Scalars and Vectors

Walk 3 m east, then 4 m north. Your feet covered 7 m, but you are only 5 m from where you started, and "5 m" alone doesn't say where you are: you need "5 m, 53° north of east". The first number is a scalar and the second is a vector, and nearly every quantity in mechanics is one or the other.

Magnitude only, or magnitude and direction

A scalar is a quantity that a single number with a unit describes completely: mass ($2.0\ \text{kg}$), time ($3.5\ \text{s}$), temperature, energy. A vector needs a size, called its magnitude, and a direction. Pushing a stalled car with $300\ \text{N}$ forward and pushing it with $300\ \text{N}$ backward have the same magnitude and opposite effects, so force must be a vector.

The kinematics quantities split like this:

Speed is the magnitude of velocity. Distance is not the magnitude of displacement, because distance counts every back-and-forth while displacement only compares the start and finish. In the opening walk, distance is $7\ \text{m}$ and the magnitude of the displacement is $5\ \text{m}$.

You draw a vector as an arrow: it points in the vector's direction, and its length is proportional to its magnitude, using whatever scale the diagram states. A velocity arrow twice as long as another means twice the speed. In print, a vector gets an arrow over its symbol, $\vec{A}$, and its magnitude is written $A$ or $|\vec{A}|$. A magnitude is never negative.

In one dimension, there are only two directions, so a sign does the job of the arrow. Pick a positive direction (say, right or up); then the opposite direction is negative. A car that moves from $x = +5\ \text{m}$ to $x = -3\ \text{m}$ has displacement $\Delta x = -3 - 5 = -8\ \text{m}$: $8\ \text{m}$ in the negative direction. The minus sign is direction, not a "smaller" quantity. Reverse your choice of positive direction and every sign flips, but the physics is unchanged.

Components and unit-vector notation

In two or three dimensions, you describe a vector by how far it reaches along each axis. Those numbers are its components. To name the directions, use the unit vectors $\hat{i}$, $\hat{j}$ and $\hat{k}$: arrows of magnitude 1 (no units) pointing along $+x$, $+y$ and $+z$. Any vector is the sum of three vectors parallel to the axes:

The components $A_x$, $A_y$, $A_z$ carry the units and the signs. A displacement $(-3.0\,\hat{i} + 4.0\,\hat{j})\ \text{m}$ means $3.0\ \text{m}$ in the $-x$ direction and $4.0\ \text{m}$ in the $+y$ direction.

Going between the two descriptions is right-triangle trigonometry. If $\theta$ is the angle measured counterclockwise from the $+x$ axis to the vector, the projections onto the axes, and the way back, are:

For the vector in the figure, $A = \sqrt{(-3.0)^2 + (4.0)^2} = 5.0$. A calculator's inverse tangent gives $\tan^{-1}\left(\frac{4.0}{-3.0}\right) = -53.1^\circ$, which points into the fourth quadrant, the wrong one. The inverse tangent can't tell $\frac{4}{-3}$ from $\frac{-4}{3}$, so it always answers between $-90^\circ$ and $90^\circ$. When $A_x$ is negative, add $180^\circ$: $\theta = 126.9^\circ$, or equivalently $53.1^\circ$ above the $-x$ axis.

The position vector $\vec{r}$ points from the origin to the object: $\vec{r} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k}$. Its magnitude is the object's distance from the origin, and dividing $\vec{r}$ by that magnitude gives the unit vector $\hat{r} = \frac{\vec{r}}{|\vec{r}|}$, which has length 1 and points the same way. For a particle at $\vec{r} = (2.0\,\hat{i} - 2.0\,\hat{j} + 1.0\,\hat{k})\ \text{m}$, the distance is $\sqrt{4 + 4 + 1} = 3.0\ \text{m}$ and

You will use $\hat{r}$ whenever a quantity points straight away from or toward a center, as gravity does in topic 2.6 (it points along $-\hat{r}$, toward the center).

Adding vectors by components

The resultant of several vectors is their vector sum. Geometrically, you place them tip to tail, and the resultant runs from the first tail to the last tip. Algebraically, you add the components axis by axis:

This works because the $\hat{i}$ parts of the arrows lie along one line, where adding is ordinary 1D addition with signs, and the same is true for $\hat{j}$ and $\hat{k}$ separately. What you can't do is add magnitudes. Two displacements of $3\ \text{m}$ and $4\ \text{m}$ have a resultant anywhere from $1\ \text{m}$ (opposite directions) to $7\ \text{m}$ (same direction); only when they are perpendicular is it $5\ \text{m}$. In general, $|\vec{A} + \vec{B}| \le A + B$, with equality only for parallel vectors.

Worked examples

Where students lose points

On the exam

A vector answer needs both parts: unit-vector components with units, or a magnitude with a direction stated relative to an axis or a compass direction ("$8.72\ \text{km}$, $53.4^\circ$ south of west"). When a question asks for a "magnitude", give a positive number and no direction. Answers in one dimension may use a sign for direction, as long as the positive direction is clear from the question or from your work.