Scalar vs Vector: What Is the Difference?

A scalar is described by a magnitude and a unit. A vector needs a magnitude and a direction. The difference that matters is how they combine: scalars add arithmetically, so 3 and 4 always make 7, while vectors add head to tail, so a 3 and a 4 make anything from 1 to 7 depending on the angle.

AP Physics: Unit 1 (topics 1.1 Scalars and Vectors in One Dimension, 1.5 Vectors and Motion in Two Dimensions). The definition sits in AP Physics 1 Unit 1, Kinematics, weighted at 10 to 15 percent of the multiple-choice section over an estimated 12 to 17 class periods, and the classification is spread across the whole course. EK 1.1.A.1 defines scalars as quantities described by magnitude only and vectors as quantities described by both magnitude and direction. EK 1.1.A.2 models vectors as arrows with lengths proportional to their magnitude; EK 1.1.A.3.i requires an arrow over the symbol; EK 1.1.A.3.ii exempts components along an axis, where the sign carries the direction; EK 1.1.B.1 assigns opposite signs to opposite directions. Topic 1.5 adds the two-dimensional treatment: EK 1.5.A.1 models a vector as the resultant of two perpendicular components, EK 1.5.A.3 gives the trigonometric relations, and EK 1.5.B.1 analyses two-dimensional motion by components. The quantities the CED explicitly calls scalars are distance and speed (1.1.A.3), translational kinetic energy (3.1.A.2), work (3.2.A.2), potential energy (3.3.A.2), rotational kinetic energy (6.1.A.3) and pressure (8.2.A.2); the ones it explicitly calls vectors are position, displacement, velocity and acceleration (1.1.A.3), force (2.2.A.1), momentum (4.1.A.2) and impulse (4.2.A.3). Suggested skills are 1.A, 2.C, 3.B and 3.C for Topic 1.1, and 1.B, 2.A, 2.D, 3.A and 3.C for Topic 1.5. AP Physics C: Mechanics keeps the same definitions and adds unit vector notation (its EK 1.1.A.4), the dot product and the cross product.

The distinction, stated once

A scalar is a magnitude with a unit. A vector is a magnitude with a unit and a direction. That is the definition, and the AP Physics 1 CED gives it in one sentence at EK 1.1.A.1: scalars are quantities described by magnitude only, while vectors are quantities described by both magnitude and direction.

The definition is easy and the consequence is where the marks are. Adding a direction to a quantity changes how it combines with another one of its kind. Two scalars of 3 and 4 add to 7 and there is nothing to discuss. Two vectors of magnitude 3 and 4 can produce a resultant anywhere from 1 to 7, and the answer is fixed by the angle between them rather than by their sizes. Every vector mistake worth making is a version of forgetting that.

The CED handles the representation side too. EK 1.1.A.2 says vectors can be visually modeled as arrows with appropriate direction and lengths proportional to their magnitude, and EK 1.1.A.3.i says vectors are notated with an arrow above the symbol for that quantity. Then EK 1.1.A.3.ii adds the exemption that matters in practice: vector notation is not required for vector components along an axis, and in one dimension the sign of the component completely describes the direction of that component.

So in one dimension a vector looks like a signed number, which is convenient and is also how the distinction hides. A negative sign on a one-dimensional velocity is a direction. A negative sign on a work value is not. Sorting out which minus means what is the middle of this page.

Each side on its own sits in the scalar and vector glossary entries. The pairs that inherit this distinction have their own pages: speed vs velocity, distance vs displacement and mass vs weight.

