3.1 Vectors and Their Components Explained with Examples

Physics is impossible without vectors. Every displacement, force, velocity, and field in your exams has both size and direction, and this guide gives you the definitive language to handle them.

Key Insights

• Scalars have magnitude only, vectors have both magnitude and direction and obey vector algebra.
• A displacement vector depends only on start and end points, not on the path taken.
• Vectors add head-to-tail to give a resultant; addition is commutative and associative.
• Any vector can be resolved into components \(a_x = a \cos\theta\) and \(a_y = a \sin\theta\) that act as its perpendicular shadows.
• Magnitude and direction can be rebuilt from components with \(a = \sqrt{a_x^2+a_y^2}\) and \(\tan\theta = a_y/a_x\).

What Is Physics?

Physics deals with quantities that have both size and direction, and it needs a special mathematical language — the language of vectors — to describe them. This language is also used in engineering, other sciences, and in common speech. If you have ever given directions such as “Go five blocks down this street and then hang a left,” you have used the language of vectors.

In fact, navigation of any sort is based on vectors, but physics and engineering also need vectors in special ways to explain phenomena involving rotation and magnetic forces, which we will get to in later chapters. In this module, we focus on the basic language of vectors: what they are, how to add them geometrically, and how to break them into components that make calculation easy. Mastering this is non-negotiable for projectile motion, forces, momentum, and fields.

Why Vectors Matter for Exams:Every A-Level, AP Physics 1/C Mechanics, and IB HL question about motion in two dimensions starts by asking you to add or resolve vectors. If you cannot convert between magnitude-angle and component form without hesitation, you will lose time on every paper.

Vectors and Scalars

A vector has magnitude as well as direction, and vectors follow certain rules of combination. A quantity that has both a magnitude and a direction and thus can be represented with a vector. Some physical quantities that are vector quantities are displacement, velocity, and acceleration.

Not all physical quantities involve a direction. Temperature, pressure, energy, mass, and time, for example, do not point in the spatial sense. We call such quantities scalars, and we deal with them by the rules of ordinary algebra. A single value, with a sign as in a temperature of \(-40^{\circ}F\), specifies a scalar.

The simplest vector quantity is displacement, or change of position. A vector that represents a displacement is called a displacement vector. If a particle changes position by moving from \(A\) to \(B\), we say that it undergoes a displacement from \(A\) to \(B\), which we represent with an arrow pointing from \(A\) to \(B\). The arrow specifies the vector graphically.

To distinguish vector symbols from other kinds of arrows, we use the outline of a triangle as the arrowhead.

A critical property: The arrows from \(A\) to \(B\), from \(A'\) to \(B'\), and from \(A''\) to \(B''\) have the same magnitude and direction. Thus, they specify identical displacement vectors and represent the same change of position for the particle. A vector can be shifted without changing its value if its length and direction are not changed.

The displacement vector tells us nothing about the actual path that the particle takes. All paths connecting the two points correspond to the same displacement vector. Displacement vectors represent only the overall effect of the motion, not the motion itself.

Feature	Scalar	Vector
Definition	Magnitude only	Magnitude + direction
Example	Temperature \(10^{\circ}C\), mass 5 kg, time 12 s	Displacement 5 m north, velocity 20 m/s east
Algebra	Ordinary arithmetic	Head-to-tail, components, sign indicates direction along axis
Representation	Number with unit	Arrow: length = magnitude, direction = direction

Three identical blue 3D arrows with same length and direction shifted in space representing same displacement vector, plus three different curved paths connecting same start and end points showing path independence

Exam Tip:When a question says “particle moves from A to B via different routes,” the displacement is identical. Don’t integrate the path length unless they ask for distance.

Exam Trap:Students think a vector changes if you move its tail. It does not. Only magnitude (length) and direction matter. You can slide a vector anywhere parallel to itself for addition.

Adding Vectors Geometrically

Suppose a particle moves from \(A\) to \(B\) and then later from \(B\) to \(C\). We can represent its overall displacement (no matter what its actual path) with two successive displacement vectors, \(AB\) and \(BC\). The net displacement of these two displacements is a single displacement from \(A\) to \(C\). We call \(AC\) the vector sum or resultant of the vectors \(AB\) and \(BC\).

