Mathematics 9709 · Pure Mathematics 3

Displacement vectors

A displacement vector is one arrow. It only says “how far” and “which way”. It does not care where it starts.
Write each one as a column vector, then add or subtract them one row at a time.
Following one arrow then the next (the triangle law) gives the same answer as adding the columns.

Two arrows with the same components are the same vector. It does not matter where you draw them. To add, put them tip-to-tail: $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$ . To turn one round, flip the sign: $\overrightarrow{BA}=-\overrightarrow{AB}$ .

The length of a vector is Pythagoras on all its components: $|\mathbf{v}|=\sqrt{x^2+y^2+z^2}$ . Every “find the distance” or “find the length” part is just this one square root.

Add by following arrows (or adding components)

Write a vector as a column $\begin{pmatrix}x\\y\end{pmatrix}$ : the change in $x$ on top, the change in $y$ below (add a third row $z$ in 3D).
It holds a direction and a length, but no fixed starting point.
To add two displacements, follow the first arrow, then start the second from where you land:

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC},\qquad \overrightarrow{BA}=-\overrightarrow{AB}$

Triangle law: follow AB then BC, land at C, so AC = AB + BC.

The length of a vector is Pythagoras on its components: $\left|\begin{pmatrix}x\\y\\z\end{pmatrix}\right|=\sqrt{x^2+y^2+z^2}$ .
That one number is called the magnitude.

Worked examples

Worked example

$\overrightarrow{AB}=\begin{pmatrix}4\\2\end{pmatrix}$ and $\overrightarrow{BC}=\begin{pmatrix}3\\7\end{pmatrix}$ : find $\overrightarrow{AC}$ and its length

Add tip-to-tail, component by component: $\overrightarrow{AC}=\begin{pmatrix}4\\2\end{pmatrix}+\begin{pmatrix}3\\7\end{pmatrix}=\begin{pmatrix}7\\9\end{pmatrix}$ .
Length is Pythagoras on the components: $|\overrightarrow{AC}|=\sqrt{7^2+9^2}=\sqrt{49+81}=\sqrt{130}$ .

Answer

\overrightarrow{AC}=\begin{pmatrix}7\\9\end{pmatrix},\quad |\overrightarrow{AC}|=\sqrt{130}

Leave it as $\sqrt{130}$ unless the question asks for a decimal.

Worked example

3D length: $\overrightarrow{AB}=2\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and $\overrightarrow{AC}=4\mathbf{i}-3\mathbf{k}$ : which step is longer?

Two steps from a 9709/32 O/N 2022 Q6 setup: comparing lengths is just Pythagoras.

Square all three components of each (a missing component counts as $0$ ): $|\overrightarrow{AB}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3$ .
$|\overrightarrow{AC}|=\sqrt{4^2+0^2+(-3)^2}=\sqrt{25}=5$ , so $\overrightarrow{AC}$ is the longer step.

Answer

|\overrightarrow{AB}|=3,\quad |\overrightarrow{AC}|=5

These lengths $3$ and $5$ come back in the angle work in §9.3, so you only find them once.

Why a closed path adds to zero

Walk round any closed shape. Start at $P$ , follow the sides, end back at $P$ . You finish where you began, so the displacements add to the zero vector:

$\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}+\overrightarrow{SP}=\mathbf{0}$

So “find the side that closes the shape” just means minus the sum of the other sides.

Where the marks go

Common mistake

\overrightarrow{AB}

means “from A to B”.

\overrightarrow{BA}=-\overrightarrow{AB}

points the other way. Always write the start point first. Check the signs too: left and down are negative. One dropped minus flips the whole arrow.

Common mistake

A length is a magnitude, so square every component. In 3D that means all three:

\sqrt{x^2+y^2+z^2}

, never just two of them.

