The top number of a column vector is the movement across and the bottom number is the movement up, with left and down counted as negative.

Going from \(A\) to \(B\) is 3 to the right and 2 down.

The column vector is \(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\)

The top number of a column vector is the movement across and the bottom number is the movement up, with left and down counted as negative.

Going from the tail of the arrow to its head is 3 to the right and 1 up.

\(\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\)

From the tail of the second arrow to its head is 4 to the left and 2 down.

\(\mathbf{b} = \begin{pmatrix} -4 \\ -2 \end{pmatrix}\)

\(A\) is the point \((-3, 5)\) and \(B\) is the point \((1, -1)\).

Going from \(A\) to \(B\) is 4 to the right and 6 down, which you can get by counting squares or by subtracting the coordinates of \(A\) from the coordinates of \(B\):

\(\overrightarrow{AB} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} - \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ -6 \end{pmatrix}\)

Subtracting the wrong way round gives \(\overrightarrow{BA}\), which is the next part.

\(\overrightarrow{BA}\) is the same journey backwards, so every part changes sign:

\(\overrightarrow{BA} = -\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\)

Translating \(D\) to \(C\) undoes the translation from \(C\) to \(D\), so \(\overrightarrow{ DC } = -\overrightarrow{ CD }\).

Every component changes sign:

\(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\)

From \(P\) to \(Q\) is 5 to the left and 3 down.

\(\mathbf{a} = \begin{pmatrix} -5 \\ -3 \end{pmatrix}\)

Travelling from \(P\) to \(R\) means going along \(\mathbf{a}\) and then along \(\mathbf{b}\), so \(\overrightarrow{PR} = \mathbf{a} + \mathbf{b}\).

\(\mathbf{b}\) is 3 to the left and 3 up, so \(\mathbf{b} = \begin{pmatrix} -3 \\ 3 \end{pmatrix}\).

Add the top numbers and add the bottom numbers:

\(\begin{pmatrix} -5 \\ -3 \end{pmatrix} + \begin{pmatrix} -3 \\ 3 \end{pmatrix} = \begin{pmatrix} -8 \\ 0 \end{pmatrix}\)

Counting straight from \(P\) to \(R\) on the grid gives the same answer, which is a good check.

Count each arrow off the grid first:

\(\mathbf{a} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}\)

Subtract the top numbers and subtract the bottom numbers:

\(\begin{pmatrix} 6 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}\)

On the diagram \(\mathbf{a} - \mathbf{b}\) is the journey from \(B\) to \(A\): back along \(\mathbf{b}\) to \(O\), then out along \(\mathbf{a}\). Subtracting a vector means travelling backwards along it, and that is why the answer comes out 4 to the right.

The arrow goes 3 to the right and 3 down.

\(\mathbf{a} = \begin{pmatrix} 3 \\ -3 \end{pmatrix}\)

Counting from \(R\) to \(S\) gives \(\overrightarrow{RS} = \begin{pmatrix} 9 \\ -9 \end{pmatrix}\).

Both parts are 3 times the parts of \(\mathbf{a}\):

\(\begin{pmatrix} 9 \\ -9 \end{pmatrix} = 3 \times \begin{pmatrix} 3 \\ -3 \end{pmatrix}\)

So \(\overrightarrow{RS} = 3\mathbf{a}\). The two arrows are parallel because one is a multiple of the other — that is what "parallel" means for vectors.

Reading \(\mathbf{a}\) off the grid, \(\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\).

Multiplying by a number multiplies each part of the column vector — the bottom one as well as the top one:

\(2\mathbf{a} = \begin{pmatrix} 2 \times 3 \\ 2 \times 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}\)

\(2\mathbf{a}\) points the same way as \(\mathbf{a}\) and is 2 times as long.

\(\mathbf{b} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\), and part (a) gave \(2\mathbf{a} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}\).

Multiply first, then subtract — doing it the other way round is the commonest slip here:

\(\begin{pmatrix} 6 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \end{pmatrix}\)

\(|\mathbf{a}|\) means the length of \(\mathbf{a}\).

Counting squares, \(\mathbf{a} = \begin{pmatrix} -12 \\ -5 \end{pmatrix}\). The arrow is the longest side of a right-angled triangle with shorter sides 12 and 5, so use Pythagoras' theorem. The signs make no difference to a length — squaring removes them.

\(|\mathbf{a}| = \sqrt{ 144 + 25 } = \sqrt{ 169 } = 13\)

There is no arrow straight from \(P\) to \(Q\), so travel there through \(O\):

\(\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ}\)

\(\overrightarrow{PO}\) is \(\overrightarrow{OP}\) backwards, so \(\overrightarrow{PO} = -3\mathbf{a}\).

\(\overrightarrow{PQ} = -3\mathbf{a} + 3\mathbf{b} = 3\mathbf{b} - 3\mathbf{a}\)

\(\overrightarrow{QP}\) is the same journey as \(\overrightarrow{PQ}\) in reverse, so change the sign of every term:

\(\overrightarrow{QP} = -\overrightarrow{PQ} = 3\mathbf{a} - 3\mathbf{b}\)