We now relabel them in the way that we will use from now on, namely, with an arrow over an italic symbol, as in (\vec{a}). If we want to indicate only the magnitude of the vector (a quantity that lacks a sign or direction), we shall use the italic symbol, as in \(a\), \(b\), and \(s\). A symbol with an overhead arrow always implies both properties of a vector, magnitude and direction.

We can represent the relation among the three vectors with the vector equation

\(\vec{s} = \vec{a} + \vec{b}\)

which says that the vector \(\vec{s}\) is the vector sum of vectors \(\vec{a}\) and \(\vec{b}\). The symbol \(+\) and the words “sum” and “add” have different meanings for vectors than they do in the usual algebra because they involve both magnitude and direction.

The procedure for adding two-dimensional vectors \(\vec{a}\) and \(\vec{b}\) geometrically is:

1. On paper, sketch vector \(\vec{a}\) to some convenient scale and at the proper angle.
2. Sketch vector \(\vec{b}\) to the same scale, with its tail at the head of vector \(\vec{a}\), again at the proper angle.
3. The vector sum \(\vec{s}\) is the vector that extends from the tail of \(\vec{a}\) to the head of \(\vec{b}\).

Properties

Vector addition, defined in this way, has two important properties.

First, the order of addition does not matter. Adding \(\vec{a}\) to \(\vec{b}\) gives the same result as adding \(\vec{b}\) to \(\vec{a}\):

\(\vec{a} + \vec{b} = \vec{b} + \vec{a} \quad \text{(commutative law)}\)

You get the same vector result for either order of adding vectors. Geometrically, this forms a parallelogram.

Second, when there are more than two vectors, we can group them in any order as we add them. Thus, if we want to add vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we can add \(\vec{a}\) and \(\vec{b}\) first and then add their vector sum to \(\vec{c}\). We can also add \(\vec{b}\) and \(\vec{c}\) first and then add that sum to \(\vec{a}\). We get the same result either way:

\((\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c}) \quad \text{(associative law)}\)

Vector Subtraction

The vector \(-\vec{b}\) is a vector with the same magnitude as \(\vec{b}\) but the opposite direction. Adding the two vectors would yield

\(\vec{b} + (-\vec{b}) = 0\)

Thus, adding \(-\vec{b}\) has the effect of subtracting \(\vec{b}\). We use this property to define the difference between two vectors: let \(\vec{d} = \vec{a} - \vec{b}\). Then

\(\vec{d} = \vec{a} - \vec{b} = \vec{a} + (-\vec{b}) \quad \text{(vector subtraction)}\)

That is, we find the difference vector \(\vec{d}\) by adding the vector \(-\vec{b}\) to the vector \(\vec{a}\). Geometrically, this is still a head-to-tail arrangement for addition: flip \(\vec{b}\), then add.

As in the usual algebra, we can move a term that includes a vector symbol from one side of a vector equation to the other, but we must change its sign. For example, if we are given \(\vec{d} = \vec{a} - \vec{b}\) and need to solve for \(\vec{a}\), we can rearrange as

\(\vec{d} + \vec{b} = \vec{a} \quad \text{or} \quad \vec{a} = \vec{d} + \vec{b}\)

Remember that although we have used displacement vectors here, the rules for addition and subtraction hold for vectors of all kinds, whether they represent velocities, accelerations, or any other vector quantity. However, we can add only vectors of the same kind. For example, we can add two displacements, or two velocities, but adding a displacement and a velocity makes no sense. In the arithmetic of scalars, that would be like trying to add 21 s and 12 m.

The Zero Vector:\(\vec{b} + (-\vec{b}) = 0\) defines the zero vector. It has no direction and zero magnitude. It is essential for equilibrium problems where sum of forces is zero.