Scalar vs vector, side by side

	Scalar	Vector
Fully specified by	A magnitude and a unit	A magnitude, a unit and a direction
CED definition	EK 1.1.A.1, magnitude only	EK 1.1.A.1, magnitude and direction
Written as	$m$ , $K$ , $T$ , plain symbols	$\vec{v}$ , $\vec{F}$ , with an arrow, from EK 1.1.A.3.i
In one dimension	A number with a unit	A signed number, where the sign is the direction, from EK 1.1.A.3.ii
Drawn as	A number on a diagram	An arrow whose length is proportional to the magnitude, from EK 1.1.A.2
Adding two of them	Ordinary arithmetic, $3 + 4 = 7$	Head to tail, $3$ and $4$ give anything from $1$ to $7$
Order of addition	Irrelevant	Irrelevant, but the angles are not
Components	It has none	Resolvable into perpendicular components, from EK 1.5.A.1
What a minus sign means	Below a reference, or leaving the system, or a type of charge	The opposite direction along the chosen axis
Multiplied by a scalar	Another scalar	A vector, same direction, scaled magnitude, reversed if the scalar is negative
Its magnitude	Itself, or its absolute value	A scalar, written $\lvert \vec{v} \rvert$ or just $v$
Can be zero with nonzero parts	Not meaningfully	Yes, two equal and opposite vectors sum to zero
AP Physics 1 examples the CED names	Distance, speed, kinetic energy, potential energy, pressure, work	Position, displacement, velocity, acceleration, force, momentum, impulse

The row about a nonzero-part zero is worth pausing on, because it has no scalar analogue. Two forces of 40 N each, pointing opposite ways, sum to a net force of exactly zero, and by EK 2.4.A.3 the system's velocity then stays constant. Nothing about the individual 40 N values went away. A vector sum can destroy information that a scalar sum cannot, and that is why a free-body diagram lists the forces separately before anything is added.

The classification, as the CED actually gives it

The CED does not print one master table of scalars and vectors. It labels quantities as it introduces them, so the classification is scattered across the units. Collected, and quoting the statements that use the words:

Called scalar quantities:

EK 1.1.A.3: distance and speed are examples of scalar quantities.
EK 3.1.A.2: translational kinetic energy is a scalar quantity.
EK 3.2.A.2: work is a scalar quantity that may be positive, negative, or zero.
EK 3.3.A.2: potential energy is a scalar quantity associated with the position of objects within a system.
EK 6.1.A.3: rotational kinetic energy is a scalar quantity.
EK 8.2.A.2: pressure is a scalar quantity.

Called vector quantities:

EK 1.1.A.3: position, displacement, velocity, and acceleration are examples of vector quantities.
EK 2.2.A.1: forces are vector quantities that describe the interactions between objects or systems.
EK 4.1.A.2: momentum is a vector quantity and has the same direction as the velocity.
EK 4.2.A.3: impulse is a vector quantity and has the same direction as the net force exerted on the system.

Two entries in that list repay a second look.

Pressure is a scalar, and it is built from a vector. EK 8.2.A.1 defines pressure as the magnitude of the perpendicular force component exerted per unit area, $P = F_\perp / A$ . The force is a vector; taking its magnitude threw the direction away; what came out is a scalar. So "it was made from a force" is not a reason for something to be a vector.

Work is a scalar that can be negative. EK 3.2.A.2 says so in the same breath as calling it a scalar, and $W = F_{\parallel} d = F d \cos\theta$ is built from two vectors and returns a plain number. That negative sign carries no direction at all; it says energy left the system. Section five below separates the meanings of a minus sign.

One quantity is worth naming for what the CED does not say. Mass is not labelled a scalar anywhere in the AP Physics 1 course framework, and neither is temperature. They are scalars, and no exam question will suggest otherwise, but if you are reciting the CED rather than physics, know which is which.

The case that separates them: addition

Take two quantities of size 3 and 4 in the same units and add them. If they are scalars there is exactly one answer. If they are vectors there are infinitely many, and the angle picks.

The rule is the law of cosines, with $\theta$ the angle between the two vectors:

\lvert \vec{A} + \vec{B} \rvert = \sqrt{A^2 + B^2 + 2 A B \cos\theta}

Run it across the range:

Angle between them	Resultant of a 3 and a 4	What it looks like
$0^\circ$ , same direction	$7.00$	The scalar answer, and the maximum
$60^\circ$	$6.08$	Mostly cooperating
$90^\circ$ , perpendicular	$5.00$	Pythagoras, the 3-4-5 case
$120^\circ$	$3.61$	Mostly fighting
$180^\circ$ , opposite	$1.00$	The minimum

Add two masses of 3.0 kg and 4.0 kg instead and the answer is 7.0 kg, once, whatever you do with the objects. That single contrast is the whole reason vectors get their own topic.

