Vectors

What a vector is

A vector is a movement with two pieces of information: how far, and in which direction. A scalar is an ordinary number with size only, so 5 kilometres is a scalar and 5 kilometres due north is a vector. Vectors are written as two numbers in a column, $\left(\begin{array}{c}x\\ y\end{array}\right)$, where the top number is the movement across and the bottom number is the movement up. Right and up are positive; left and down are negative.

A vector also has a name. A single bold letter such as $𝐚$ names one vector, and a directed line segment names the movement from one point to another: $\overrightarrow{AB}$ means the vector from $A$ to $B$. The arrow is the important part, because $\overrightarrow{BA}$ is the same journey in the opposite direction.

Two vectors are equal when they have the same length and the same direction. That means the same column of numbers, wherever the arrows happen to be drawn on the page: a vector is a movement, not a position.

Writing a translation as a vector

A translation moves every point of a shape by the same vector. If a point $A$ moves to a point $B$, the translation is the vector $\overrightarrow{AB}$, and you find it by subtracting: $\overrightarrow{AB}=B-A$, taking the coordinates in the same order both times.

For example, if $A$ is $(1,4)$ and $B$ is $(5,2)$, then $\overrightarrow{AB}=\left(\begin{array}{c}5-1\\ 2-4\end{array}\right)=\left(\begin{array}{c}4\\ -2\end{array}\right)$: 4 to the right and 2 down. Every other point of the shape moves by exactly the same column, which is the check that your vector is right.

• $A(2,3)$ to $B(7,3)$: $\overrightarrow{AB}=\left(\begin{array}{c}5\\ 0\end{array}\right)$, a movement to the right with no vertical change
• $A(4,6)$ to $B(4,1)$: $\overrightarrow{AB}=\left(\begin{array}{c}0\\ -5\end{array}\right)$, straight down
• The vector of a translation is the same for every point of the shape, so any pair of matching points gives the same column - a useful check on your subtraction

Adding and subtracting vectors

Adding vectors is following one movement and then another. If you go from $A$ to $B$ and then from $B$ to $C$, you end up at $C$, so $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. This is the rule that solves almost every vector question, and it works because the two arrows are joined head to tail.

In a column, addition is done one line at a time: add the top numbers together and add the bottom numbers together. Subtraction works the same way, one line at a time, and $\overrightarrow{BA}=-\overrightarrow{AB}$, because going back along the same path is the same movement with every sign changed.

Going all the way round a closed shape brings you back to where you started, so the vectors of a closed path add up to the zero vector $\left(\begin{array}{c}0\\ 0\end{array}\right)$. That is a quick way to check a path you have built up yourself.

Multiplying a vector by a scalar

Multiplying a vector by a number, called a scalar, stretches it. $3𝐚$ means three copies of $𝐚$ laid end to end, so both components are multiplied by 3: if $𝐚=\left(\begin{array}{c}3\\ -1\end{array}\right)$ then $3𝐚=\left(\begin{array}{c}9\\ -3\end{array}\right)$. A negative scalar reverses the direction as well as changing the length, so $-𝐚$ is the same length as $𝐚$ but points the opposite way.

A scalar multiple always produces a parallel vector, in the same direction if the scalar is positive and in the opposite direction if it is negative. That gives a reliable test: if $𝐪=k𝐩$ for some number $k$, the two vectors are parallel, and $k$ must come out the same from the top line and the bottom line. For $𝐩=\left(\begin{array}{c}2\\ 3\end{array}\right)$ and $𝐪=\left(\begin{array}{c}-6\\ -9\end{array}\right)$, the top line gives $-6÷2=-3$ and the bottom gives $-9÷3=-3$, so $𝐪=-3𝐩$ and the vectors are parallel.

You can combine both operations in one expression. $3𝐚-2𝐛$ means multiply $𝐚$ by 3, multiply $𝐛$ by 2, then subtract line by line, which is exactly how a vector question with two given vectors is marked.

• $2\left(\begin{array}{c}3\\ -1\end{array}\right)=\left(\begin{array}{c}6\\ -2\end{array}\right)$
• $\frac{1}{2}\left(\begin{array}{c}8\\ 6\end{array}\right)=\left(\begin{array}{c}4\\ 3\end{array}\right)$
• $-3\left(\begin{array}{c}2\\ 3\end{array}\right)=\left(\begin{array}{c}-6\\ -9\end{array}\right)$
• The scalar multiplies both numbers, never just the top one

The magnitude of a vector

The magnitude, or modulus, of a vector is its length. It is written with straight bars, $|𝐚|$ or $|\overrightarrow{AB}|$, and the bars are what tell you a length is wanted rather than the vector itself. The two components of a vector are at right angles, so the length is the hypotenuse of a right-angled triangle and Pythagoras gives the rule: $\left|\right(\begin{array}{c}x\\ y\end{array}\left)\right|=\sqrt{x^2+y^2}$.

Square both components, add, then take the square root. The squaring removes the signs, so a vector and its negative always have the same magnitude: $\left(\begin{array}{c}-3\\ -4\end{array}\right)$ is as long as $\left(\begin{array}{c}3\\ 4\end{array}\right)$, both of length 5. Leave the answer as a surd such as $\sqrt{65}$ unless the question asks for a decimal, and never write a magnitude with a minus sign - a length cannot be negative.

The same rule finds the distance between two points. Work out the vector between them first, then take its magnitude: for $A(1,2)$ and $B(7,10)$, $\overrightarrow{AB}=\left(\begin{array}{c}6\\ 8\end{array}\right)$ and $|\overrightarrow{AB}|=\sqrt{36+64}=10$.

Using vectors to describe a path

Most vector questions are about a path through a shape, and the shape you meet most is the triangle $OAB$, where $O$ is the origin. If $\overrightarrow{OA}=𝐚$ and $\overrightarrow{AB}=𝐛$, then the direct route from $O$ to $B$ is $\overrightarrow{OB}=𝐚+𝐛$, because following the arrows from $O$ to $A$ and then from $A$ to $B$ is the same as going straight from $O$ to $B$.

A position vector is the vector from the origin to a point, so $\overrightarrow{OA}$ is the position vector of $A$. The vector between any two points is the second position vector minus the first: $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$. Rearranged, that is the same statement as $\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$, which is why it is worth remembering the path rule rather than a formula.

Midpoints and ratios are just fractions of a vector. If $M$ is the midpoint of $AB$ then $\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}$, and if $P$ divides $AB$ in the ratio $1:2$ then $\overrightarrow{AP}=\frac{1}{3}\overrightarrow{AB}$, because the whole side is split into three equal parts. Write the path you need as a chain of vectors you know, then substitute.

• With $𝐚=\left(\begin{array}{c}1\\ 3\end{array}\right)$ and $𝐛=\left(\begin{array}{c}5\\ 1\end{array}\right)$: $\overrightarrow{OB}=𝐚+𝐛=\left(\begin{array}{c}6\\ 4\end{array}\right)$
• The midpoint $M$ of $OB$ has $\overrightarrow{OM}=\frac{1}{2}(𝐚+𝐛)=\left(\begin{array}{c}3\\ 2\end{array}\right)$
• $\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=\left(\begin{array}{c}3\\ 2\end{array}\right)-\left(\begin{array}{c}1\\ 3\end{array}\right)=\left(\begin{array}{c}2\\ -1\end{array}\right)$
• All the way round the triangle: $\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BO}=\left(\begin{array}{c}0\\ 0\end{array}\right)$