Maths4U
← Back to Chapter
⬅ Displacement or translation vectors The scalar product ➡
Displacement or translation vectors Position vectors The scalar product The vector equation of a line Intersection of two lines Review

Vectors — Position vectors

Pick what you’d like to study:

📘 Notes

Vectors — Position Vectors

A position vector describes the position of a point relative to the origin. In three dimensions, it tells us how far the point is from the origin in the \(x\)-, \(y\)-, and \(z\)-directions.

Position vectors are very important in Year 13 vector work because they are used to describe points, find displacement vectors, and write equations of lines in space.

Key definitions and formulae

1. Position vector of a point

If a point \(P\) has coordinates \((x,y,z)\), then its position vector is the vector from the origin \(O\) to \(P\).

\[ \overrightarrow{OP}= \begin{pmatrix} x\\ y\\ z \end{pmatrix}. \]

This vector gives the exact position of the point relative to the origin.

2. Using bold vector notation

Sometimes the position vector of \(P\) is written as

\[ \mathbf{p}= \begin{pmatrix} x\\ y\\ z \end{pmatrix}. \]

So \(\mathbf{p}\) and \(\overrightarrow{OP}\) can represent the same vector.

3. Displacement vector between two points

If points \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\), then the vector from \(A\) to \(B\) is

\[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a}. \]

If

\[ \mathbf{a}= \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} x_2\\ y_2\\ z_2 \end{pmatrix}, \]

then

\[ \overrightarrow{AB}= \begin{pmatrix} x_2-x_1\\ y_2-y_1\\ z_2-z_1 \end{pmatrix}. \]

4. Midpoint position vector

If points \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\), then the midpoint \(M\) of \(AB\) has position vector

\[ \overrightarrow{OM}=\frac{\mathbf{a}+\mathbf{b}}{2}. \]

This result is often used in vector geometry questions.

Worked examples

Example 1: Write down a position vector

Point \(P\) has coordinates \((3,-2,5)\). Write down the position vector of \(P\).

Solution

The position vector is the vector from the origin to the point.

\[ \overrightarrow{OP}= \begin{pmatrix} 3\\ -2\\ 5 \end{pmatrix}. \]
So the position vector of \(P\) is \[ \begin{pmatrix} 3\\ -2\\ 5 \end{pmatrix}. \]

Example 2: Find a displacement vector using position vectors

The position vectors of points \(A\) and \(B\) are

\[ \mathbf{a}= \begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} 7\\ 3\\ -2 \end{pmatrix}. \]

Find \(\overrightarrow{AB}\).

Solution

Use

\[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a}. \]
\[ \overrightarrow{AB}= \begin{pmatrix} 7\\ 3\\ -2 \end{pmatrix} - \begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} = \begin{pmatrix} 5\\ 4\\ -6 \end{pmatrix}. \]
Therefore, \[ \overrightarrow{AB}= \begin{pmatrix} 5\\ 4\\ -6 \end{pmatrix}. \]

Example 3: Find the coordinates of a point from its position vector

The position vector of point \(Q\) is

\[ \overrightarrow{OQ}= \begin{pmatrix} -4\\ 6\\ 1 \end{pmatrix}. \]

Find the coordinates of \(Q\).

Solution

The coordinates of the point are read directly from the position vector.

\[ Q(-4,6,1). \]
The coordinates of \(Q\) are \((-4,6,1)\).

Example 4: Find the midpoint using position vectors

Points \(A\) and \(B\) have position vectors

\[ \mathbf{a}= \begin{pmatrix} 1\\ 4\\ -3 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} 5\\ -2\\ 7 \end{pmatrix}. \]

Find the position vector of the midpoint \(M\) of \(AB\).

Solution

Use

\[ \overrightarrow{OM}=\frac{\mathbf{a}+\mathbf{b}}{2}. \]
\[ \overrightarrow{OM} = \frac{1}{2} \left( \begin{pmatrix} 1\\ 4\\ -3 \end{pmatrix} + \begin{pmatrix} 5\\ -2\\ 7 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 6\\ 2\\ 4 \end{pmatrix} = \begin{pmatrix} 3\\ 1\\ 2 \end{pmatrix}. \]
The midpoint has position vector \[ \begin{pmatrix} 3\\ 1\\ 2 \end{pmatrix}. \]

Standard methods

Finding a position vector

If you know the coordinates of a point, write the same numbers as a column vector.

\[ P(x,y,z) \quad \Longrightarrow \quad \overrightarrow{OP}= \begin{pmatrix} x\\ y\\ z \end{pmatrix}. \]

Finding the vector from one point to another

Subtract the position vector of the starting point from the position vector of the finishing point.

\[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a}. \]

Finding a midpoint

Add the two position vectors and divide by \(2\).

\[ \overrightarrow{OM}=\frac{\mathbf{a}+\mathbf{b}}{2}. \]

Common mistakes and exam tips

Common mistakes

  • Confusing a point such as \((2,3,4)\) with a vector such as \(\begin{pmatrix}2\\3\\4\end{pmatrix}\).
  • Reversing the subtraction when finding \(\overrightarrow{AB}\).
  • Forgetting that position vectors are always measured from the origin.
  • Making sign errors when coordinates are negative.

Exam tips

  • Write vectors clearly as column vectors.
  • Use finish minus start for \(\overrightarrow{AB}\).
  • Check that the coordinates of a point match the entries of its position vector in the same order.
  • In geometry questions, label points and vectors carefully before starting algebra.

Summary

\[ \overrightarrow{OP}= \begin{pmatrix} x\\ y\\ z \end{pmatrix} \] \[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a} \] \[ \overrightarrow{OM}=\frac{\mathbf{a}+\mathbf{b}}{2} \]

A position vector gives the position of a point relative to the origin, and it is the starting point for most Year 13 vector methods in three dimensions.

Open Full Notes
⚡ Practice Questions

0/0 mastered, 0 attempted

0%
▶ Start Practice 🔁 Review All Questions