Two boundaries fall out of the table and both are worth carrying into an exam:

The resultant of two vectors can never exceed the sum of their magnitudes, and can never fall below the difference of their magnitudes. For a 3 and a 4 that means $1 \leq \lvert \vec{A} + \vec{B} \rvert \leq 7$ . If a resultant lands outside that window, the arithmetic is wrong.
The perpendicular case, $90^\circ$ , is the one that gives $\sqrt{3^2 + 4^2}$ , and it is the case the AP course is built around. EK 1.5.A.1 says vectors can be mathematically modeled as the resultant of two perpendicular components, and EK 1.5.A.3 gives the trigonometry for splitting them, listing $\sin\theta = a/c$ , $\cos\theta = b/c$ , $\tan\theta = a/b$ and $a^2 + b^2 = c^2$ as the relevant equations.

The practical method is therefore always the same: resolve every vector into perpendicular components, add the components as signed scalars along each axis, then reassemble. EK 1.5.B.1 states it for motion, saying two-dimensional motion can be analyzed using one-dimensional kinematic relationships if the motion is separated into components. How to find net force walks the routine.

What a minus sign means, which is not always a direction

This is the error that survives longest, because in one dimension a vector genuinely does look like a signed number, and the habit generalises where it should not.

On a vector component, a minus sign is a direction. EK 1.1.B.1 says that when determining a vector sum in a given one-dimensional coordinate system, opposite directions are denoted by opposite signs. So $v_x = -12\ \text{m/s}$ means 12 m/s in whichever direction you declared negative, and $F_x = -8\ \text{N}$ means 8 N the other way. The magnitude is 12 and 8; the sign is geometry.

On a scalar, a minus sign means one of several other things, and you have to know which.

Negative work. By EK 3.2.A.2 work may be positive, negative or zero, and negative work means energy was transferred out of the system by that force. Friction on a sliding block does negative work. Nothing points anywhere.
Negative gravitational potential energy. $U_G = -G m_1 m_2 / r$ is negative for every finite separation, because EK 3.3.A.3 says the definition of zero potential energy for a given system is a decision made by the observer, and this choice puts the zero at infinite separation. The minus sign records where the zero was placed.
A negative change. $\Delta U_g = mg\Delta y$ comes out negative when an object descends. That is a decrease in a scalar, not a downward-pointing energy.
Negative charge. In Units 10 onward, the sign of a charge is a type rather than a direction or a size.

And two things a minus sign never means: a negative magnitude, since magnitudes are absolute values and cannot be negative, and a negative speed, distance, mass, or kinetic energy, since each of those is a magnitude or is built from squares. $K = \frac{1}{2}mv^2$ cannot come out negative because $v$ is squared, so a negative kinetic energy is always an arithmetic slip.

The test that resolves any case: ask whether flipping your coordinate axis would flip the sign. Reverse the positive direction and every vector component changes sign. Reverse it and the work done by friction is still negative, the mass is still positive, and the kinetic energy is unchanged. Signs that survive an axis flip are not directions.

Notation, and what changes in AP Physics C

The notation is not decoration; on the exam it is part of the answer, and the two courses ask for different amounts of it.

In AP Physics 1, Topic 1.1 is titled Scalars and Vectors in One Dimension, and the one-dimensional restriction is real. EK 1.1.B is describe a vector sum in one dimension, and full two-dimensional work waits for Topic 1.5, Vectors and Motion in Two Dimensions, which is about resolving a vector into perpendicular components rather than about a general vector algebra. There is no dot product and no cross product in the AP Physics 1 course framework, and none on its equation sheet.

