Vectors and vector geometry

A vector describes a movement with both size and direction. This note covers column vectors, adding, subtracting and scaling them, finding the length of a vector, and using vectors to prove geometric facts such as parallel lines and collinear points. Vectors are used to describe forces, velocities and movements in navigation and computer graphics.

Key ideas

A vector has a magnitude (length) and a direction. A scalar has size only. In print a vector is written in bold, such as a, or as \(\overrightarrow{AB}\) for the movement from point \(A\) to point \(B\). By hand, underline the letter: \(\underline{a}\).

A column vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) says move \(x\) units right (left if negative) and \(y\) units up (down if negative). Two vectors are equal when they have the same length and direction, wherever they are drawn.

• Adding: add the top numbers and add the bottom numbers. On a diagram, place the vectors head to tail. The single vector from the start to the end is the resultant.
• Subtracting: subtract components. Note that \(-\mathbf{a}\) has the same length as a but points the opposite way, and \(\overrightarrow{BA} = -\overrightarrow{AB}\).
• Multiplying by a scalar: multiply both components by the number. \(2\mathbf{a}\) is twice as long and in the same direction. A negative multiplier reverses the direction.
• Finding a vector between points: \(\overrightarrow{AB}\) = (position of \(B\)) − (position of \(A\)). If \(A=(2,5)\) and \(B=(6,2)\), then \(\overrightarrow{AB}=\begin{pmatrix}4\\-3\end{pmatrix}\).

Vectors are useful because the same movement can be drawn anywhere. A vector from \(A\) to \(B\) is the same vector as one of equal length and direction starting at \(C\). Only the size and direction matter, not the starting point. This is why two opposite sides of a parallelogram are described by the same vector.

When a question gives you a diagram, the first job is to label every vector you already know. Then look for a route from your start point to your end point that uses only those vectors. Each step along the route is either with an arrow (add the vector) or against an arrow (subtract the vector). For example, to go from \(B\) to \(A\) when you know \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\), you travel back to \(O\) and then out to \(A\), giving \(\overrightarrow{BA}=-\mathbf{b}+\mathbf{a}=\mathbf{a}-\mathbf{b}\).

The same ideas apply to coordinates. If a point has coordinates \((x,y)\), the vector from the origin to that point is \(\begin{pmatrix}x\\y\end{pmatrix}\). This is called its position vector, and it lets you treat points and movements in the same way.

Magnitude (Higher and useful for all): the length of \(\begin{pmatrix} x \\ y \end{pmatrix}\) is found with Pythagoras' theorem. The magnitude is written \(|\mathbf{a}|\), and \(|\mathbf{a}| = \sqrt{x^2+y^2}\).

Parallel vectors (Higher): two vectors are parallel when one is a scalar multiple of the other. For example, \(\begin{pmatrix}6\\-9\end{pmatrix} = 3\begin{pmatrix}2\\-3\end{pmatrix}\), so the two vectors are parallel.

Vector geometry (Higher): to prove a result, write the route between two points as a sum of known vectors, taking any path along the diagram. Then simplify.

• If \(\overrightarrow{AB} = k\overrightarrow{CD}\), the lines \(AB\) and \(CD\) are parallel, and \(|AB| = |k| \times |CD|\).
• If \(\overrightarrow{AB} = k\overrightarrow{BC}\), the lines are parallel and share the point \(B\), so \(A\), \(B\), \(C\) lie on one straight line (they are collinear).
• If \(M\) is the midpoint of \(AB\), then \(\overrightarrow{AM} = \tfrac{1}{2}\overrightarrow{AB}\).
• If a point \(C\) divides \(AB\) in the ratio \(m:n\), then \(\overrightarrow{AC} = \dfrac{m}{m+n}\overrightarrow{AB}\).

Definitions and formulae

Worked examples

Common errors

• Subtracting in the wrong order. \(\overrightarrow{AB}\) is "end minus start", so it is \(\mathbf{b}-\mathbf{a}\), not \(\mathbf{a}-\mathbf{b}\).
• Writing a vector with a sign error in only one component when multiplying by a negative number. Both components change sign.
• Forgetting that \(\overrightarrow{BA}=-\overrightarrow{AB}\) when a route goes against the arrow on the diagram.
• Writing a proof that stops at \(\overrightarrow{OD}=3\overrightarrow{OC}\) without saying why this means the points are collinear. State that the vectors are parallel and share a point.
• Treating the magnitude as the sum of the components. It is \(\sqrt{x^2+y^2}\).

Exam tip: In a vector geometry question, write the route you take, such as \(\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}\), before you substitute. Always give your final answer in its simplest form, and use the same letters as the question.

Check yourself

1. Given \(\mathbf{p}=\begin{pmatrix}5\\-2\end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix}1\\3\end{pmatrix}\), find \(\mathbf{p}-2\mathbf{q}\).
   Answer

   \(2\mathbf{q}=\begin{pmatrix}2\\6\end{pmatrix}\), so \(\mathbf{p}-2\mathbf{q}=\begin{pmatrix}5-2\\-2-6\end{pmatrix}=\begin{pmatrix}3\\-8\end{pmatrix}\).
2. Find the magnitude of \(\begin{pmatrix}-5\\12\end{pmatrix}\).
   Answer

   \(\sqrt{(-5)^2+12^2}=\sqrt{25+144}=\sqrt{169}=13\).
3. (Higher) Are \(\begin{pmatrix}6\\-9\end{pmatrix}\) and \(\begin{pmatrix}4\\-6\end{pmatrix}\) parallel?
   Answer

   Yes. \(\begin{pmatrix}6\\-9\end{pmatrix}=1.5\begin{pmatrix}4\\-6\end{pmatrix}\), because \(1.5\times4=6\) and \(1.5\times(-6)=-9\). One is a scalar multiple of the other.
4. (Higher) \(OABC\) is a parallelogram with \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OC}=\mathbf{c}\). Find \(\overrightarrow{AC}\) and \(\overrightarrow{OB}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\).
   Answer

   \(\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}=\mathbf{c}-\mathbf{a}\). Since \(\overrightarrow{AB}=\overrightarrow{OC}=\mathbf{c}\), we get \(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\mathbf{a}+\mathbf{c}\).