Head-to-tail vector addition showing light blue vector a and dark blue vector b forming green resultant s, plus inset showing commutative law a+b=b+a, and anti-parallel pair b and -b

Exam Tip:For Checkpoint-style questions: If magnitudes of \(\vec{a}\) and \(\vec{b}\) are 3 m and 4 m and \(\vec{c} = \vec{a} + \vec{b}\), maximum possible \(|\vec{c}| = 3+4 = 7\) m when \(\vec{a}\) and \(\vec{b}\) are parallel. Minimum \(|\vec{c}| = |4-3| = 1\) m when antiparallel. Any other angle gives \(1 < c < 7\).

Exam Trap:Never add magnitudes directly unless vectors are collinear and same direction. \(|\vec{a} + \vec{b}| \neq |\vec{a}| + |\vec{b}|\) in general. You must add components or use law of cosines.

Components of Vectors

Adding vectors geometrically can be tedious. A neater and easier technique involves algebra but requires that the vectors be placed on a rectangular coordinate system. The \(x\) and \(y\) axes are usually drawn in the plane of the page, and the \(z\) axis comes directly out of the page at the origin; we ignore it for now and deal only with two-dimensional vectors.

A component of a vector is the projection of the vector on an axis. For example, \(a_x\) is the component of vector \(\vec{a}\) on (or along) the \(x\) axis and \(a_y\) is the component along the \(y\) axis. To find the projection of a vector along an axis, we draw perpendicular lines from the two ends of the vector to the axis, as shown. The projection of a vector on an \(x\) axis is its \(x\) component, and similarly the projection on the \(y\) axis is the \(y\) component. The process of finding the components of a vector is called resolving the vector.

A component of a vector has the same direction (along an axis) as the vector. In general, \(a_x\) and \(a_y\) are both positive because \(\vec{a}\) extends in the positive direction of both axes. The small arrowheads on the components indicate their direction. If we were to reverse vector \(\vec{a}\), then both components would be negative and their arrowheads would point toward negative \(x\) and \(y\). Resolving vector \(\vec{b}\) yields a positive component \(b_x\) and a negative component \(b_y\).

In general, a vector has three components, although for the case where the component along the \(z\) axis is zero, we deal with two. As we show, if you shift a vector without changing its direction, its components do not change.

Finding the Components

We can find the components of \(\vec{a}\) geometrically from the right triangle there:

\(a_x = a \cos\theta \quad \text{and} \quad a_y = a \sin\theta\)

where \(\theta\) is the angle that the vector \(\vec{a}\) makes with the positive direction of the \(x\) axis, and \(a\) is the magnitude of \(\vec{a}\). The components and the vector form a right triangle whose hypotenuse is the magnitude of the vector.

The components and the vector form the legs of a right triangle whose hypotenuse is the magnitude of the vector.

Once a vector has been resolved into its components along a set of axes, the components themselves can be used in place of the vector. For example, \(\vec{a}\) is given (completely determined) by \(a\) and \(\theta\). It can also be given by its components \(a_x\) and \(a_y\). Both pairs of values contain the same information. If we know a vector in component notation \((a_x, a_y)\) and want it in magnitude-angle notation \((a, \theta)\), we can use the equations

\(a = \sqrt{a_x^2 + a_y^2}\) and \(\tan\theta = \frac{a_y}{a_x}\)

to transform it. Note: \(\tan\theta\) alone is ambiguous because \(\theta\) and \(\theta+180^{\circ}\) have same tangent. Always check signs of \(a_x\) and \(a_y\) to pick correct quadrant.

In the more general three-dimensional case, we need a magnitude and two angles (say, \(a\), \(\theta\), and \(\phi\)) or three components \((a_x, a_y, a_z)\) to specify a vector.

Magenta 3D arrow vector a at angle theta above x-axis with dashed perpendicular projections forming right triangle, showing x-component ax and y-component ay as purple arrows, with formulas ax=a cosθ, ay=a sinθ, a=sqrt(ax²+ay²), tanθ=ay/ax

Component Notation vs Magnitude-Angle Notation:Component notation \((a_x, a_y)\) is best for adding vectors: add x’s, add y’s. Magnitude-angle notation \((a, \theta)\) is best for visualizing. Convert to components, add, then convert back. This is the AP/IB/A-Level method.

