MHB Find Point E in Harmonic Range with A, B, C using Cross Ratio -1

[Poirot1]

A=$\begin{bmatrix}1\\1\\0\end{bmatrix}$, B=$\begin{bmatrix}3\\1\\-1\end{bmatrix}$

C=$\begin{bmatrix}5\\3\\-1\end{bmatrix}$, D=$\begin{bmatrix}4\\0\\-2\end{bmatrix}$

Find the coordinates of a point E such that A,B,C,E forms a harmonic range. This means that the cross ratio is -1.

 

[Sudharaka]

A=$\begin{bmatrix}1\\1\\0\end{bmatrix}$, B=$\begin{bmatrix}3\\1\\-1\end{bmatrix}$

C=$\begin{bmatrix}5\\3\\-1\end{bmatrix}$, D=$\begin{bmatrix}4\\0\\-2\end{bmatrix}$

Find the coordinates of a point E such that A,B,C,E forms a harmonic range. This means that the cross ratio is -1.

Hi Poirot, :)

Let, \(E\equiv (x,y,z)\). Write down the cross ratio considering the \(x\), \(y\) and \(z\) coordinates of the points \(A\), \(B\), \(C\) and \(E\). For example,

\[(A,B,C,E)=\frac{AC}{BC}: \frac{AE}{AB}=\frac{1-5}{3-5}:\frac{1-x}{1-3}\]

Since \(A\), \(B\), \(C\) and \(E\) are in harmonic range, \((A,B,C,E)=-1\). Therefore,

\[(A,B,C,E)=\frac{1-5}{3-5}:\frac{1-x}{1-3}=-1\]

Find \(x\). Similarly you can also find \(y\) and \(z\). More information about Cross ratios can be found http://www.ping.be/math/cross.htm#Dividing-ratio.

Kind Regards,
Sudharaka.

 

[Opalg]

A=$\begin{bmatrix}1\\1\\0\end{bmatrix}$, B=$\begin{bmatrix}3\\1\\-1\end{bmatrix}$

C=$\begin{bmatrix}5\\3\\-1\end{bmatrix}$, D=$\begin{bmatrix}4\\0\\-2\end{bmatrix}$

Find the coordinates of a point E such that A,B,C,E forms a harmonic range. This means that the cross ratio is -1.

There is something puzzling about this question. A harmonic range is normally only defined for four collinear points. If you have three collinear points then you can find a fourth point on the line making up a harmonic range, using the method described by Sudharaka. But in this problem the three given points are not collinear. You can easily see this because the points $A$ and $B$ have the same $y$-coordinate 1, but $C$ has $y$-coordinate 3. If the points were collinear that could not happen.