§ 7: answers

O27
a) Each plane through the normal of a plane α is perpendicular to α.
b) Two lines l and m through the origin in ℜ³are perpendicular if and only if m belongs to the plane α whereof l is a normal.
c) If l is a normal of α and m a normal of β, then l lies in β if and only if m lies in α.
d) If n runs through the plane γ, then the plane whereof n is a normal runs through the set of planes through the normal of γ.

O28 Beneath we have drawn three hyperbolic lines through P (in the circle model of the hyperbolic plane) that don't intersect l.

O29 First the definitions of sinh and cosh: sinh(x) = (exp(x)-exp(-x))/2, cosh(x) = (exp(x)+exp(-x))/2.
The points (cosh(t),sinh(t)) run through one half of the hyperbola x²-y²=1, like (cos(t),sin(t)) run through the ellipse (circle) x²+y²=1.
Furthermore, d/dx sinh(x) = cosh(x) and d/dx cosh(x) = sinh(x), like d/dx sin(x) = cos(x) and d/dx cos(x) = -sin(x).
Now let f(x)=cosh(x) (x not negative). Because f ' isn't negative, and f(0)=1, f is at least 1 and a bijection from [0,∞) onto [1,∞). The inverse mapping is called arccosh.
Now we have to show: |p₃q₃-p₁q₁-p₂q₂| ≥ √(p₃²-p₁²-p₂²) √(q₃²-q₁²-q₂²) and ... = ... ⇔ p = q.
Define a symmetrical bilinear form by p.q := p₃q₃-p₁q₁-p₂q₂.
(λp+q).(λp+q) is positive if λp+q is lying inside k, 0 if λp+q is lying on k, and negative if λp+q is lying outside k. So if we choose two distinct fixed hyperbolic points p and q, then λ²p.p+2λp.q+q.q has two distinct zeros, and then the discriminant is positive.

O30 Let P = λ(0,0,1) and Q = λ(a,0,1). Then d(P,Q) = arccosh(1/√(1-a²)) → ∞ als a ↑ 1.

O31 Consider P² with in it the circle x²+y²=z². Let X=(x_o,y_o,z_o).
The polar line of P is the line with equation xx_o+yy_o=zz_o.
We have: P is on the polar line of Q ⇔ x_Px_Q+y_Py_Q=z_Pz_Q ⇔ Q is on the polar line of P.
Furthermore: if X lies on the circle, then the polar line of X is tangent to the circle in X.
So, in the picture, P lies on the polar lines of both intersection points of l and the circle, so l is the polar line of P.
Hence, the equation of l is xx_P+yy_P=zz_P.
Now let n be a hyperbolic line "through P". Then the equation of n has the form ax+by=cz where ax_P+by_P=cz_P.
Since l has projective coordinates (x_P,y_P,-z_P), and n has projective coordinates (a,b,-c), whilst cz_P-ax_P-by_P=0, we find l ⊥ m.