Scalars and vectors in one and two dimensions

Three moves underlie this topic. First, recognize whether a quantity is a scalar (magnitude only) or a vector (magnitude plus direction). Second, in 1D, read direction from the sign of a signed component. Third, in 2D, read direction from the unit vectors $\hat{\text{ı}}$ and $\hat{\text{ȷ}}$ (and $\hat{k}$ in 3D).

Every equation from Topic 1.2 onward depends on getting these distinctions right. If a scalar quietly picks up a direction, or a vector loses its unit vectors, the math after that point will run but produce wrong physics. The diagnostic at the end of this topic is built around exactly the places those errors leak in.

For students arriving from algebra-based Physics 1, the 1D material is familiar territory. The new piece is unit-vector notation in 2D, where direction lives in the unit vectors rather than in a $\pm$ sign. Spend the time on §4: the translation table there is the bridge from one notation to the other.

A scalar is a quantity described by magnitude only: a number and a unit, with no direction attached. A vector is described by both a magnitude AND a direction.

The canonical examples for Unit 1:

• Scalarsmass, time, distance, speed, temperature, kinetic energy.
• Vectorsposition, displacement, velocity, acceleration, force, momentum.

A vector is drawn as an arrow whose length is proportional to its magnitude and whose direction shows its direction. Twice the magnitude, twice the length. In symbols, vectors are written with an arrow ($\vec{v}$, $\vec{r}$, $\vec{a}$), and their magnitudes (which are scalars) are written with bars ($|\vec{v}|$, $|\vec{r}|$, $|\vec{a}|$). The magnitude of a velocity is a speed; the magnitude of a displacement is the straight-line distance from start to end, equal to the distance traveled only when the path has no reversal or turn.

Five pieces of notation worth fixing right now:

• $\vec{v}$ denotes the full velocity vector (magnitude AND direction together).
• $|\vec{v}|$ denotes the magnitude of $\vec{v}$, also called the speed. This is a scalar.
• $v_x$, $v_y$, $v_z$ (no arrow) denote the signed components along the $x$, $y$, $z$ axes. These are scalars; their signs carry the direction along each axis.
• $\hat{\text{ı}}$, $\hat{\text{ȷ}}$, $\hat{k}$ are unit vectors: vectors of magnitude $1$ along $+x$, $+y$, $+z$. They carry direction only.
• $\vec{v} = v_x\hat{\text{ı}} + v_y\hat{\text{ȷ}} + v_z\hat{k}$ assembles the full vector from its signed components and the unit vectors.

The component is a scalar with a sign. The unit vector is direction with magnitude 1. The product of the two is a piece of the vector. The sum of those pieces is the vector. Component, unit vector, and full vector are three different objects.

In 1D, there are only two directions to choose from, so a sign is enough to encode direction. Every problem starts the same way:

• Pick one direction and call it positive. The other direction is automatically negative.
• Every quantity in the problem (position, displacement, velocity, acceleration) carries a sign relative to that choice.
• Stick with the convention for the whole problem. Switching mid-stream is one of the most common ways to break the math.

Once a convention is fixed, the sign of a component tells you direction, and the magnitude (absolute value) tells you speed (or distance, or force magnitude, etc.). Concretely:

Two things that look related but aren't:

• The sign of $v_x$ is direction, not size. $v_x = -50$ m/s is faster than $v_x = +20$ m/s, because $|-50| > |+20|$.
• A negative displacement isn't a non-event. $\Delta x = -8$ m is a real displacement of $8$ m in the $-\hat{\text{ı}}$ direction. Negative isn't "stopped" or "impossible"; it's just "the other way."

Worked example. A cart starts at $x_0 = +4$ m, with east taken as $+\hat{\text{ı}}$. It rolls $9$ m west, then $3$ m east. What is its final position and its displacement?

Track signed values with the convention in hand. The first leg is $-9$ m (west is negative). The second is $+3$ m. Total displacement: $\Delta x = (-9) + (+3) = -6$ m. Final position: $x_f = x_0 + \Delta x = (+4) + (-6) = -2$ m, or $\vec{r}_f = -2\hat{\text{ı}}$ m. The cart ended $2$ m west of the origin, after a net displacement of $6$ m westward.

In 1D, sign IS direction. Magnitude is the absolute value of the signed component. Never compare signed values to decide which is larger in magnitude.

In 2D, a $\pm$ sign isn't enough to encode direction. Now there are infinitely many directions, and the language has to handle all of them at once. The trick is to break every vector into two pieces, one along each axis, and tag each piece with a unit vector.

A unit vector is a vector of magnitude $1$ pointing along an axis. The three standard ones:

• $\hat{\text{ı}}$ points along $+x$. Magnitude: $1$. Direction: $+x$.
• $\hat{\text{ȷ}}$ points along $+y$. Magnitude: $1$. Direction: $+y$.
• $\hat{k}$ points along $+z$. Magnitude: $1$. Direction: $+z$. (In Physics C: Mechanics we mostly stay in 2D, so $\hat{k}$ shows up sparingly.)

