1 Two-dimensional coordinate frames

Figure 21 shows a two-dimensional coordinate frame. Any point $P$ in the $xy$ plane can be defined in terms of its $x$ and $y$ coordinates.

A unit vector pointing in the positive direction of the $x$ -axis is denoted by $\underset{̲}{i}$ . (Note that it is common practice to write this particular unit vector without the hat $\stackrel{̂}{}$ .) It follows that any vector in the direction of the $x$ -axis will be a multiple of $\underset{̲}{i}$ . Figure 22 shows vectors $\underset{̲}{i}$ , $2\underset{̲}{i}$ , $5\underset{̲}{i}$ and $-3\underset{̲}{i}$ . In general a vector of length $\ell$ in the direction of the $x$ -axis can be written $\ell \underset{̲}{i}$ .

Similarly, a unit vector pointing in the positive $y$ -axis is denoted by $\underset{̲}{j}$ . So any vector in the direction of the $y$ -axis will be a multiple of $\underset{̲}{j}$ . Figure 23 shows $\underset{̲}{j}$ , $4\underset{̲}{j}$ and $-2\underset{̲}{j}$ . In general a vector of length $\ell$ in the direction of the $y$ -axis can be written $\ell \underset{̲}{j}$ .

Example 3

Draw the vectors $5\underset{̲}{i}$ and $4\underset{̲}{j}$ . Use your diagram and the triangle law of addition to add these two vectors together. First draw the vectors $5\underset{̲}{i}$ and $4\underset{̲}{j}$ . Then, by translating the vectors so that they lie head to tail, find the vector sum $5\underset{̲}{i}+4\underset{̲}{j}$ .

Solution

We now generalise the situation in Example 3. Consider Figure 25.

It shows a vector $\underset{̲}{r}=\overrightarrow{AB}$ . We can regard $\underset{̲}{r}$ as being the resultant of the two vectors $\overrightarrow{AC}=a\underset{̲}{i}$ , and $\overrightarrow{CB}=b\underset{̲}{j}$ . From the triangle law of vector addition

We conclude that any vector in the $xy$ plane can be expressed in the form $\underset{̲}{r}=a\underset{̲}{i}+b\underset{̲}{j}$ . The numbers $a$ and $b$ are called the components of $\underset{̲}{r}$ in the $x$ and $y$ directions. Sometimes, for emphasis, we will use $a_x$ and $a_y$ instead of $a$ and $b$ to denote the components in the $x$ - and $y$ -directions respectively. In that case we would write $\underset{̲}{r}=a_x\underset{̲}{i}+a_y\underset{̲}{j}$ .

1.1 Column vector notation

An alternative, useful, and often briefer notation is to write the vector $\underset{̲}{r}=a\underset{̲}{i}+b\underset{̲}{j}$ in column vector notation as

$\underset{̲}{r}=\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \end{array}\right)$

Task!

1. Draw an $xy$ plane and show the vectors $\underset{̲}{p}=2\underset{̲}{i}+3\underset{̲}{j}$ , and $\underset{̲}{q}=5\underset{̲}{i}+\underset{̲}{j}$ .
2. Express $\underset{̲}{p}$ and $\underset{̲}{q}$ using column vector notation.
3. By translating one of the vectors apply the triangle law to show the sum $\underset{̲}{p}+\underset{̲}{q}$ .
4. Express the resultant $\underset{̲}{p}+\underset{̲}{q}$ in terms of $\underset{̲}{i}$ and $\underset{̲}{j}$ .

1. Draw the $xy$ plane and the required vectors.

   Answer

   (They can be drawn from any point in the plane)

2. The column vector form of $\underset{̲}{p}$ is $\left(\begin{array}{c}2\\ 3\end{array}\right)$ . Write down the column vector form of $\underset{̲}{q}$ :

   Answer

   $\underset{̲}{p}=\left(\begin{array}{c}2\\ 3\end{array}\right)$ $\underset{̲}{q}=\left(\begin{array}{c}5\\ 1\end{array}\right)$

3. Translate one of the vectors in part (1) so that they lie head to tail, completing the third side of the triangle to give the resultant $\underset{̲}{p}+\underset{̲}{q}$ :

   Answer

   Note that the vectors have not been drawn to scale.

   

4. By studying your diagram note that the resultant has two components $7\underset{̲}{i}$ , horizontally, and $4\underset{̲}{j}$ vertically. Hence write down an expression for $\underset{̲}{p}+\underset{̲}{q}$ :

   Answer

   $7\underset{̲}{i}+4\underset{̲}{j}$

It is very important to note from the last task that vectors in Cartesian form can be added by simply adding their respective $\underset{̲}{i}$ and $\underset{̲}{j}$ components.

Thus, if $\underset{̲}{a}=a_x\underset{̲}{i}+a_y\underset{̲}{j}$ and $\underset{̲}{b}=b_x\underset{̲}{i}+b_y\underset{̲}{j}$ then

$\underset{̲}{a}+\underset{̲}{b}=\left(a_x+b_x\right)\underset{̲}{i}+\left(a_y+b_y\right)\underset{̲}{j}$

A similar, and obvious, rule applies when subtracting:

$\underset{̲}{a}-\underset{̲}{b}=\left(a_x-b_x\right)\underset{̲}{i}+\left(a_y-b_y\right)\underset{̲}{j}$ .

Task!

If $\underset{̲}{a}=9\underset{̲}{i}+7\underset{̲}{j}$ and $\underset{̲}{b}=8\underset{̲}{i}+3\underset{̲}{j}$ find

1. $\underset{̲}{a}+\underset{̲}{b}$
2. $\underset{̲}{a}-\underset{̲}{b}$

Answer

Now consider the special case when $\underset{̲}{r}$ represents the vector from the origin $O$ to the point $P\left(a,b\right)$ . This vector is known as the position vector of $P$ and is shown in Figure 26.