Conversion	Formula	When to Use
Magnitude-angle → Components	\(a_x = a\cos\theta\), \(a_y = a\sin\theta\)	Before adding multiple vectors
Components → Magnitude	\(a = \sqrt{a_x^2 + a_y^2}\)	To report final answer as size
Components → Angle	\(\theta = \tan^{-1}(a_y/a_x)\) + quadrant check	To report direction

Exam Tip:For Checkpoint 3.1.2 on proper head-to-tail combinations: Only methods where \(a_x\) and \(a_y\) are placed head-to-tail to form \(\vec{a}\) as the hypotenuse are valid. If both components start at origin and \(\vec{a}\) is not the hypotenuse, it is wrong. Look for right triangle with \(\vec{a}\) as hypotenuse.

Sample Problem Deep Dive: Spelunking Displacement

For two decades spelunking teams crawled through 200 km of Mammoth Cave and Flint Ridge cave system seeking a connection. The team finally found it, caving for 12 hours from Austin Entrance in Flint Ridge system to Echo River in Mammoth Cave, traveling a net 2.6 km westward, 3.9 km southward, and 25 m upward. What were magnitude and angle of displacement from start to finish?

We have components of a three-dimensional vector, and we need the magnitude and two angles.

Horizontal legs 2.6 km west and 3.9 km south form legs of horizontal right triangle. Horizontal displacement magnitude \(d_h\) by Pythagorean theorem:

\(\begin{aligned} d_h &= \sqrt{(2.6\,\text{km})^2 + (3.9\,\text{km})^2} \\ &= 4.69\,\text{km} \end{aligned}\)

Also from horizontal triangle, this horizontal is directed south of due west by angle \(\theta_h\):

\(\begin{aligned} \tan\theta_h &= \frac{3.9}{2.6} \implies \theta_h \\ &= \tan^{-1}1.5 \\ &= 56^{\circ} \end{aligned}\)

So \(56^{\circ}\) south of west.

Include vertical component 25 m = 0.025 km. Vertical component and horizontal displacement form another right triangle. Overall displacement magnitude \(d\):

\(\begin{aligned} d &= \sqrt{(4.69)^2 + (0.025)^2} \\ &= 4.69\,\text{km} \approx 4.7\,\text{km} \end{aligned}\)

Angle upward from horizontal:

\(\begin{aligned} \theta_v &= \tan^{-1}\frac{0.025}{4.69} \\ &= 0.3^{\circ} \end{aligned}\)

Net vertical motion was insignificant compared to horizontal, but had to climb up and down countless times to get through the cave.

Problem-Solving Tactics: Angles, Trig Functions, and Inverse Trig Functions

Angles—Degrees and Radians

Angles measured relative to positive direction of \(x\) axis are positive if measured counterclockwise and negative if measured clockwise. For example, \(210^{\circ}\) and \(-150^{\circ}\) are the same angle.

Angles may be measured in degrees and radians. To relate the two measures, recall that a full circle is \(360^{\circ}\) and \(2\pi\) rad. To convert, say, \(40^{\circ}\) to radians, write

\(40^{\circ} \times \dfrac{2\pi\,\text{rad}}{360^{\circ}} = 0.70\,\text{rad}\)

Trig Functions

You need to know definitions of sine, cosine, tangent—they are part of language of science and engineering. They are given in a form that does not depend on how the triangle is labeled:

\(\begin{aligned} \sin\theta &= \frac{\text{leg opposite }\theta}{\text{hypotenuse}} \\[18pt] \cos\theta &= \frac{\text{leg adjacent to }\theta}{\text{hypotenuse}} \\[18pt] \tan\theta &= \frac{\text{leg opposite }\theta}{\text{leg adjacent to }\theta} \end{aligned}\)

You should also be able to sketch how trig functions vary with angle, to judge whether calculator result is reasonable. Knowing signs of functions in various quadrants can be of help.

