Components encode directed change along coordinate axes.
- In two dimensions, \(\mathbf{v}=\langle a,b\rangle\).
- The magnitude is \(\|\mathbf{v}\|=\sqrt{a^2+b^2}\).
- Magnitude is always nonnegative.
- Direction angles are measured from the positive \(x\)-axis in standard position.
- Quadrant information must be checked when using inverse tangent.
- A unit vector has magnitude \(1\).
- A nonzero vector is normalized by dividing by its magnitude.
- \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are standard coordinate unit vectors.
- A vector can be reconstructed from magnitude and direction.
Where We Go Next
Components let us describe individual vectors precisely. The next step is to combine directed quantities.
The next lesson, Vector Addition and Subtraction, develops the geometric head-to-tail and parallelogram interpretations of vector addition and connects them to component-wise algebra. Subtraction will then appear as addition of an opposite vector.