On the AP Physics 1 equation sheet, every vector arrow sits in the force, momentum and centre-of-mass lines of the translational group. The sheet prints $\vec{p} = m\vec{v}$ , $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m \frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ , $\vec{J} = \vec{F}_{\text{avg}} \Delta t = \Delta \vec{p}$ , $\vec{F}_s = -k\Delta \vec{x}$ , $\vec{a}_{\text{sys}}$ , $\vec{x}_{\text{cm}}$ , $\vec{v}_{\text{cm}}$ and the magnitudes $\lvert \vec{F}_g \rvert$ and $\lvert \vec{F}_f \rvert$ . The three kinematic equations carry no arrows, because they are written for components along an axis, exactly as EK 1.1.A.3.ii permits. The whole rotational and fluids group carries no arrows either: torque appears as $\tau = r_{\perp} F = r F \sin\theta$ and angular momentum as $L = I\omega$ , both as magnitudes.

In AP Physics C: Mechanics, the same two quantities are printed as vector products. That sheet prints $\vec{\tau} = \vec{r} \times \vec{F}$ and $\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$ , and it carries a separate Vectors table listing $\vec{A} \cdot \vec{B} = AB\cos\theta$ , $\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta$ , $\vec{r} = (A\hat{i} + B\hat{j} + C\hat{k})$ , $\vec{C} = \vec{A} + \vec{B}$ and $\vec{C} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$ . Its Topic 1.1 is titled simply Scalars and Vectors, with no one-dimensional qualifier, and EK 1.1.A.4 there adds that vectors can be expressed in unit vector notation or as a magnitude and a direction. Work also changes shape: AP Physics 1 prints $W = F_{\parallel} d = F d \cos\theta$ while C: Mechanics prints $W = \int_a^b \vec{F} \cdot d\vec{r}$ , the same scalar built from vectors, now with the dot product doing the projecting.

So the pattern across the two courses: the scalar-and-vector distinction is identical, and the machinery for handling vectors is not. AP Physics 1 handles directions with signs and perpendicular components. C: Mechanics adds unit vectors, dot products and cross products. The C: Mechanics sheet and the AP Physics 1 sheet are side by side if you want to compare them line for line.

Where it costs a mark

Adding vector magnitudes. Two 5 N forces do not make a 10 N net force unless they point the same way. This is the single most expensive habit on the list, because it produces a plausible number.
Giving a vector answer with no direction. A force, a velocity, an acceleration, a momentum or an impulse needs a direction, and in one dimension a sign counts only once the axis has been declared.
Not declaring the axis, or flipping it partway. EK 1.1.B.1 makes signs meaningful only relative to a chosen coordinate system. Pick one, write it down, hold it to the end.
Reporting a negative magnitude, speed, distance or kinetic energy. All are magnitudes or are built from squares. If a minus appears, it belongs to a component, not to the magnitude.
Treating a negative work as a direction. EK 3.2.A.2 puts work among the scalars. Negative work means energy left the system.
Assuming a quantity is a vector because it was built from vectors. Work comes from a force and a displacement and is a scalar. Pressure comes from a force and an area and is a scalar by EK 8.2.A.2. Kinetic energy comes from a velocity and is a scalar by EK 3.1.A.2.
Drawing components as extra force arrows on a free-body diagram. EK 2.2.B.3 says forces exerted on an object are represented as vectors originating from the representation of the center of mass, such as a dot. Draw each force once, then resolve it in the algebra rather than on the diagram. The rules are in how to draw a free-body diagram.
Losing a component to a cosine that should have been a sine. EK 1.5.A.3 gives $\sin\theta = a/c$ and $\cos\theta = b/c$ against a labelled right triangle, and which one you want depends on which side of the angle your component sits. Sketch the triangle every time; it costs five seconds.
Concluding that zero net force means no forces. A vector sum of zero is compatible with large individual vectors, which is what translational equilibrium is: EK 2.4.A.2 calls it a configuration of forces such that the net force exerted on a system is zero.

A 3 and a 4, added five ways

Two forces act on an object at a point, one of magnitude $3.0\ \text{N}$ and one of magnitude $4.0\ \text{N}$ . Find the magnitude of the net force when the angle between them is $0^\circ$ , $60^\circ$ , $90^\circ$ , $120^\circ$ and $180^\circ$ . Then find the total mass of a 3.0 kg block and a 4.0 kg block for comparison. The object has a mass of $2.0\ \text{kg}$ ; find its acceleration in the perpendicular case.

