Fluency exercise

Solution

A square is enclosed by the lines:

1. $\mathbf{r}=\begin{pmatrix}0\\2\end{pmatrix} + a\begin{pmatrix}1\\1\end{pmatrix}$
1. $\mathbf{r}=\begin{pmatrix}0\\6\end{pmatrix} + b\begin{pmatrix}-1\\1\end{pmatrix}$
1. $\mathbf{r}=\begin{pmatrix}2\\1\end{pmatrix} + c\begin{pmatrix}2\\2\end{pmatrix}$
1. $\mathbf{r}=\begin{pmatrix}1\\2\end{pmatrix} + d\begin{pmatrix}1\\-1\end{pmatrix}$

Can you convince yourself that the adjacent edges are perpendicular?

We need to consider the direction vector of each line. Lines (1) and (2) do not have the same direction vector and so they will cross at some point. If we now think about the behaviour of each line as it moves away from the crossing point we can show that they are perpendicular using the diagram below.

Can you use a similar argument to convince yourself about the other adjacent edges of the square?

Find the area of the square.

To do this we need to identify the positions of some of the vertices. We could aim to find the length of one side by finding two adjacent vertices, or we could find the length of the diagonal by identifying vertices opposite each other. It doesn’t matter which pair we choose to find so let’s say we look for vertices A and B shown on the diagram below.

To find the position of point A, we need to find the point that is common to equations (1) and (4), (they need to have the same position vector $\mathbf{r}$). With that in mind, we can write down that $a=1+d$ and $2+a=2-d$. Solving these equations simultaneously we find that $a=\frac{1}{2}$ and $d=-\frac{1}{2}$.

Substituting these values into either equation (1) or equation (4) we find the position vector of point A to be $\mathbf{r}=\begin{pmatrix}\frac{1}{2}\\\frac{5}{2}\end{pmatrix}$.

Repeating the process to find the position of point B, from equations (1) and (2) we can write down that $a=-b$ and $2 + a = 6 + b$. Solving these equations simultaneously we find that $a=2$ and $b=-2$.

Substituting these values into either equation (1) or equation (2) we find the position vector of point B to be $\mathbf{r}=\begin{pmatrix}2\\4\end{pmatrix}$.

We might have found the position of vertex B in another way. The line represented by vector equation (1) could have been constructed by starting at the position vector $\begin{pmatrix}0\\2\end{pmatrix}$ and ‘stepping out’ the direction vector as shown below. If the same process was then repeated to construct the line represented by vector equation (2) then we notice that the two lines intersect at point B with position vector $\begin{pmatrix}2\\4\end{pmatrix}$.

If the process is repeated to construct the lines represented by vector equations (3) and (4) then we notice that this pair intersect at $\begin{pmatrix}2\\1\end{pmatrix}$, forming the vertex opposite B in the enclosed square. These two vertices are aligned vertically and the line joining them forms the diagonal, length 3, of the enclosed square.

Now that we have the position vectors of vertices A and B we could find the length of the line AB using Pythagoras.

However, you might notice that the area of the square will be given by $AB^2$ and we can write this down directly, $AB^2 = 2 \times \left(\frac{3}{2}\right)^2 = \frac{9}{2}.$

You could also have noticed that the right-angled triangle joining points A and B is isosceles. This tells us that the square could be divided up into four of these triangles.

We can now show that the area of the square is again equal to $4 \times \frac{1}{2}\left(\frac{3}{2}\right)^2 = \frac{9}{2}.$

What can we now say about the positions of the other vertices and the diagonals of the square?

Find a vector equation for each of the two diagonals.

In order to write down a vector equation of a line we need to know the position of a point on that line and a direction vector. For the vertical diagonal of the square we know that point B, with position vector $\begin{pmatrix}2\\4\end{pmatrix}$ lies on the line and that the direction vector is $\begin{pmatrix}0\\1\end{pmatrix}$.

A vector equation of the vertical diagonal of the square is therefore $\mathbf{r}=\begin{pmatrix}2\\4\end{pmatrix} + p\begin{pmatrix}0\\1\end{pmatrix}.$

We could also have written $\mathbf{r}=\begin{pmatrix}2\\1\end{pmatrix} + p\begin{pmatrix}0\\1\end{pmatrix},$ or any vector equation of the form $\mathbf{r}=\begin{pmatrix}2\\s\end{pmatrix} + p\begin{pmatrix}0\\q\end{pmatrix}$ for any non-zero $q$.

The horizontal diagonal could be written as any vector equation of the form $\mathbf{r}=\begin{pmatrix}s\\\frac{5}{2}\end{pmatrix} + p\begin{pmatrix}q\\0\end{pmatrix}$ for any non-zero $q$.