Revise Vectors: properties, operations and applications for IB Mathematics: Applications and Interpretation (3.5)

A vector is often pictured as an arrow: its length shows the magnitude and the way it points shows the direction. Numerically we describe it by its components — the change it makes in each coordinate direction. In two dimensions that means a horizontal component and a vertical component; in three dimensions we add a component along the third axis. The same vector can be written three equivalent ways, and you should be fluent moving between them.

Column form: $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$. \n Base-vector form: $\mathbf{v} = x\mathbf{i} + y\mathbf{j}$, where $\mathbf{i}$ and $\mathbf{j}$ are the unit vectors along the $x$- and $y$-axes. \n In three dimensions: $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$.

For instance $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ means 3 units right and 2 units down, and is the same object as $3\mathbf{i} - 2\mathbf{j}$. A position vector is the special case that starts at the origin: the position vector of the point $P(3, -2)$ is $\vec{OP} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$, so its components are just the coordinates. A displacement vector between two points need not start at the origin, and you find it by subtracting position vectors — the single most useful move in the whole topic.

Displacement from $A$ to $B$: $\vec{AB} = \mathbf{b} - \mathbf{a}$ ('destination minus start'), where $\mathbf{a}$ and $\mathbf{b}$ are the position vectors of $A$ and $B$.

Vectors are added, subtracted and scaled component by component, and each operation has a clean geometric meaning. Adding $\mathbf{a} + \mathbf{b}$ is following $\mathbf{a}$ and then $\mathbf{b}$ nose-to-tail; the resultant is the direct arrow from the start of the first to the end of the second. Subtracting $\mathbf{a} - \mathbf{b}$ is adding the reverse of $\mathbf{b}$. Multiplying by a scalar $k$ stretches the vector by a factor $|k|$ and, if $k$ is negative, flips its direction.

The magnitude (or modulus) of a vector is its length — a non-negative scalar written $|\mathbf{v}|$. It comes straight from Pythagoras on the components, and it is what a question means when it asks for a distance, a speed, or 'how far'. Once you have the magnitude you can also build the unit vector in the same direction, a vector of length exactly 1, by dividing the vector by its own magnitude. Unit vectors are how you record a pure direction.

Magnitude: $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2}$ in 2D, and $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ in 3D. \n Unit vector in the direction of $\mathbf{v}$: $\hat{\mathbf{v}} = \dfrac{1}{|\mathbf{v}|}\mathbf{v}$.

The scalar product multiplies two vectors to give a single number, and it is the tool for angles. It can be computed two ways that always agree: from components, or from the two magnitudes and the angle between the vectors. Setting the two forms equal and rearranging gives a formula for the angle — and, as a special case, a one-line test for perpendicularity.

Component form: $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 (+\, a_3 b_3)$. \n Geometric form: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$. \n Angle between vectors: $\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$. \n Perpendicular test: $\mathbf{a} \perp \mathbf{b} \iff \mathbf{a} \cdot \mathbf{b} = 0$.

The sign of the dot product already tells a story before you take an arccos: positive means an acute angle, zero means a right angle, and negative means an obtuse angle. That is why perpendicularity needs no angle calculation at all — if the components multiply and add to zero, the vectors meet at $90^\circ$, because $\cos 90^\circ = 0$.

In Paper 1 the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards the correct approach — the right formula with the candidate's numbers substituted — while an accuracy mark (A) rewards a correct value and depends on the method mark it follows. Follow-through (FT) means an error made early does not have to cost you the marks that depend on it, provided each later step is carried out correctly on your own figures. But that protection only exists if the method is written down. Study how every mark below is earned by a specific line.

Vectors earn their keep in context. Displacement problems use $\vec{AB} = \mathbf{b} - \mathbf{a}$ to find the path between two positions, and the magnitude of that displacement is the straight-line distance. Velocity problems add a craft's velocity to a wind or current to get the true resultant, whose magnitude is the actual speed. Force problems add several forces acting on a body; when the resultant is the zero vector the body is in equilibrium, and a right angle between two forces is confirmed by a zero scalar product. In every case the routine is the same: model each quantity as a vector, combine with the right operation, then take a magnitude or an angle as the question demands.

Everything here scales up. The addition and magnitude you practised in two dimensions work identically in three, where the same $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ and the same dot-product angle formula describe lines and planes in space. The scalar product is the seed of the vector equation of a line and of measuring how one direction projects onto another, ideas that run through kinematics and modelling. And the discipline you have built — model each quantity as a vector, combine with the right operation, decide whether the answer is a length or an angle, and show every line so method and follow-through protect the result — is precisely what every vectors question on the exam rewards.