Extended. A vector is a movement: so far across, so far up. Write it as a column, add vectors by following one after another, and multiply by a number to stretch it along the same line.
A column vector \(\begin{pmatrix}x\\y\end{pmatrix}\) means \(x\) across and \(y\) up. Add or subtract vectors by combining the top numbers and the bottom numbers separately; \(k\mathbf{a}\) multiplies both numbers by \(k\) and is parallel to \(\mathbf{a}\).
Fig. 1: following a = (3, 1) and then b = (1, 2) tip to tail gives the single journey a + b = (4, 3)
Section 01
A vector is a journey.
In the lab, drag the head and the tail of the vector. Watch the column vector update: the top number is the movement across, the bottom number the movement up. The faint copies show that moving a vector without turning or stretching it does not change it.
Notation: \(\overrightarrow{AB}\) is the vector from \(A\) to \(B\); a single vector is printed in bold, \(\mathbf{a}\) (handwrite it underlined, a̲).
\(P\) is the point \((-1, 5)\) and \(\overrightarrow{PQ} = \begin{pmatrix}2\\-3\end{pmatrix}\). Find the coordinates of \(Q\).
Start at \(P\) and make the journey: 2 across, 3 down.
\((-1 + 2, 5 - 3) = (1, 2)\).
Answer: \(Q(1, 2)\)
Examiner trap
Writing 'start minus end'. \(\overrightarrow{AB}\) is \(B - A\): the journey FROM \(A\) TO \(B\). Getting it backwards flips both signs and gives \(\overrightarrow{BA}\).
Section 02
Adding and subtracting.
In the lab, place \(\mathbf{b}\) so its tail sits on the head of \(\mathbf{a}\). The single arrow from the start to the finish is \(\mathbf{a} + \mathbf{b}\). Switch to \(\mathbf{a} - \mathbf{b}\) and see that it is \(\mathbf{a}\) followed by \(\mathbf{b}\) reversed.
Sign slips when subtracting a negative: \(5 - (-3) = 8\), not 2. Write the subtraction out in full for each number before you simplify.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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\(A\) is \((2, 3)\) and \(B\) is \((5, 1)\). What is \(\overrightarrow{AB}\)?
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\(\mathbf{a} = \begin{pmatrix}2\\-1\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-3\\4\end{pmatrix}\). What is \(\mathbf{a} + \mathbf{b}\)?
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What is \(3\begin{pmatrix}-2\\5\end{pmatrix}\)?
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Which vector is parallel to \(\begin{pmatrix}2\\-3\end{pmatrix}\)?
Section 03
Multiplying by a scalar.
In the lab, multiply a vector by 2, by \(\frac{1}{2}\) and by \(-1\). The result always lies along the same line as the original: it is parallel. A negative multiplier also reverses the direction.
\(k\mathbf{a}\) multiplies BOTH numbers by \(k\): \(3\begin{pmatrix}2\\-1\end{pmatrix} = \begin{pmatrix}6\\-3\end{pmatrix}\).
Parallel vectors are multiples of each other: \(\begin{pmatrix}-4\\6\end{pmatrix} = -2\begin{pmatrix}2\\-3\end{pmatrix}\).
Combine by doing the multiplications first, then adding or subtracting.
Find \(m\) and \(k\) when \(3\begin{pmatrix}2\\k\end{pmatrix} + \begin{pmatrix}m\\4\end{pmatrix} = \begin{pmatrix}1\\10\end{pmatrix}\).
Top numbers: \(6 + m = 1\), so \(m = -5\).
Bottom numbers: \(3k + 4 = 10\), so \(3k = 6\) and \(k = 2\).
Answer: \(m = -5\), \(k = 2\)
Worked example Spotting parallel vectors
Show that \(\begin{pmatrix}6\\-4\end{pmatrix}\) and \(\begin{pmatrix}-9\\6\end{pmatrix}\) are parallel.
Look for a multiplier: \(6 \times (-1.5) = -9\) and \(-4 \times (-1.5) = 6\).
The same multiplier works for both numbers.
Answer: \(\begin{pmatrix}-9\\6\end{pmatrix} = -1.5\begin{pmatrix}6\\-4\end{pmatrix}\), so they are parallel (in opposite directions).
Examiner trap
Multiplying only the top number. \(3\begin{pmatrix}2\\k\end{pmatrix}\) is \(\begin{pmatrix}6\\3k\end{pmatrix}\): the scalar multiplies every component.
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