I am a bit struggling with the following exercise.
Let Q be a non-degenerate quadric in $\mathbb{RP}^3$, $l \subset \mathbb{RP}^3$ a line and l' $\subset \mathbb{RP}^3$ its polar. Show that all polar planes of points on l pass throught l'.
I tried to prove this algebraically. Let X=[x] be a point on l and let b be the symmetric bilinearform describing Q. By defnition of the polar, for each point P=[p] on l' it follows that b(x,p)=0. This implies that P is contained in the polar plane $P_X$ of X.
So for each X $\in$ l it follows l' $\subset P_X$. But this does not agree with my sketches where the polar l' is orthogonal to l and intersects the polar planes in exactly one point. So is my proof wrong or is my understanding what the polar line is flawed?
Let us set apart the fact that we are in projective space $\mathbb{RP}^3$ because the issue can be set in ordinary Euclidean space $\mathbb{R}^3$.
The point where I do not agree with you is that the polar line of a given line is not orthogonal in general to the initial line.
Let us look at an example. Take for our quadric the ellipsoid with equation :
$$x^2+4y^2+z^2=1$$
Besides, let us take $\ell$ with parametric equations :
$$\begin{cases}x=2t+1\\y=t\\z=3t-1\end{cases}$$
the polar line $\ell^{\circ}$ of $\ell$ is defined as the set of points $(x,y,z)$ such that:
$$\forall t \in \mathbb{R}, \ \ \ \ (2t+1)x+4ty+(3t-1)z-1=0.$$
Collecting constant terms and coefficients of $t$, we get the equivalent system:
$$\begin{cases}x-z-1=0\\ 2x+4y+3z=0\end{cases}$$
whose solution is straight line $\ell^{\circ}$ with equations:
$$\begin{cases}x=z+1\\y=\dfrac{-5z-2}{4}\\z=1z+0\end{cases}$$
In particular, a directing vector of this straight line $\ell^{\circ}$ is $(1,-5/4,1)$ which is not orthogonal in the Euclidean sense with respect to the directing vector : $(2,1,3)$ of $\ell.$
Nevertheless, these two former vectors can be considered as orthogonal with respect to the quadratic form induced by the ellipsoid:
$$\begin{pmatrix}1 & -5/4 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 4 & 0\\0& 0& 1\end{pmatrix}\begin{pmatrix}2\\1\\3\end{pmatrix}=0.$$