Use the law of cosines for a vector sum, with $\theta$ the angle between the two vectors: $\lvert \vec{F}_{\text{net}} \rvert = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ . Here $A^2 + B^2 = 9 + 16 = 25$ and $2AB = 24$ in every row, so only the cosine changes.
$\theta = 0^\circ$ : $\cos 0^\circ = 1$ , so $\sqrt{25 + 24} = \sqrt{49} = 7.00\ \text{N}$ . This is the largest value available and it is also the answer scalars would give.
$\theta = 60^\circ$ : $\cos 60^\circ = 0.5$ , so $\sqrt{25 + 12} = \sqrt{37} = 6.08\ \text{N}$ .
$\theta = 90^\circ$ : $\cos 90^\circ = 0$ , so $\sqrt{25 + 0} = \sqrt{25} = 5.00\ \text{N}$ . This is the 3-4-5 triangle, and it is the case EK 1.5.A.1 builds the course on.
$\theta = 120^\circ$ : $\cos 120^\circ = -0.5$ , so $\sqrt{25 - 12} = \sqrt{13} = 3.61\ \text{N}$ .
$\theta = 180^\circ$ : $\cos 180^\circ = -1$ , so $\sqrt{25 - 24} = \sqrt{1} = 1.00\ \text{N}$ , the difference of the magnitudes, and the smallest value available.
Now the scalars. A 3.0 kg block and a 4.0 kg block have a combined mass of $7.0\ \text{kg}$ . There is no angle to specify, no range of possible answers, and no second question to ask. Same two numbers, one answer instead of a continuum.
The acceleration in the perpendicular case: $\vec{a}_{\text{sys}} = \vec{F}_{\text{net}} / m_{\text{sys}} = 5.00\ \text{N} / 2.0\ \text{kg} = 2.5\ \text{m/s}^2$ , and it points along the resultant, at $\arctan(3.0/4.0) = 36.9^\circ$ from the 4.0 N force. A magnitude alone would not have answered the question.
Check the bounds. Every result sits between $\lvert 4.0 - 3.0 \rvert = 1.0\ \text{N}$ and $4.0 + 3.0 = 7.0\ \text{N}$ , and both bounds were actually reached, at $180^\circ$ and $0^\circ$ .

Net force magnitudes 7.00 N, 6.08 N, 5.00 N, 3.61 N, 1.00 N at $0^\circ$ , $60^\circ$ , $90^\circ$ , $120^\circ$ and $180^\circ$ . The two masses total 7.0 kg with no angle involved. In the perpendicular case the acceleration is 2.5 m/s $^2$ at $36.9^\circ$ from the larger force. The same pair of numbers gives one scalar answer and a range of five vector answers, which is the entire distinction made arithmetic.

Two vectors in, one scalar out

A crate is pulled $8.0\ \text{m}$ horizontally across a floor. Find the work done on it by a $50\ \text{N}$ force in three cases: directed $30^\circ$ above the direction of motion, directed perpendicular to the motion, and directed $150^\circ$ from the direction of motion. Then find the total work done by the first and third forces if both act over the same displacement.

The relevant printed relation is $W = F_{\parallel} d = F d \cos\theta$ , where $\theta$ is the angle between the force and the displacement of the point of application. Both inputs are vectors; the output, by EK 3.2.A.2, is a scalar that may be positive, negative or zero.
Common factor first: $Fd = (50\ \text{N})(8.0\ \text{m}) = 400\ \text{J}$ , so each case is $400\cos\theta$ .
$\theta = 30^\circ$ : $\cos 30^\circ = 0.8660$ , so $W = (400)(0.8660) = 346\ \text{J}$ . Positive, so energy went into the system.
$\theta = 90^\circ$ : $\cos 90^\circ = 0$ , so $W = 0$ . Both vectors are nonzero and the scalar they produce is zero. EK 3.2.A.3.ii is the reason: a force component perpendicular to the displacement of the center of mass can change the direction of the motion without changing the kinetic energy.
$\theta = 150^\circ$ : $\cos 150^\circ = -0.8660$ , so $W = -346\ \text{J}$ . Negative, so energy left the system. That minus sign is not a direction. Flip your coordinate axis and it stays negative, which is the test from the section above.
Total work from the first and third forces together: work is a scalar, so it adds arithmetically. $346 + (-346) = 0\ \text{J}$ , and by the work-energy theorem in EK 3.2.A.4 the crate's kinetic energy is unchanged.
Compare with what the two forces do as vectors. A 50 N force at $30^\circ$ and a 50 N force at $150^\circ$ have a vector sum of magnitude $\sqrt{2500 + 2500 + 2(2500)\cos 120^\circ} = \sqrt{5000 - 2500} = 50\ \text{N}$ , pointing perpendicular to the motion. So the two forces have a nonzero vector sum and a zero scalar sum of works, in the same situation, at the same time.
That last line is the point of the example. The vector question and the scalar question have different answers, and neither is the other one's check.

