Scalars and vectors

A scalar is a quantity determined by magnitude alone. Examples include mass, time, temperature and speed. A vector needs both magnitude and direction; displacement, velocity and force are vectors.

A vector can be shown geometrically as a directed line segment. The segment’s length represents magnitude, while the arrow gives its direction. Location is irrelevant, so equal vectors can be translated provided they aren’t rotated or resized.

Components and base vectors

In three dimensions, write

In two dimensions, leave out the third component.

A component is a signed scalar measuring how much of a vector acts along a chosen coordinate direction. A unit vector has magnitude $1$, so it carries direction without scale. The standard base vectors are $\boldsymbol{i}$, $\boldsymbol{j}$ and $\boldsymbol{k}$: unit vectors used to construct every vector in the coordinate system.

The zero vector $\boldsymbol{0}$ has zero in every component. Its magnitude is zero, and it has no unique direction. By contrast, $-\boldsymbol{v}$ has the same magnitude as $\boldsymbol{v}$ but points in the opposite direction.

Magnitude and vector arithmetic

For a three-dimensional vector,

This is Pythagoras extended into three dimensions.

Add or subtract vectors component by component. For a geometric approach, place them head-to-tail. The vector running from the first tail to the final head is the resultant, the single vector equal to their sum. Subtraction works by adding the opposite vector, so $\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v})$, where $\boldsymbol{u}$ is another vector in the same units and coordinate system as $\boldsymbol{v}$.

Multiplying by a scalar changes a vector’s scale. In $k\boldsymbol{v}$, where $k$ is a dimensionless scalar multiplier, the new magnitude is $|k||\boldsymbol{v}|$. If $k$ is positive, the direction stays the same; if it is negative, the direction reverses. When $k=0$, the result is $\boldsymbol{0}$. Non-zero vectors that are scalar multiples of each other are parallel.

Position vectors

A position vector runs from the origin to a point. If $A$ is a point, then

For points whose position vectors are $\boldsymbol{a}$ and $\boldsymbol{b}$,

Watch the order: destination minus starting point.

Rescaling and normalizing

To rescale a vector is to multiply it by a scalar, changing its magnitude by a chosen factor. To normalize a non-zero vector, divide it by its magnitude. This gives the unit vector

The zero vector can’t be normalized because division by zero is undefined.

If an object must travel at speed $s$ in the direction of a non-zero vector $\boldsymbol{d}$, use

Normalize first, then multiply by the required speed. Reversing these steps will usually give the wrong stated speed.

Vector addition and resultants support displacement and force calculations in physics. The same mathematics allows accurate tracking and targeting. This dual use raises an ethical question: reliable location knowledge may create obligations about how it is applied and whose benefit it serves.

Building the equation

A direction vector of a line is any non-zero vector parallel to that line. Suppose the line passes through a fixed point with position vector $\boldsymbol{a}$ and has direction vector $\boldsymbol{b}$. Its vector equation is

where $\boldsymbol{r}$ is the position vector of any point on the line in coordinate units. Here, $\boldsymbol{a}$ is the position vector of a fixed point on the line in the same units, $\boldsymbol{b}$ is a non-zero direction vector in coordinate units and $\lambda$ is a dimensionless real parameter.

As $\lambda$ takes every real value, $\boldsymbol{r}$ traces the whole line. When $\lambda=0$, the point is $\boldsymbol{a}$. Positive and negative values move in opposite directions. Any non-zero scalar multiple of $\boldsymbol{b}$ gives the same line, though the parameter values will change.

When the position vectors of two distinct points are known, their difference gives a suitable direction vector. So, for a line through points with position vectors $\boldsymbol{p}$ and $\boldsymbol{q}$, one choice is $\boldsymbol{q}-\boldsymbol{p}$, where $\boldsymbol{p}$ and $\boldsymbol{q}$ are position vectors in the same coordinate units.

Parametric form

Suppose

where $x_0$, $y_0$ and $z_0$ are the coordinates of a fixed point in coordinate units. The values $l$, $m$ and $n$ are the corresponding components of the direction vector, also in coordinate units. Equating components gives the parametric equations

where $x$, $y$ and $z$ are the coordinates of an arbitrary point on the line in coordinate units. In two dimensions, leave out the equation involving $z$.

To check whether a point lies on the line, substitute its coordinates into the component equations. One common value of $\lambda$ must satisfy every equation. If different components give different values, the point isn’t on the line.

Symbolic vector equations are often easier to work with than perspective drawings of three-dimensional lines. Algebra keeps incidence and direction exact, even when a sketch distorts them. This raises a broader question about mathematical knowledge: if symbols give us access to spaces that are difficult, or impossible, to visualize, does that still count as genuine understanding?

