Vectors in two dimensions

What a vector is Extended

A vector is a quantity with both a size and a direction, such as a movement of 3 km due north. A quantity with size only, such as a mass of 5 kg, is a scalar.

In two dimensions a vector can be written as a column vector $\left(\begin{array}{c}x\\ y\end{array}\right)$:

• the top number $x$ is the movement across, positive to the right and negative to the left;
• the bottom number $y$ is the movement up, positive upwards and negative downwards.

So $\left(\begin{array}{c}-2\\ 5\end{array}\right)$ means 2 to the left and 5 up. It does not matter where the movement starts: any arrow 2 left and 5 up represents the same vector.

Vectors are written in three ways, and you need to recognise all of them:

• as a column vector, $\left(\begin{array}{c}4\\ -1\end{array}\right)$;
• as $\overrightarrow{AB}$, the vector from the point A to the point B;
• as a single bold letter, $\mathbf{a}$. In handwriting you cannot write bold, so underline the letter instead.

Two vectors are equal if they have the same size and the same direction, so equal column vectors have equal components. The vector $-\mathbf{a}$ has the same size as $\mathbf{a}$ but the opposite direction: $\overrightarrow{BA}=-\overrightarrow{AB}$.

Vectors and translations Extended

A translation moves every point by the same vector, so a vector is exactly what describes a translation. "Translation by $\left(\begin{array}{c}3\\ -2\end{array}\right)$" means every point moves 3 right and 2 down.

To find the vector $\overrightarrow{AB}$ from coordinates, subtract the coordinates of the start from the coordinates of the end:
$$\overrightarrow{AB}=\left(\begin{array}{c}{x}_B-x_A\\ y_B-y_A\end{array}\right)$$

For example, from $A(1,4)$ to $B(6,-2)$: $\overrightarrow{AB}=\left(\begin{array}{c}6-1\\ -2-4\end{array}\right)=\left(\begin{array}{c}5\\ -6\end{array}\right)$.

Working the other way, if you know a starting point and a vector, add the vector to the coordinates to find where you finish. If $P$ is $(3,1)$ and $\overrightarrow{PQ}=\left(\begin{array}{c}-4\\ 2\end{array}\right)$, then $Q$ is $(3-4,1+2)=(-1,3)$.

Adding and subtracting vectors Extended

To add column vectors, add the top components and add the bottom components:
$$\left(\begin{array}{c}3\\ 1\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)=\left(\begin{array}{c}4\\ 4\end{array}\right)$$

Why this works: $\mathbf{a}+\mathbf{b}$ means "do the movement $\mathbf{a}$, then the movement $\mathbf{b}$". The total movement across is the two across movements added, and the same for up. On a diagram, draw $\mathbf{b}$ starting where $\mathbf{a}$ finishes (nose to tail); $\mathbf{a}+\mathbf{b}$ goes straight from the start of $\mathbf{a}$ to the end of $\mathbf{b}$. This single vector is called the resultant.

To subtract, take away component by component. Subtracting $\mathbf{b}$ is the same as adding $-\mathbf{b}$, the movement $\mathbf{b}$ in reverse:
$$\left(\begin{array}{c}5\\ -2\end{array}\right)-\left(\begin{array}{c}-1\\ 4\end{array}\right)=\left(\begin{array}{c}5-(-1)\\ -2-4\end{array}\right)=\left(\begin{array}{c}6\\ -6\end{array}\right)$$

Watch the double negatives: $5-(-1)=6$. Order matters for subtraction: $\mathbf{a}-\mathbf{b}$ is the negative of $\mathbf{b}-\mathbf{a}$.

Here $\mathbf{a}=\left(\begin{array}{c}3\\ 1\end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{c}1\\ 3\end{array}\right)$. Following $\mathbf{a}$ then $\mathbf{b}$ ends at the same place as the single movement $\mathbf{a}+\mathbf{b}=\left(\begin{array}{c}4\\ 4\end{array}\right)$.

Multiplying a vector by a scalar Extended

Multiplying a vector by a number (a scalar) multiplies each component by that number:
$$3\left(\begin{array}{c}2\\ -1\end{array}\right)=\left(\begin{array}{c}6\\ -3\end{array}\right)$$

The result $3\mathbf{a}$ points in the same direction as $\mathbf{a}$ and is 3 times as long. In general, $k\mathbf{a}$ is parallel to $\mathbf{a}$:

• if $k>0$, it points the same way and is $k$ times as long;
• if $k<0$, it points the opposite way; $-2\mathbf{a}$ is twice as long as $\mathbf{a}$ and reversed;
• $\frac{1}{2}\mathbf{a}$ is half as long, in the same direction.

This gives the test for parallel vectors: two vectors are parallel when one is a number times the other. $\left(\begin{array}{c}6\\ -9\end{array}\right)$ and $\left(\begin{array}{c}-4\\ 6\end{array}\right)$ are parallel because both are multiples of $\left(\begin{array}{c}2\\ -3\end{array}\right)$: $\left(\begin{array}{c}6\\ -9\end{array}\right)=-\frac{3}{2}\left(\begin{array}{c}-4\\ 6\end{array}\right)$.

Expressions such as $2\mathbf{a}-3\mathbf{b}$ are worked out in two steps: multiply each vector by its number, then add or subtract.

Unknown numbers in vector equations Extended

Because equal vectors have equal components, a vector equation is really two ordinary equations, one for the top and one for the bottom.

• Solving for a vector. Treat it like an ordinary equation: $2\mathbf{x}+\left(\begin{array}{c}1\\ 5\end{array}\right)=\left(\begin{array}{c}9\\ -3\end{array}\right)$ gives $2\mathbf{x}=\left(\begin{array}{c}8\\ -8\end{array}\right)$, so $\mathbf{x}=\left(\begin{array}{c}4\\ -4\end{array}\right)$.
• Finding a number that makes vectors parallel. If $\left(\begin{array}{c}k\\ 6\end{array}\right)$ is parallel to $\left(\begin{array}{c}3\\ 2\end{array}\right)$, the multiplier is $6÷2=3$, so $k=3\times 3=9$.
• Writing one vector as a combination of two others. To find $m$ and $n$ with $m\mathbf{a}+n\mathbf{b}=\mathbf{c}$, write one equation for each component and solve them simultaneously.

Always check your answer by substituting back into the original vector equation.