Work +346 J at $30^\circ$ , 0 J at $90^\circ$ , and -346 J at $150^\circ$ . Together the first and third do 0 J of total work, while their vector sum is a nonzero 50 N force perpendicular to the motion. Work is a scalar even though it is assembled from two vectors, and its sign records the direction of energy flow rather than a direction in space.

Signs in one dimension, and what happens when you flip the axis

Three horizontal forces act on a $2.0\ \text{kg}$ object on a frictionless surface: $12\ \text{N}$ to the right, $5.0\ \text{N}$ to the left, and $3.0\ \text{N}$ to the left. Find the net force and the acceleration, first taking right as positive and then taking left as positive, and compare the answers with the sum of the three magnitudes.

Convention one: right is positive. Then the three forces are $+12\ \text{N}$ , $-5.0\ \text{N}$ and $-3.0\ \text{N}$ , following EK 1.1.B.1, which says opposite directions are denoted by opposite signs.
Net force: $F_{\text{net},x} = +12 - 5.0 - 3.0 = +4.0\ \text{N}$ , so 4.0 N to the right. By EK 2.4.A.1 the net force is the vector sum of all forces exerted on the system, which in one dimension is this signed sum.
Acceleration: $a_x = F_{\text{net},x} / m = +4.0\ \text{N} / 2.0\ \text{kg} = +2.0\ \text{m/s}^2$ , so $2.0\ \text{m/s}^2$ to the right.
Convention two: left is positive. Now the same three forces are $-12\ \text{N}$ , $+5.0\ \text{N}$ and $+3.0\ \text{N}$ . Every sign flipped, because the axis did.
Net force in convention two: $-12 + 5.0 + 3.0 = -4.0\ \text{N}$ , and the acceleration is $-2.0\ \text{m/s}^2$ . Both numbers changed sign and the physical answer did not: negative in a left-positive system still means 4.0 N and $2.0\ \text{m/s}^2$ to the right.
Now the wrong calculation, for contrast. Adding the three magnitudes gives $12 + 5.0 + 3.0 = 20\ \text{N}$ , which would make the acceleration $10\ \text{m/s}^2$ . That is five times too large, and it is what treating the forces as scalars produces.
One more check on what survives the axis flip. The magnitude of the net force, $4.0\ \text{N}$ , is the same in both conventions, as is the object's mass, and so is any kinetic energy it acquires: after $3.0\ \text{s}$ from rest, $v = at = 6.0\ \text{m/s}$ in magnitude and $K = \frac{1}{2}(2.0)(6.0)^2 = 36\ \text{J}$ in either convention. Vector components depend on the axis you chose. Scalars do not.

Net force 4.0 N to the right, acceleration 2.0 m/s $^2$ to the right, written as $+4.0\ \text{N}$ and $+2.0\ \text{m/s}^2$ with right positive and as $-4.0\ \text{N}$ and $-2.0\ \text{m/s}^2$ with left positive. Adding the three magnitudes gives 20 N and an acceleration five times too large. The magnitude 4.0 N and the kinetic energy of 36 J after 3.0 s are the same in both conventions, which is the practical test for whether a sign is a direction.

Frequently asked questions

What is the difference between a scalar and a vector?

