Question

Hi
Recall that we can thought of a projective plane as an upper hemisphere. We define the binary operation on a projective plane as follows: Let x and y be any two points belong to the projective plane, think of the projective plane as an upper hemisphere. Then take the antipodal point of y which is in the lower hemispher and identified with y. Draw a line through x and the antipodal point of y. Then draw a line parallel to that line with same length and the origin is the mid point, x*y is then defined as the endpoint of this line (Since the end points are antipodal so they are equal in RP).
My question is
Is it true that x*(y*z)=(x*y)*(x*z) for x,y,z in RP
If yes, then can you prove it for me please

thank you in advance

Answer 1

Answer on Question 37613 - Math - Geometry

Recall that we can thought of a projective plane as an upper hemisphere. We define the binary operation on a projective plane as follows: Let $x$ and $y$ be any two points belong to the projective plane, think of the projective plane as an upper hemisphere. Then take the antipodal point of $y$ which is in the lower hemispher and identified with $y$ . Draw a line through $x$ and the antipodal point of $y$ . Then draw a line parallel to that line with same length and the origin is the mid point, $x^*y$ is then defined as the endpoint of this line (Since the end points are antipodal so they are equal in RP).

My question is

Is it true that $x^{*}(y^{*}z) = (x^{*}y)^{*}(x^{*}z)$ for $x, y, z$ in RP

Solving

Assume that it is true. Then this equation holds for any three points that belong to the upper hemisphere. Let points $x, y, z$ has coordinates $(0, 0, 1), \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right), \left( \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ respectively. And for simplicity we assume that center of the sphere at the origin of coordinates and radios equals 1. Then equation of the sphere is $x^2 + y^2 + z^2 = 1$ .

Let's find for this points $x^{*}(y^{*}z)$ and $(x^{*}y)^{*}(x^{*}z)$ .

$x^{*}y$ : -y has coordinates $\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$ . We should find coordinates of point Q. The point -y devides the segment xQ in the ratio of 2 to 1. Using the formula of dividing

the segment in the given ratio we get

$$- \frac {1}{\sqrt {3}} = \frac {0 + 2 * x}{3} = \frac {2 x}{3}$$ $$x = - \frac {3}{2 \sqrt {3}}$$

Similarly we get

$$y = - \frac {3}{2 \sqrt {3}}, z = - \frac {3 + \sqrt {3}}{2 \sqrt {3}}.$$

Equation of the line QO is

$$\left\{ \begin{array}{l} x = \frac {3}{2 \sqrt {3}} t \\ y = \frac {3}{2 \sqrt {3}} t \\ z = \frac {3 + \sqrt {3}}{2 \sqrt {3}} t \end{array} \right.$$

To find the coordinates of the point $x^{*}y$ we should substitute this

into equation of the sphere

$$\left(\frac {3}{2 \sqrt {3}} t\right) ^ {2} + \left(\frac {3}{2 \sqrt {3}} t\right) ^ {2} + \left(\frac {3 + \sqrt {3}}{2 \sqrt {3}} t\right) ^ {2} = 1$$ $$t = \frac {1}{\sqrt {5 + \sqrt {3}}}$$ $$x ^ {*} y: \left(\frac {3}{2 \sqrt {1 5 + 3 \sqrt {3}}}, \frac {3}{2 \sqrt {1 5 + 3 \sqrt {3}}}, \frac {3 + \sqrt {3}}{2 \sqrt {1 5 + 3 \sqrt {3}}}\right).$$

$x^{*}z$ : -z has coordinates $(- \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}})$ . Similarly to the previous case we should find coordinates of point $Q'$ .

$$- \frac {1}{\sqrt {3}} = \frac {0 + 2 * x}{3} = \frac {2 x}{3}, \quad x = - \frac {3}{2 \sqrt {3}}$$ $$\begin{array}{l} \frac{1}{\sqrt{3}} = \frac{0 + 2 \cdot y}{3} = \frac{2y}{3}, \quad y = \frac{3}{2\sqrt{3}} \\ - \frac{1}{\sqrt{3}} = \frac{1 + 2z}{3}, \quad z = - \frac{3 + \sqrt{3}}{2\sqrt{3}}. \end{array}$$

Equation of the line $Q'O$ is

$$\left\{ \begin{array}{l} x = \frac{3}{2\sqrt{3}} t \\ y = - \frac{3}{2\sqrt{3}} t \\ z = \frac{3 + \sqrt{3}}{2\sqrt{3}} t \end{array} \right.$$

To find the coordinates of the point $x^{*}z$ we should substitute this into equation of the sphere

$$\left(\frac{3}{2\sqrt{3}} t\right)^{2} + \left(- \frac{3}{2\sqrt{3}} t\right)^{2} + \left(\frac{3 + \sqrt{3}}{2\sqrt{3}} t\right)^{2} = 1$$ $$t = \frac{1}{\sqrt{5 + \sqrt{3}}}$$ $$x^{*}z: \left(\frac{3}{2\sqrt{15 + 3\sqrt{3}}}, -\frac{3}{2\sqrt{15 + 3\sqrt{3}}}, \frac{3 + \sqrt{3}}{2\sqrt{15 + 3\sqrt{3}}}\right)$$

$y^{*}z$: -z has coordinates $\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$. Similarly to the previous cases we should find coordinates of point $Q''$.

