Learn7357Paper 1 · PureVectorsPosition vectors, magnitude and unit vectors

Vectors · Subtopic 1 of 1

Position Vectors, Magnitude and Unit Vectors

Vectors are arrows that carry a size and a direction. Learn to move between points with displacement vectors, measure them with three-dimensional Pythagoras, scale them to length $1$ , and use them to find midpoints, ratios and missing corners of shapes.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Write vectors in $\mathbf{i},\mathbf{j},\mathbf{k}$ form and in column form, and convert between them.
Find the displacement vector $\overrightarrow{AB}$ from two position vectors.
Find the magnitude of a vector and the unit vector in its direction.
Use position vectors to find midpoints, points dividing a line in a ratio, and the fourth vertex of a parallelogram.

Pause and think

A drone takes off from $O$ and flies $3$ m east, $4$ m north and $12$ m up. How far is it from $O$ in a straight line?

Section 2 of 14

Core idea

Start hereFlight paths in 3D

An air-traffic controller tracks planes by three coordinates: east, north and up. A plane's movement is a vector: size and direction together. Vectors let us describe and calculate with positions and movements in three dimensions as easily as in two.

The big idea

$\mathbf a=x\mathbf i+y\mathbf j+z\mathbf k$ ; $|\mathbf a|=\sqrt{x^2+y^2+z^2}$ . $\overrightarrow{AB}=\mathbf b-\mathbf a$ . Parallel vectors are scalar multiples.

1. Components and magnitude

The vector a = 3i + 4j, its length and its unit vector

Length by Pythagoras:

\sqrt{3^2+4^2}=5

. The unit vector points the same way but has length 1: divide by 5.

\mathbf a=\begin{pmatrix}x\\y\\z\end{pmatrix}=x\mathbf i+y\mathbf j+z\mathbf k,\qquad|\mathbf a|=\sqrt{x^2+y^2+z^2},\qquad\hat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}

$\mathbf i,\mathbf j,\mathbf k$: unit vectors along the $x$ , $y$ , $z$ axes.
$\hat{\mathbf a}$: unit vector (length 1) in the direction of a.

Magnitude in 3D (Pythagoras twice)

$\text{base diagonal}^2=x^2+y^2$
Pythagoras in the floor.
$|\mathbf a|^2=(x^2+y^2)+z^2$
Pythagoras with the vertical.

2. Position vectors and routes

Adding vectors: head to tail

\mathbf a

, then

\mathbf b

. The single arrow from start to finish is

\mathbf a+\mathbf b

. In components, just add:

\begin{pmatrix}3\\1\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix}=\begin{pmatrix}4\\4\end{pmatrix}

Worked example $A(1,2,3)$ , $B(4,-2,3)$

Find $\overrightarrow{AB}$ , $|AB|$ and the midpoint.

Goal: Three results.
Strategy: $\mathbf b-\mathbf a$ .

$\overrightarrow{AB}=(3,-4,0)$
Subtract.
$|AB|=\sqrt{9+16+0}=5$
Magnitude.
$\tfrac12(\mathbf a+\mathbf b)=(2.5,0,3)$
Average.

Answer

(3,-4,0)

;

5

;

(2.5,0,3)

✓ Check

A+\overrightarrow{AB}=(4,-2,3)=B

✓.

Key takeawayParallel and collinear

\mathbf u\parallel\mathbf v\iff\mathbf u=k\mathbf v

A,B,C

collinear if

\overrightarrow{AB}=k\overrightarrow{BC}

(shared point

B

Pause and think

Find $k$ if $(k,6,-3)\parallel(2,4,-2)$ .

Summary

Components; magnitude by Pythagoras.
$\overrightarrow{AB}=\mathbf b-\mathbf a$ .
Parallel = multiple.

Position vector: Vector from the origin to a point.
Unit vector: Length 1.

HistoryCarved into a bridge. On 16 October 1843, walking along the Royal Canal in Dublin, William Rowan Hamilton suddenly saw how to multiply in three dimensions. He scratched

i^2=j^2=k^2=ijk=-1

into the stone of Broom Bridge. His "quaternions" gave us the letters

\mathbf{i}

\mathbf{j}

\mathbf{k}

; Josiah Gibbs and Oliver Heaviside later simplified them into the vectors used today.

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