Inverse Trig Functions

When inverse trig functions \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) are taken on a calculator, you must consider reasonableness of answer you get, because there is usually another possible answer that calculator does not give. The range of operation for a calculator in taking each inverse is indicated. For example, \(\sin^{-1}0.5\) has associated angles \(30^{\circ}\) (displayed) and \(150^{\circ}\). To see both values, draw a horizontal line through 0.5 on sine curve and note where it cuts. How to distinguish correct answer? It is the one that seems more reasonable for given situation.

Four quadrants with vectors showing sign of x and y components, angle theta positive counterclockwise and negative clockwise, plus signs table for sin cos tan by quadrant and small sine cosine tangent curves

Measuring Vector Angles

The equations for \(\cos\theta\) and \(\sin\theta\) and for \(\tan\theta\) are valid only if the angle is measured from the positive direction of the \(x\) axis. If it is measured relative to some other direction, then the trig functions may have to be interchanged and the ratio may have to be inverted. A safer method is to convert the angle to one measured from the positive direction of the \(x\) axis.

The Quadrant Trap:Calculator \(\tan^{-1}(a_y/a_x)\) returns \(-90^{\circ}\) to \(+90^{\circ}\) only. If \(a_x\) is negative, you must add \(180^{\circ}\). Always draw the components to see quadrant: \(a_x>0, a_y>0\) Quadrant I; \(a_x<0, a_y>0\) Quadrant II; \(a_x<0, a_y<0\) Quadrant III; \(a_x>0, a_y<0\) Quadrant IV.

Exam Tip:In WileyPLUS and AP, system expects you to report angle direction like “ \(56^{\circ}\) south of west” or “ \(34^{\circ}\) above +x”. Positive if counterclockwise from +x, negative if clockwise. Be explicit.

Exam Trap:Mixing degrees and radians in same equation. Check mode on calculator. For AP Physics C, radians are mandatory for angular motion but degrees are fine for vector angles—keep consistent.

Practice Questions

MCQs (Multiple Choice Questions)

01. Which is a vector?

1. Temperature 300 K
2. Mass 5 kg
3. Displacement 5 m north
4. Time 10 s

Correct: C
Explanation: Vectors have magnitude + direction. Displacement north is vector; others are scalars.

02. Two displacement vectors have same length but start at different points. They are:

1. Different vectors
2. Same vector if same direction
3. Always zero
4. Cannot be compared

Correct: B
Explanation: Vector equality depends only on magnitude and direction, not starting point. Shifting without changing length/direction leaves vector unchanged.

03. If \(\vec{a} + \vec{b} = \vec{b} + \vec{a}\), this illustrates:

1. Associative law
2. Commutative law
3. Distributive law
4. Zero vector

Correct: B
Explanation: Order of addition does not matter, commutative.

04. Magnitudes 3 m and 4 m add to give \(\vec{c}\). Maximum \(|\vec{c}|\) is:

1. 1 m
2. 5 m
3. 7 m
4. 12 m

Correct: C
Explanation: Max when parallel: \(3+4=7\) m.

05. Minimum \(|\vec{c}|\) for same vectors is:

1. 0 m
2. 1 m
3. 3 m
4. 5 m

Correct: B
Explanation: Min when antiparallel: \(|4-3|=1\) m.

06. Vector \(\vec{a}\) magnitude 10 at \(60^{\circ}\) above +x. \(a_x\) is:

1. 5
2. 8.66
3. 10
4. 0

Correct: A
Explanation: \(a_x = a\cos\theta =10\cos60=10*0.5=5\).

07. Components \(a_x = -3\), \(a_y = +4\). Vector is in:

1. Quadrant I
2. Quadrant II
3. Quadrant III
4. Quadrant IV

Correct: B
Explanation: Negative x, positive y = Quadrant II. Angle \(>90^{\circ}\) and \(<180^{\circ}\).

