Two-thirds of the way from A to B

\overrightarrow{OP} = \mathbf{a} + \tfrac{2}{3}(\mathbf{b} - \mathbf{a})

\mathbf{b} - \mathbf{a} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}, \quad \tfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \begin{pmatrix} 4 \\ 4 \end{pmatrix}

\overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}

start at A, then travel ⅔ of the journey

Vectors

A vector is a journey, not a place.

It has a length and a direction and nothing else — slide it anywhere and it's the same vector. This one goes 4 across and 3 up.

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Magnitude

\left|\begin{pmatrix} a \\ b \end{pmatrix}\right| = \sqrt{a^2 + b^2}

Its components are $4$ and $3$ , so its length is $\sqrt{4^2 + 3^2} = 5$ . A vector's magnitude is Pythagoras on its components.

Addition

\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}

Follow $\mathbf{a} = (3,\,1)$ , then $\mathbf{b} = (1,\,2)$ from its tip, and you land on $(4,\,3)$ — the vector we started with. Take $\mathbf{b}$ first and you reach the same corner: order doesn't matter.

Scalar multiples

k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}

$2\mathbf{a}$ keeps the direction and doubles the length; $-\mathbf{a}$ points exactly backwards. Every multiple of $\mathbf{a}$ is parallel to it — which is how you prove lines are parallel.

Dividing a line

\overrightarrow{OP} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b} - \mathbf{a})

The journey from $A$ to $B$ is $\mathbf{b} - \mathbf{a}$ : end minus start. Go two-thirds of the way and you're at $P(6,\,5)$ , which divides $AB$ in the ratio $2 : 1$ .

Chapter 15

Every vector problem is a journey.

Displacement and distance, showing points are collinear, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›