Post by Admin on Oct 13, 2016 7:10:37 GMT
Gary
This exercise is posed near the beginning of paragraph 2 on page 305.
The result we are being asked to show is analogous to the result in Euclidean and spherical geometry:
There is only one line through $p$ orthogonal to the line $L$, and this line can be constructed as the line through $p$ and its reflection in $L$.
We can refer to figure 24 on page 304. We must also remember the definition of reflection as described in the paragraph immediately following (38) on page 303.
If we have a point $p$ on the map, then since $\mathcal{I}_L(p)=\mathfrak{R}_L(p)$, we can construct $\mathfrak{R}_L(p)$ as the inverse point to $p$ in $L$. We can now find the centre of the h-line through these two points by constructing the perpendicular bisector to the line joining them and finding its point of intersection with the horizon. If we now draw the unique h-line through the points $p$ and $\mathfrak{R}_L(p)$, symmetric with respect to $L$, it must cut $L$ at right angles (see page 129 and figure 6a).
So we have shown that $M$ may be constructed as the unique h-line through $p$ and its reflection $\mathfrak{R}_L(p)$ in $L$.
Vasco
This exercise is posed near the beginning of paragraph 2 on page 305.
The result we are being asked to show is analogous to the result in Euclidean and spherical geometry:
There is only one line through $p$ orthogonal to the line $L$, and this line can be constructed as the line through $p$ and its reflection in $L$.
We can refer to figure 24 on page 304. We must also remember the definition of reflection as described in the paragraph immediately following (38) on page 303.
If we have a point $p$ on the map, then since $\mathcal{I}_L(p)=\mathfrak{R}_L(p)$, we can construct $\mathfrak{R}_L(p)$ as the inverse point to $p$ in $L$. We can now find the centre of the h-line through these two points by constructing the perpendicular bisector to the line joining them and finding its point of intersection with the horizon. If we now draw the unique h-line through the points $p$ and $\mathfrak{R}_L(p)$, symmetric with respect to $L$, it must cut $L$ at right angles (see page 129 and figure 6a).
So we have shown that $M$ may be constructed as the unique h-line through $p$ and its reflection $\mathfrak{R}_L(p)$ in $L$.
Vasco