08. \(\tan\theta = a_y/a_x\). If \(a_x<0\), correct \(\theta\) is:

1. Calculator value alone
2. Calculator value + \(180^{\circ}\)
3. Calculator value *2
4. Always \(45^{\circ}\)

Correct: B
Explanation: \(\tan^{-1}\) range \(-90\) to \(90\); for \(a_x<0\), add \(180^{\circ}\) to get Quadrant II or III.

09. You can add:

1. Displacement + velocity
2. Force + force
3. Time + displacement
4. Mass + velocity

Correct: B
Explanation: Only vectors of same kind can be added physically.

10. Displacement depends on:

1. Actual path taken
2. Only start and end points
3. Time taken
4. Speed

Correct: B
Explanation: Displacement vector represents overall effect, not path. Many paths give same displacement.

FRQs (Free Response Questions)

FRQ 1: (AP 1 style)
A hiker walks 2.6 km west and 3.9 km south, then climbs 0.025 km up. Find resultant displacement magnitude and its direction south of west and above horizontal.

Rubric:

• \(d_h = \sqrt{2.6^2+3.9^2}=4.69\) km (2 pts)
• \(\theta_h = \tan^{-1}(3.9/2.6)=56^{\circ}\) south of west (2 pts)
• \(d = \sqrt{4.69^2+0.025^2}\approx4.7\) km (2 pts)
• \(\theta_v = \tan^{-1}(0.025/4.69)=0.3^{\circ}\) upward (2 pts)

FRQ 2: (IB HL style)
Two forces \(\vec{F}_1 = 6\) N at \(0^{\circ}\) and \(\vec{F}_2 = 8\) N at \(90^{\circ}\) act on particle. Find net force magnitude and direction. If a third force balances them, what is it?

Rubric:

• Components \(F_{1x}=6\), \(F_{2y}=8\) (1 pt)
• resultant \(F =10\) N (1 pt)
• direction \(\tan^{-1}(8/6)=53^{\circ}\) from x (2 pts)
• equilibrium force \(-10\) N opposite (2 pts)

FRQ 3: (A-Level)
Explain why vectors \(\vec{a} = 5\) m east and \(\vec{b}=5\) m north are not equal, and show geometrically how to find \(\vec{a}+\vec{b}\).

Rubric:

• Equality requires same magnitude and direction (2 pts)
• they differ in direction (1 pt)
• head-to-tail diagram description (2 pts)
• resultant \(7.07\) m NE \(45^{\circ}\) (1 pt)

FRQ 4: (College Physics)
A vector has components \(a_x= -4\), \(a_y= -3\). Find magnitude, angle from +x, and express \(-\vec{a}\).

Rubric:

• Magnitude 5 (1 pt)
• angle \(\tan^{-1}(-3/-4)=36.9^{\circ}\) but in QIII so \(180+36.9=216.9^{\circ}\) or \(-143.1^{\circ}\) (3 pts)
• \(-\vec{a}\) components \(+4,+3\), same magnitude opposite direction (2 pts)

FRQ 5: (USAPhO)
Show that \((\vec{a}+\vec{b})+\vec{c} = \vec{a}+(\vec{b}+\vec{c})\) geometrically and why converting to radians matters when using \(\sin\theta\approx\theta\) approximation.

Rubric:

• Associative grouping explanation with diagram (3 pts)
• small angle approximation valid only in radians because derivative of sin depends on radian measure (3 pts)

Frequently Asked Questions

What is the difference between a scalar and a vector?

A scalar has only magnitude like temperature or mass. A vector has both magnitude and direction like displacement or velocity and follows vector addition rules, not ordinary algebra.

How do you add vectors geometrically?

Place vectors head-to-tail: draw first vector, place second’s tail at first’s head, resultant goes from first tail to second head. Order doesn’t matter due to commutative law.

What is the formula for vector components?

If vector magnitude is \(a\) and angle \(\theta\) from positive x-axis, \(a_x = a\cos\theta\) and \(a_y = a\sin\theta\). Components are the perpendicular projections onto axes.

How do you find magnitude from components?

Use Pythagoras: \(a = \sqrt{a_x^2 + a_y^2}\). Direction is \(\tan\theta = a_y/a_x\), but check quadrant signs to get correct angle.