A scalar is fully described by a magnitude and a unit. A vector needs a magnitude, a unit and a direction. The AP Physics 1 CED states this at essential knowledge 1.1.A.1: scalars are quantities described by magnitude only, while vectors are quantities described by both magnitude and direction. The consequence that matters in calculations is that they combine differently. Scalars add by ordinary arithmetic, so three plus four is always seven. Vectors add head to tail, so a vector of magnitude three and a vector of magnitude four can produce a resultant anywhere between one and seven, with the angle between them deciding.

Which quantities in AP Physics 1 are vectors?

The CED labels quantities as it introduces them. It names position, displacement, velocity and acceleration as vectors at essential knowledge 1.1.A.3; forces at essential knowledge 2.2.A.1; momentum at essential knowledge 4.1.A.2, adding that it has the same direction as the velocity; and impulse at essential knowledge 4.2.A.3, adding that it has the same direction as the net force. On the scalar side it names distance and speed at 1.1.A.3, translational kinetic energy at 3.1.A.2, work at 3.2.A.2, potential energy at 3.3.A.2, rotational kinetic energy at 6.1.A.3 and pressure at 8.2.A.2. Mass, time and temperature are scalars although the course framework does not use the word for them.

Is a negative scalar just a vector pointing backwards?

No, and this is the most persistent version of the confusion. On a vector component a minus sign is a direction, because essential knowledge 1.1.B.1 says opposite directions are denoted by opposite signs in a chosen one-dimensional coordinate system. On a scalar a minus sign means something else entirely: negative work means energy left the system, a negative change in potential energy means that energy decreased, and a negative charge is a type rather than a direction. The test is to imagine reversing your positive axis. Every vector component changes sign; a negative work stays negative, so it was never a direction.

Why is work a scalar if force and displacement are both vectors?

Because the operation that combines them throws the direction away. Work is the force multiplied by the displacement multiplied by the cosine of the angle between them, and the result is a single number with no direction left in it. The AP Physics 1 CED states outright at essential knowledge 3.2.A.2 that work is a scalar quantity that may be positive, negative or zero. Pressure is the same story: essential knowledge 8.2.A.1 defines it as the magnitude of the perpendicular force component per unit area, and 8.2.A.2 calls it a scalar. So being built from vectors does not make a quantity a vector.

Do I need to draw arrows over vector symbols on the AP exam?

Not for components along an axis. Essential knowledge 1.1.A.3.i says vectors are notated with an arrow above the symbol, but 1.1.A.3.ii then states that vector notation is not required for vector components along an axis, because in one dimension the sign of the component completely describes its direction. The AP Physics 1 equation sheet follows the same practice: it prints arrows on the momentum, impulse, spring force and net force lines, and none on the three kinematic equations, which are written for components. What you do have to supply is a direction for every vector answer, whether by a sign against a stated axis or in words.

Can two vectors add to give zero?

Yes, whenever they are equal in magnitude and opposite in direction. This has no scalar equivalent: two nonzero masses can never sum to zero. It is also the basis of equilibrium in mechanics, since essential knowledge 2.4.A.2 defines translational equilibrium as a configuration of forces such that the net force exerted on a system is zero, and essential knowledge 2.4.A.3 says Newton's first law then keeps the system's velocity constant. Note what has not happened: the individual forces are still acting and still belong on a free-body diagram. A zero vector sum hides its parts, which is why you list the forces before you add them.

How does AP Physics C treat vectors differently from AP Physics 1?

The definition is identical and the machinery is not. AP Physics 1 titles its topic Scalars and Vectors in One Dimension and handles directions using signs and perpendicular components, with no dot or cross product anywhere in its course framework or on its equation sheet. AP Physics C: Mechanics titles the same topic Scalars and Vectors, adds unit vector notation with i, j and k at essential knowledge 1.1.A.4, and its equation sheet carries a Vectors table with the dot product as AB cosine theta and the magnitude of the cross product as AB sine theta. That sheet also prints torque as r cross F and angular momentum as r cross p, where the AP Physics 1 sheet prints both as magnitudes.