# [SOLVED]Cross Ratio

#### Poirot

##### Banned
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

##### Well-known member
MHB Math Helper
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 >>here<<.

Kind Regards,
Sudharaka.

#### Opalg

##### MHB Oldtimer
Staff member
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.