Now you try

You are given $\overrightarrow{PQ}=\begin{pmatrix}5\\2\end{pmatrix}$ , $\overrightarrow{QR}=\begin{pmatrix}1\\4\end{pmatrix}$ and $\overrightarrow{RS}=\begin{pmatrix}-4\\1\end{pmatrix}$ . Find the $\overrightarrow{SP}$ that closes the quadrilateral $PQRS$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

A closed path adds to zero, so $\overrightarrow{SP}=-(\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS})=-\begin{pmatrix}2\\7\end{pmatrix}$ :

Answer

\overrightarrow{SP}=\begin{pmatrix}-2\\-7\end{pmatrix}

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Displacement vectors

A displacement vector is one arrow. It only says “how far” and “which way”. It does not care where it starts.
Write each one as a column vector, then add or subtract them one row at a time.
Following one arrow then the next (the triangle law) gives the same answer as adding the columns.

The length of a vector is Pythagoras on all its components: $|\mathbf{v}|=\sqrt{x^2+y^2+z^2}$ . Every “find the distance” or “find the length” part is just this one square root.

Add by following arrows (or adding components)

Write a vector as a column $\begin{pmatrix}x\\y\end{pmatrix}$ : the change in $x$ on top, the change in $y$ below (add a third row $z$ in 3D).
It holds a direction and a length, but no fixed starting point.
To add two displacements, follow the first arrow, then start the second from where you land:

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC},\qquad \overrightarrow{BA}=-\overrightarrow{AB}$

Triangle law: follow AB then BC, land at C, so AC = AB + BC.

The length of a vector is Pythagoras on its components: $\left|\begin{pmatrix}x\\y\\z\end{pmatrix}\right|=\sqrt{x^2+y^2+z^2}$ .
That one number is called the magnitude.

Worked examples

Worked example

$\overrightarrow{AB}=\begin{pmatrix}4\\2\end{pmatrix}$ and $\overrightarrow{BC}=\begin{pmatrix}3\\7\end{pmatrix}$ : find $\overrightarrow{AC}$ and its length

Add tip-to-tail, component by component: $\overrightarrow{AC}=\begin{pmatrix}4\\2\end{pmatrix}+\begin{pmatrix}3\\7\end{pmatrix}=\begin{pmatrix}7\\9\end{pmatrix}$ .
Length is Pythagoras on the components: $|\overrightarrow{AC}|=\sqrt{7^2+9^2}=\sqrt{49+81}=\sqrt{130}$ .

Answer

\overrightarrow{AC}=\begin{pmatrix}7\\9\end{pmatrix},\quad |\overrightarrow{AC}|=\sqrt{130}

Leave it as $\sqrt{130}$ unless the question asks for a decimal.

Worked example

3D length: $\overrightarrow{AB}=2\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and $\overrightarrow{AC}=4\mathbf{i}-3\mathbf{k}$ : which step is longer?

Two steps from a 9709/32 O/N 2022 Q6 setup: comparing lengths is just Pythagoras.

Square all three components of each (a missing component counts as $0$ ): $|\overrightarrow{AB}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3$ .
$|\overrightarrow{AC}|=\sqrt{4^2+0^2+(-3)^2}=\sqrt{25}=5$ , so $\overrightarrow{AC}$ is the longer step.

Answer

|\overrightarrow{AB}|=3,\quad |\overrightarrow{AC}|=5

These lengths $3$ and $5$ come back in the angle work in §9.3, so you only find them once.

Why a closed path adds to zero

Walk round any closed shape. Start at $P$ , follow the sides, end back at $P$ . You finish where you began, so the displacements add to the zero vector:

$\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}+\overrightarrow{SP}=\mathbf{0}$

So “find the side that closes the shape” just means minus the sum of the other sides.

Where the marks go

Common mistake

\overrightarrow{AB}

means “from A to B”.

\overrightarrow{BA}=-\overrightarrow{AB}

points the other way. Always write the start point first. Check the signs too: left and down are negative. One dropped minus flips the whole arrow.

Common mistake

A length is a magnitude, so square every component. In 3D that means all three:

\sqrt{x^2+y^2+z^2}

, never just two of them.

Now you try

Your turn— tap to reveal the worked answer (9709-style)

A closed path adds to zero, so $\overrightarrow{SP}=-(\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS})=-\begin{pmatrix}2\\7\end{pmatrix}$ :

Answer

\overrightarrow{SP}=\begin{pmatrix}-2\\-7\end{pmatrix}