Projective plane and points in $\mathbb{A}^2 = \{z=1\}$

I'm learning projective geometry and need help with the following exercise :

Let $p_1, p_2$ be two points of the projective plane $\mathbb{RP}^2$ with homogeneous coordinates $[1 : 1 : 1]$ and $[-2 : 1 : 3]$ respectively. $(1)$ Find the points of the plane $\mathbb{A}^2 = \{z = 1\}$ which belongs to the equivalence classes of $p_1$ and $p_2$ respectively. $(2)$ Are they unique?

Here are my thoughts :

$(1)$ The equivalence classe of the point $p_1$ are all the points of $\mathbb{R}^3 \setminus \{0\}$ which satisfies the equivalence relation $(x, y, z) \sim (1, 1, 1) \iff \exists k \neq 0 \in \mathbb{R} : (x, y, z) = k(1, 1, 1)$, i.e. the equivalence classe of $p_1$ is the line passing through the origin and the point $(1, 1, 1)$. This line intersects the plane $A^2 = \{z=1\}$ at the point $(1, 1, 1)$.

Similarly, the equivalence classe of the point $p_2$ is the line passing through the origin and the point $(-2, 1, 3)$. This line intersects the plane $A^2 = \{z=1\}$ at the point $(-\frac{2}{3}, \frac{1}{3}, 1)$.

$(2)$ Concerning uniqueness I would say that since a line in $\mathbb{R^3}$ passing through the origin intersects the plane $\mathbb{A}^2 = \{z = 1\}$ at a unique point then the points we found in $(1)$ are unique.

I still don't know how to correctly write an answer to this exercise. I feel my answer/explanation is "too messy" and doesn't really answer the questions being asked. I would like some help to improve it and correct my errors if I have made any.

• I don't see any issue. Your answer looks good for me. – Huy Dang Jan 2 '18 at 21:51

For a slightly different angle, you could take a more algorithmic approach: describe what you did to come up with $(-\frac23,\frac13,1)$, namely divide by the third coordinate. This makes uniqueness more obvious: there is a single third coordinate, and a single result for a division of two numbers.