Vectors in space

Points and axes in space

To locate a point in space, fix an origin $O$ and three mutually perpendicular axes that pass through it: the $x$-axis, the $y$-axis and the $z$-axis. Each point $P$ is described by a triple of numbers, its coordinates $P(x,y,z)$: how far you must move along each axis to reach it from $O$. The set of all points is written ${\mathbb{R}}^3$.

Fixed vector and free vector

Two points $A$ and $B$ determine an arrow that goes from $A$ (the origin) to $B$ (the end point). That arrow is the fixed vector $\overrightarrow{AB}$. It is characterised by three pieces of information:

• its magnitude, which is the length of the segment $AB$ and is written $|\overrightarrow{AB}|$;
• its direction, that of the line $AB$ (or any line parallel to it);
• its sense, which goes from the origin to the end point.

The reverse path, $\overrightarrow{BA}$, is the opposite vector, written $-\overrightarrow{AB}$. If origin and end point coincide you get the zero vector, $\overrightarrow{AA}=0$.

Two fixed vectors are equal when they have the same magnitude, the same direction and the same sense, even if they are drawn in different places. Grouping all the fixed vectors that are equal to each other gives a free vector: a displacement with no particular starting point, named with a letter in bold, $\mathbf{u}$, and which can be drawn starting at any point you like. The one that starts at the origin $O$ and ends at a point $P$ is the position vector of $P$, $\overrightarrow{OP}$.

Components of a vector

A free vector is described by a column of three numbers, its components:

$$\mathbf{u}=\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right)$$

Each component is the displacement along one axis: positive if it goes in the direction of the axis and negative if it goes the opposite way.

To go from $A$ to $B$ you travel three segments, one parallel to each axis, and each segment is a component. This is the same as in the plane (you can revise it in vectors in the plane), with one extra component.

Since $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are the position vectors, whose components are the coordinates of $A$ and $B$, it also holds that $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$. And if the origin and the components are known, the end point is found by adding: $B=A+\overrightarrow{AB}$. By the same reasoning, if the end point $B$ and the vector are known, the origin is $A=B-\overrightarrow{AB}$.

Two vectors are equal if and only if they have the same components. Therefore, if $ABCD$ is a parallelogram with its vertices in that order, $\overrightarrow{AB}=\overrightarrow{DC}$, and this equality lets you find an unknown vertex.

Magnitude of a vector

Applying Pythagoras' theorem twice in succession (first on the base of the box and then with the height) gives the length of the diagonal.

The magnitude is never negative and equals $0$ only for the zero vector. A vector of magnitude $1$ is a unit vector. Dividing a vector $\mathbf{u}\ne 0$ by its magnitude gives the unit vector that points in the same direction as $\mathbf{u}$:

$$\hat{\mathbf{u}}=\frac{1}{|\mathbf{u}|}\mathbf{u}$$

For example, $\mathbf{u}=\left(\begin{array}{c}2\\ 3\\ 6\end{array}\right)$ has magnitude $\sqrt{4+9+36}=7$ and its unit vector is $\frac{1}{7}\left(\begin{array}{c}2\\ 3\\ 6\end{array}\right)$.

Addition, subtraction and scalar multiplication

All three operations are carried out component by component.

• Addition. It means chaining two displacements: if $\mathbf{u}=\overrightarrow{AB}$ and $\mathbf{v}=\overrightarrow{BC}$, then $\mathbf{u}+\mathbf{v}=\overrightarrow{AC}$. This equality $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ is known as Chasles' relation and holds for any three points.
• Subtraction. $\mathbf{u}-\mathbf{v}=\mathbf{u}+(-\mathbf{v})$. Notice that $\mathbf{u}+(-\mathbf{u})=0$, where $0=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$ is the zero vector of ${\mathbb{R}}^3$.
• Scalar multiplication. The vector $k\mathbf{u}$ keeps the direction of $\mathbf{u}$, has the same sense if $k>0$ and the opposite sense if $k<0$, and its magnitude is multiplied by $|k|$:

$$|k\mathbf{u}|=|k||\mathbf{u}|$$

Addition is commutative and scalar multiplication is distributive: $k(\mathbf{u}+\mathbf{v})=k\mathbf{u}+k\mathbf{v}$.

Parallel (collinear) vectors

Two vectors are parallel when one is a multiple of the other. They are also said to be collinear, because when drawn from the same point they lie on a single line.

If some component of $\mathbf{v}$ is $0$, you cannot divide: then the corresponding component of $\mathbf{u}$ must also be $0$, and the value of $k$ is found from the others. A safe method is to calculate $k$ using a component that is not zero and check the others with it. Be careful: in space it is not enough to compare two of the three ratios.

Three points $A$, $B$ and $C$ are collinear exactly when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel. And if a vector parallel to another has an unknown parameter, the proportionality of the components gives equations to calculate it.

Midpoint of a segment

The midpoint $M$ satisfies $\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}$. Equating components gives an average of coordinates.

If you know the midpoint $M$ and one endpoint $A$, the other endpoint is found by solving: $B=2M-A$. More generally, the point $P$ of the segment with $\overrightarrow{AP}=t\overrightarrow{AB}$ (with $0\le t\le 1$) is $P=A+t\overrightarrow{AB}$.

Base vectors: $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$

The unit vectors in the direction of each axis are

$$\mathbf{i}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\mathbf{j}=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\mathbf{k}=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$$

For example, $\left(\begin{array}{c}2\\ 0\\ -5\end{array}\right)=2\mathbf{i}-5\mathbf{k}$: the missing component is $0$. In IB problems the notation changes from one part to another without warning, but the operations are done in the same way: $(\mathbf{i}+2\mathbf{j}-\mathbf{k})+(3\mathbf{j}+2\mathbf{k})=\mathbf{i}+5\mathbf{j}+\mathbf{k}$.