What is the maximum and minimum resultant of two vectors?

Maximum when parallel: sum of magnitudes. Minimum when antiparallel: absolute difference of magnitudes. Other angles give intermediate values.

Can you add a displacement and a velocity?

No. Only vectors of same physical kind can be added. Adding displacement (meters) to velocity (m/s) is dimensionally inconsistent, like adding seconds and meters.

Why can you slide a vector without changing it?

A vector is defined only by length and direction. Moving its starting point parallel to itself preserves both, so it remains the same vector and represents same change.

What does negative component mean?

It means projection points opposite positive axis direction. Positive \(a_x\) points toward +x, negative toward -x. Sign indicates direction along that axis.

When is \(\theta\) positive or negative?

Angles from +x are positive if measured counterclockwise, negative if clockwise. \(210^{\circ}\) and \(-150^{\circ}\) point same way.

How do you subtract vectors?

Subtracting \(\vec{b}\) is adding \(-\vec{b}\), where \(-\vec{b}\) has same magnitude but opposite direction. So \(\vec{a}-\vec{b} = \vec{a}+(-\vec{b})\) head-to-tail.

Exam Mapping

Concept	A-Level Physics (AQA/OCR/Edexcel)	AP Physics 1 & C	IB Physics HL	College Physics 1	USAPhO/IPhO
Scalar vs Vector distinction	Foundations, SI units, vector nature of force, velocity	Unit 1 Kinematics, vector vs scalar	Topic 1.3 Vectors & scalars	Ch 1 Measurement	Basic tool
Displacement vector path independence	Motion graphs, displacement-time	Displacement vs distance	Displacement in kinematics	Displacement definition	Vector concept
Head-to-tail addition, resultant	Forces as vectors, resultant force	Adding forces, net displacement	Vector addition, resultant	Graphical addition	Vector addition
Commutative & associative laws	Parallelogram law	Vector properties	Algebraic properties	Laws of addition	Formal properties
Negative vector & subtraction	Relative velocity	Relative motion	Vector subtraction	Subtraction method	Relative vectors
Resolving into components \(a_x=a\cos\theta\), \(a_y=a\sin\theta\)	Resolving forces, projectile	Projectile motion, inclined planes	Projectile, components	Components method	Essential method
Magnitude from components \(a=\sqrt{a_x^2+a_y^2}\), \(\tan\theta=a_y/a_x\)	Finding resultant	Finding net force/velocity	Converting notations	Magnitude-angle conversion	Core skill
Quadrant & inverse trig trap	Angle measurement	Direction reporting	Bearings	Angle quadrant	Calculator trap
3D vectors	Not required	Not in AP 1, optional in C	HL extension	3D extension	Required
Degrees ⇌ radians	Radians in SHM	Radians in rotation	Radians	Conversion	Small-angle approx

Conclusion

You have now built the complete foundation: scalars versus vectors, why a displacement vector cares only about start and finish, how to add any two vectors head-to-tail to get a resultant that obeys commutative \(\vec{a}+\vec{b}=\vec{b}+\vec{a}\) and associative laws, how to create and use \(-\vec{b}\) for subtraction \(\vec{d}=\vec{a}-\vec{b}=\vec{a}+(-\vec{b})\), and how to resolve a vector into its faithful shadows \(a_x = a\cos\theta\), \(a_y = a\sin\theta\) and rebuild it with \(a=\sqrt{a_x^2+a_y^2}\) and \(\tan\theta=a_y/a_x\). With the spelunking example you saw how horizontal and vertical right triangles combine to give a 4.7 km displacement at \(56^{\circ}\) south of west and \(0.3^{\circ}\) upward, and you learned the quadrant and calculator checks that save marks in every exam. Keep practicing conversion between notations—components for adding, magnitude-angle for understanding—and vector problems become routine.

References

Fundamentals of Physics by Halliday, Resnick, and Walker, 12th Edition, Volume 1, Chapter 3 – Vectors and Their Components – primary source of truth for definitions, equations, and methodology.