$$\begin{array}{l} - \frac{1}{\sqrt{3}} = \frac{\frac{1}{\sqrt{3}} + 2 \cdot x}{3} = \frac{1 + 2\sqrt{3}x}{3}, \quad x = - \frac{2}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} = \frac{\frac{1}{\sqrt{3}} + 2 \cdot y}{3} = \frac{1 + 2\sqrt{3}y}{3}, \quad y = \frac{1}{\sqrt{3}} \\ - \frac{1}{\sqrt{3}} = \frac{\frac{1}{\sqrt{3}} + 2 \cdot z}{3} = \frac{1 + 2\sqrt{3}z}{3}, \quad z = - \frac{2}{\sqrt{3}} \\ \end{array}$$

Equation of the line $Q''O$ is

$$\left\{ \begin{array}{l} x = - \frac{2}{\sqrt{3}} t \\ y = \frac{1}{\sqrt{3}} t \\ z = - \frac{2}{\sqrt{3}} t \end{array} \right.$$

To find the coordinates of the point $x^{*}z$ we should substitute this into equation of the sphere

$$\left(- \frac{2}{\sqrt{3}} t\right)^{2} + \left(\frac{1}{\sqrt{3}} t\right)^{2} + \left(- \frac{2}{\sqrt{3}} t\right)^{2} = 1$$ $$t = \frac{1}{\sqrt{3}}$$ $$y^{*}z: \left(-\frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\right)$$

Now we can find $x^{*}(y^{*}z)$.

$$y^{*}z \text{ is } \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$$

The coordinates of the endpoint $K$ is

$$\frac{2}{3} = \frac{0 + 2 \cdot x}{3} = \frac{2x}{3}, \quad x = 1$$ $$\begin{array}{l} - \frac {1}{3} = \frac {2 y}{3}, \quad y = - \frac {1}{2} \\ - \frac {2}{3} = \frac {1 + 2 z}{3}, \quad z = - \frac {3}{2} \\ \end{array}$$

Equation of the line KO is

$$\left\{ \begin{array}{l} x = - t \\ y = \frac {1}{2} t \\ z = \frac {3}{2} t \end{array} \right.$$

To find the coordinates of the point $x^{*}(y^{*}z)$ we should substitute this into equation of the sphere

$$(- t) ^ {2} + \left(\frac {1}{2} t\right) ^ {2} + \left(\frac {3}{2} t\right) ^ {2} = 1$$ $$t = \frac {2}{\sqrt {14}}$$ $$x ^ {*} (y ^ {*} z): \left(- \frac {2}{\sqrt {14}}, \frac {1}{\sqrt {14}}, \frac {3}{\sqrt {14}}\right).$$

It is remain to find $(x^{*}y)^{*}(x^{*}z)$

$$-(x ^ {*} z) \text{ has coordinates} \left(- \frac {3}{2 \sqrt {15 + 3 \sqrt {3}}}, \frac {3}{2 \sqrt {15 + 3 \sqrt {3}}}, - \frac {3 + \sqrt {3}}{2 \sqrt {15 + 3 \sqrt {3}}}\right)$$

Let us find the point $K'$

$$\begin{array}{l} - \frac {3}{2 \sqrt {15 + 3 \sqrt {3}}} = \frac {\frac {3}{2 \sqrt {15 + 3 \sqrt {3}}} + 2 x}{3}, \quad x = - \frac {3}{\sqrt {15 + 3 \sqrt {3}}} \\ \frac {3}{2 \sqrt {15 + 3 \sqrt {3}}} = \frac {\frac {3}{2 \sqrt {15 + 3 \sqrt {3}}} + 2 y}{3}, \quad y = \frac {3}{2 \sqrt {15 + 3 \sqrt {3}}} \\ - \frac {3 + \sqrt {3}}{2 \sqrt {15 + 3 \sqrt {3}}} = \frac {\frac {3 + \sqrt {3}}{2 \sqrt {15 + 3 \sqrt {3}}} + 2 z}{3}, \quad z = \frac {\sqrt {3} + 1}{\sqrt {5 + \sqrt {3}}} \\ \end{array}$$

Equation of the line $K^{\prime}O$ is

$$\left\{ \begin{array}{l} x = - \frac {3}{\sqrt {15 + 3 \sqrt {3}}} t \\ y = \frac {3}{2 \sqrt {15 + 3 \sqrt {3}}} t \\ z = \frac {3 + \sqrt {3}}{\sqrt {15 + 3 \sqrt {3}}} t \end{array} \right.$$

To find the coordinates of the point $(x^{*}y)^{*}(x^{*}z)$ we should substitute this into equation of the sphere

$$\begin{array}{l} \left(- \frac {3}{\sqrt {15 + 3 \sqrt {3}}} t\right) ^ {2} + \left(\frac {3}{2 \sqrt {15 + 3 \sqrt {3}}} t\right) ^ {2} + \left(\frac {3 + \sqrt {3}}{\sqrt {15 + 3 \sqrt {3}}} t\right) ^ {2} = 1 \\ \frac {9}{15 + 3 \sqrt {3}} t ^ {2} + \frac {9}{4 (15 + 3 \sqrt {3})} t ^ {2} + \frac {\left(3 + \sqrt {3}\right) ^ {2}}{15 + 3 \sqrt {3}} t ^ {2} = 1 \\ (93 + 24 \sqrt {3}) t ^ {2} = 4 (15 + 3 \sqrt {3}) \\ t = \sqrt {\frac {4 (15 + 3 \sqrt {3})}{(93 + 24 \sqrt {3})}} \\ \end{array}$$ $$(x^*y)^*(x^*z): \left(-\frac{6}{\sqrt{(93+24\sqrt{3})}}, \frac{3}{\sqrt{(93+24\sqrt{3})}}, \frac{2(3+\sqrt{3})}{\sqrt{(93+24\sqrt{3})}}\right)$$

Now we can see that $x^*(y^*z) \neq (x^*y)^*(x^*z)$.

**Answer**: it is no true.