Constant-velocity motion

Kinematics describes motion mathematically using quantities such as position, velocity, time and distance. In two or three dimensions, constant-velocity motion follows

This vector equation describes a line, with time as the parameter. Only physically relevant times are used, usually $t\geq 0$.

The object follows a straight-line path parallel to $\boldsymbol{v}$. Its speed is $|\boldsymbol{v}|$ in $\text{m s}^{-1}$. After a non-negative time $t$, it has travelled a distance $|\boldsymbol{v}|t$ in metres.

To determine whether two objects meet, set their position vectors equal at the same time, then solve all component equations simultaneously. Their geometric paths may intersect without a collision. For a collision, both objects must be at the intersection point at one common time.

Relative position and closest approach

The relative position of $B$ from $A$ is the displacement vector directed from $A$ to $B$:

The magnitude of this vector gives their separation.

For calculations, define

Minimizing $D(t)^2$ is normally easier than minimizing $D(t)$: both reach their minimum at the same allowable time, but squaring avoids the square root. Differentiate the scalar function, solve for a stationary time and check that it belongs to the stated domain. If time is restricted to an interval, compare any valid stationary points with the endpoints.

Variable velocity in two dimensions

For variable motion, velocity becomes a vector-valued function of time. One possible model is

The horizontal velocity stays constant, while the vertical velocity changes linearly.

Differentiate each position component to get the corresponding velocity component. To recover position, integrate the velocity components and include an integration constant in each one. Initial-position information fixes these constants. This connects vector kinematics directly to calculus.

Projectile motion is variable two-dimensional motion in which the horizontal velocity is constant while the vertical velocity changes uniformly under gravity, if air resistance is neglected. Circular motion follows a circle, with the velocity direction changing continuously. Even when speed is constant, velocity isn’t constant.

A time shift replaces a function by

When $h>0$, the same motion takes place $h$ seconds later. This is the vector form of a phase or horizontal translation: the sign inside the brackets acts opposite to the apparent direction of the shift.

Scalar product

The scalar product takes two vectors and returns a scalar that measures how closely their directions align. For

calculate the scalar product using

Its units are the product of the units of the two vectors. The geometric form is

Rearranging gives

The ratio entered into $\cos^{-1}$ is dimensionless. Check that the calculator is in the required angle mode.

A positive value of $\boldsymbol{u}\cdot\boldsymbol{w}$ gives an acute angle, while a negative value gives an obtuse angle. Two non-zero vectors are perpendicular exactly when

A line has no fixed arrow direction. Its direction vectors may therefore give an obtuse angle even when the acute angle between the lines is required. If this happens, take the supplement or use the absolute value of the scalar product before applying inverse cosine.

Vector product

The vector product, also called the cross product, takes two three-dimensional vectors and returns a vector perpendicular to both. Calculate it from

Geometrically,

Its direction comes from the right-hand screw rule. Curl the fingers of the right hand from the first vector towards the second through the smaller angle; the thumb points in the direction of the product. Order matters. Reversing the vectors reverses the answer.

Only calculation and geometric interpretation are required here. Generalized cross-product identities and their proofs are not.

Areas from the vector product

The magnitude

is the area, in square units, of the parallelogram spanned by the two vectors. It is base times perpendicular height. The corresponding triangle has area

When the cross product is the zero vector, the vectors are parallel and the area is zero.

Parallel and perpendicular scalar components

The scalar component of $\boldsymbol{u}$ in the direction of $\boldsymbol{w}$ is the signed length of $\boldsymbol{u}$ resolved along $\boldsymbol{w}$:

It can be negative because its sign records orientation relative to $\boldsymbol{w}$.

Within the plane of the two vectors, the magnitude of the component perpendicular to $\boldsymbol{w}$ is

Applications and interpretation

In computer graphics, the scalar product can compare a surface normal with the direction of incoming light. Stronger alignment may correspond to greater illumination. It also helps with perspective calculations when three-dimensional information is projected onto a two-dimensional viewing plane.

For mechanics, the work done by a constant force is

Only the component of force parallel to the displacement contributes to work. Resolving forces into components is also useful when analysing loads and strains in structures.

Crossing a lever-arm vector with a force vector gives the torque magnitude, while the product's direction gives the axis and sense of rotation. Cross products and hand rules also appear in models of magnetic forces and fields.

Understanding these operations goes beyond obtaining the correct numerical answer. You also need to recognize the type of quantity returned, interpret its sign or direction and check that the result matches the geometry or physical situation.