The unit vectors carry direction. A signed coefficient times a unit vector carries both magnitude and direction. Sum the pieces and you have the full vector:

$\vec{v} = v_x\hat{\text{ı}} + v_y\hat{\text{ȷ}}$

Here $v_x$ and $v_y$ are the signed components: $v_x = +3$ means the $x$-piece points along $+\hat{\text{ı}}$ at magnitude $3$; $v_y = -4$ means the $y$-piece points along $-\hat{\text{ȷ}}$ at magnitude $4$. The full vector $\vec{v} = 3\hat{\text{ı}} - 4\hat{\text{ȷ}}$ m/s points into the fourth quadrant (positive $x$, negative $y$), and its magnitude (the speed) is found by Pythagoras:

$|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2 + 4^2} = 5\text{ m/s}$

The magnitude formula is the most-broken step in this section. Two things to track. First, you square the components, so signs go away inside the radical. Second, you sum the squares, not the signed components themselves. Anyone who writes $|\vec{v}| = v_x + v_y$ is summing components, which gives a (signed) scalar but not a magnitude.

Translation: Physics 1 form → Physics C form. If you've seen this material in algebra-based form, every Physics 1 statement has a Physics C counterpart. The convention here: $+\hat{\text{ı}}$ is east, $+\hat{\text{ȷ}}$ is north.

The two forms hold the same physics. The Physics 1 form names directions in words; the Physics C form names them by which unit vector they multiply. Once a convention is fixed ($+\hat{\text{ı}}$ = east, etc.), the translation is mechanical.

Worked example. A drone's velocity is $\vec{v} = 6\hat{\text{ı}} - 8\hat{\text{ȷ}}$ m/s. What is its speed? What is its direction relative to $+\hat{\text{ı}}$?

Speed is magnitude: $|\vec{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ m/s. The signs of the components vanish inside the squares.

Direction comes from the components. The $x$-piece is $+6$ (along $+\hat{\text{ı}}$), the $y$-piece is $-8$ (along $-\hat{\text{ȷ}}$). The drone is moving into the fourth quadrant. The angle below $+\hat{\text{ı}}$ is $\theta = \arctan(8/6) = 53°$.

In 2D, direction lives in the unit vectors. Components carry magnitude and sign; unit vectors carry direction; the full vector is the sum. Magnitude is found from the components by Pythagoras, never by summing them.

Two ways to add vectors. The geometric way is head-to-tail: place the tail of the second arrow at the head of the first, draw a resultant from the start of the first to the end of the second. The algebraic way is component-wise: add the $\hat{\text{ı}}$ parts, add the $\hat{\text{ȷ}}$ parts, done. The two methods give the same answer, and in Physics C the component method is almost always the faster path.

• To add $\vec{a} = a_x\hat{\text{ı}} + a_y\hat{\text{ȷ}}$ and $\vec{b} = b_x\hat{\text{ı}} + b_y\hat{\text{ȷ}}$, compute $\vec{a} + \vec{b} = (a_x + b_x)\hat{\text{ı}} + (a_y + b_y)\hat{\text{ȷ}}$.
• Components stay in their own column. The $\hat{\text{ı}}$ pieces don't talk to the $\hat{\text{ȷ}}$ pieces. There is no cross term.
• Magnitudes only sum directly when the two vectors are parallel and point the same way. In every other case, the sum is found component by component, and the magnitude of the sum is then found by Pythagoras on the new components.

Worked example. A hiker walks $\vec{r}_1 = 3\hat{\text{ı}} + 4\hat{\text{ȷ}}$ km, then $\vec{r}_2 = 5\hat{\text{ı}} - 1\hat{\text{ȷ}}$ km. What is the net displacement, its magnitude, and its direction?

Net displacement: add the components.

$\Delta\vec{r} = \vec{r}_1 + \vec{r}_2 = (3 + 5)\hat{\text{ı}} + (4 + (-1))\hat{\text{ȷ}} = 8\hat{\text{ı}} + 3\hat{\text{ȷ}}\text{ km}$

Magnitude: $|\Delta\vec{r}| = \sqrt{8^2 + 3^2} = \sqrt{64 + 9} = \sqrt{73} \approx 8.5$ km.

Direction relative to $+\hat{\text{ı}}$: $\theta = \arctan(3/8) \approx 21°$ north of east.

Reading the wrong move on this problem helps fix the right one. A student who reports $|\Delta\vec{r}| = 3 + 4 + 5 + (-1) = 11$ km has summed signed components and called the result a magnitude. A student who reports $|\Delta\vec{r}| = |\vec{r}_1| + |\vec{r}_2| = \sqrt{25} + \sqrt{26} \approx 10.1$ km has summed the two individual magnitudes, which only works if $\vec{r}_1$ and $\vec{r}_2$ point the same way. They don't. The right answer comes from component-wise addition followed by one Pythagoras at the end.

Add component-wise: $\hat{\text{ı}}$ parts together, $\hat{\text{ȷ}}$ parts together. Compute the magnitude of the sum at the end with Pythagoras. Never sum individual magnitudes unless the vectors are parallel and same-direction.

Each one collapses a distinction the notation exists to protect: scalar from vector, sign from size, component from magnitude.