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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.

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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.

- Feb 5, 2012

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Hi Poirot,

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.

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.

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- Feb 7, 2012

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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.

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.