Unlike most vectors, position vectors cannot be freely translated. Because they indicate the position of a point they are fixed vectors in the sense that the tail of a position vector is always located at the origin.

Example 4

State the position vectors of the points with coordinates

1. $P\left(2,4\right)$ ,
2. $Q\left(-1,5\right)$ ,
3. $R\left(-1,-7\right)$ ,
4. $S\left(8,-4\right)$ .

Solution

1. $2\underset{̲}{i}+4\underset{̲}{j}$ .
2. $-\underset{̲}{i}+5\underset{̲}{j}$ .
3. $-\underset{̲}{i}-7\underset{̲}{j}$ .
4. $8\underset{̲}{i}-4\underset{̲}{j}$ .

Example 5

Sketch the position vectors ${\underset{̲}{r}}_1=3\underset{̲}{i}+4\underset{̲}{j}$ , ${\underset{̲}{r}}_2=-2\underset{̲}{i}+5\underset{̲}{j}$ , ${\underset{̲}{r}}_3=-3\underset{̲}{i}-2\underset{̲}{j}$ .

Solution

The vectors are shown below. Note that all position vectors start at the origin.

Figure 27

The modulus of any vector $\underset{̲}{r}$ is equal to its length. As we have noted earlier, the modulus of $\underset{̲}{r}$ is usually denoted by $\left|\underset{̲}{r}\right|$ . When $\underset{̲}{r}=a\underset{̲}{i}+b\underset{̲}{j}$ the modulus can be obtained using Pythagoras’ theorem. If $\underset{̲}{r}$ is the position vector of point $P$ then the modulus is, clearly, the distance of $P$ from the origin.

Example 6

Find the modulus of each of the vectors shown in Example 5.

Solution

1. $\left|{\underset{̲}{r}}_1\right|=|3\underset{̲}{i}+4\underset{̲}{j}|$ = $\sqrt{3^2+4^2}=\sqrt{25}=5$ .
2. $\left|{\underset{̲}{r}}_2\right|=\sqrt{{\left(-2\right)}^2+5^2}=\sqrt{4+25}=\sqrt{29}$ .
3. $\left|{\underset{̲}{r}}_3\right|=\sqrt{{\left(-3\right)}^2+{\left(-2\right)}^2}=\sqrt{9+4}=\sqrt{13}$

Task!

Point $A$ has coordinates $\left(3,5\right)$ . Point $B$ has coordinates $\left(7,8\right)$ .

1. Draw a diagram which shows points $A$ and $B$ and draw the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ :

   Answer

   

2. State the position vectors of $A$ and $B$ :

   Answer

   $\overrightarrow{OA}=\underset{̲}{a}=3\underset{̲}{i}+5\underset{̲}{j}$ , $\overrightarrow{OB}=\underset{̲}{b}=7\underset{̲}{i}+8\underset{̲}{j}$ (c) Referring to your figure and using the triangle law you can write $\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$ so that $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ . Hence write down an expression for $\overrightarrow{AB}$ in terms of the unit vectors $\underset{̲}{i}$ and $\underset{̲}{j}$ :

   Answer

   $\overrightarrow{AB}=\left(7\underset{̲}{i}+8\underset{̲}{j}\right)-\left(3\underset{̲}{i}+5\underset{̲}{j}\right)=4\underset{̲}{i}+3\underset{̲}{j}$ (d) Calculate the length of $\overrightarrow{AB}=|4\underset{̲}{i}+3\underset{̲}{j}|$ :

   Answer

   $\left|\overrightarrow{AB}\right|=\sqrt{4^2+3^2}=\sqrt{25}=5$

Exercises

1. Explain the distinction between a position vector, and a more general free vector.
2. What is meant by the symbols $\underset{̲}{i}$ and $\underset{̲}{j}$ ?
3. State the position vectors of the points with coordinates
   1. $P\left(4,7\right)$ ,
   2. $Q\left(-3,5\right)$ ,
   3. $R\left(0,3\right)$ ,
   4. $S\left(-1,0\right)$
4. State the coordinates of the point $P$ if its position vector is:
   1. $3\underset{̲}{i}-7\underset{̲}{j}$ ,
   2. $-4\underset{̲}{i}$ ,
   3. $-0.5\underset{̲}{i}+13\underset{̲}{j}$ ,
   4. $a\underset{̲}{i}+b\underset{̲}{j}$
5. Find the modulus of each of the following vectors:
   1. $\underset{̲}{r}=7\underset{̲}{i}+3\underset{̲}{j}$ ,
   2. $\underset{̲}{r}=17\underset{̲}{i}$ ,
   3. $\underset{̲}{r}=2\underset{̲}{i}-3\underset{̲}{j}$ ,
   4. $\underset{̲}{r}=-3\underset{̲}{j}$ ,
   5. $\underset{̲}{r}=a\underset{̲}{i}+b\underset{̲}{j}$ ,
   6. $\underset{̲}{r}=a\underset{̲}{i}-b\underset{̲}{j}$
6. Point $P$ has coordinates $\left(7,8\right)$ . Point $Q$ has coordinates $\left(-2,4\right)$ .
   1. Draw a sketch showing vectors $\overrightarrow{OP}$ , $\overrightarrow{OQ}$
   2. State the position vectors of $P$ and $Q$ ,
   3. Find an expression for $\overrightarrow{PQ}$ ,
   4. Find $\left|\overrightarrow{PQ}\right|$ .

Answer