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My textbook, Multiple View Geometry in Computer Vision by Hartley and Zisserman says the following:

Ideal points and the line at infinity. Homogeneous vectors $\mathbf{x} = (x_1, x_2, x_3)^T$ such that $x_3 \not= 0$ correspond to finite points in $\mathbb{R}^2$. One may augment $\mathbb{R}^2$ by adding points with last coordinate $x_3 = 0$. The resulting space is the set of all homogeneous $3$-vectors, namely the projective space $\mathbb{P}^2$. The points with last coordinate $x_3 = 0$ are known as ideal points, or points at infinity. The set of all ideal points may be written $(x_1, x_2, 0)^T$, with a particular point specified by the ratio $x_1 : x_2$. Note that this set lies on a single line, the line at infinity, denoted by the vector $\mathbf{l}_{\infty} = (0, 0, 1)^T$. Indeed, one verifies that $(0, 0, 1)(x_1, x_2, 0)^T = 0$.

Using result 2.2 one finds that a line $\mathbf{l} = (a, b, c)^T$ intersects $\mathbf{l}_{\infty}$ in the ideal point $(b, -a, 0)^T$ (since $(b, -a, 0)\mathbf{l} = 0$). A line $\mathbf{l}' = (a, b, c')^T$ parallel to $\mathbf{l}$ intersects $\mathbf{l}_{\infty}$ in the same ideal point $(b, -a, 0)^T$ irrespective of the value of $c'$. In inhomogeneous notation, $(b, -a)^T$ is a vector tangent to the line, and orthogonal to the line normal $(a, b)$, and so represents the line's direction. As the line's direction varies the ideal point $(b, -a, 0)^T$ varies over $\mathbf{l}_\infty$. For these reasons the line at infinity can be thought of as the set of directions of lines in the plane.

My confusion is with regards to this part:

In inhomogeneous notation, $(b, -a)^T$ is a vector tangent to the line, and orthogonal to the line normal $(a, b)$, and so represents the line's direction. As the line's direction varies the ideal point $(b, -a, 0)^T$ varies over $\mathbf{l}_\infty$. For these reasons the line at infinity can be thought of as the set of directions of lines in the plane.

Specifically, it is not clear to me as to precisely which line the author is referring to:

In inhomogeneous notation, $(b, -a)^T$ is a vector tangent to the line, and orthogonal to the line normal $(a, b)$, and so represents the line's direction.

If my understanding is correct, the inhomogeneous vector $(b, -a)^T$ is normal to the line $\mathbf{l}$ (and $\mathbf{l}'$) described by the vector $(a, b, c)$ (or $(a, b, c')$ in the case of $\mathbf{l}'$), and the inhomogeneous vector $(a, b)^T$ is the tangent to the same lines (and so represents the direction of the lines). However, this is phrased differently from what the author has written, and, given the vagueness of this part, it is unclear what line the author is precisely referring to.

I would greatly appreciate it if people could please take the time to clarify this.

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No, the inhomogeneous vector $(a,b)$ is normal to the line $\mathbf l = (a,b,c)^T$. Go back to how this homogeneous vector representation of a line is defined: The line is the set of all points $\mathbf x$ such that $\mathbf l^T\mathbf x=0$. Expanded, this equation is $ax+by+c=0$, which is just the (I hope) familiar point-normal equation of the line.

The point of this passage is that each ideal point corresponds to a unique direction in the plane; all lines go in this direction intersect at that ideal point.

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  • $\begingroup$ Thanks for the answer. I was drawing upon the (what I understand to be) geometric fact that the normal to any vector $(a, b)$ is $(-b, a)$. Is this not valid in projective space (is it only valid in Euclidean space)? $\endgroup$ Commented Nov 14, 2019 at 17:54
  • $\begingroup$ “Normal” isn’t an invariant of projective transformations, so, strictly speaking, no. However, in this passage you’re augmenting the Euclidean plane, so those inhomogeneous vectors are indeed orthogonal. The mistake you made was in thinking that $(a,b)$ was parallel to the line $(a,b,c)^T$. $\endgroup$
    – amd
    Commented Nov 14, 2019 at 17:59
  • $\begingroup$ I don't understand. 1. The text says that $(b, -a)$ is tangent to the line $(a, b, c)^T$; how can this be true if, as we claim, $(b, -a)$ is normal to $(a, b, c)^T$? 2. And, with regards to what you said in your answer, the author says that the line is the set of all points $\mathbf{x} = (x_1, x_2, 0)^T$ such that $\mathbf{\mathbf{l}_\infty}\mathbf{x} = (0, 0, 1)(x_1, x_2, 0)^T = 0$. You're using the line $\mathbf l = (a,b,c)^T$ and stating that $\mathbf l^T\mathbf x=0$, but this isn't what the author said? [...] $\endgroup$ Commented Nov 14, 2019 at 18:28
  • $\begingroup$ [...] 3. Furthermore, how is it that the vector $(b, -a)^T$ be tangent to $(a, b, c)^T$? This isn't clear to me. $\endgroup$ Commented Nov 14, 2019 at 18:32
  • $\begingroup$ And to add another point : 4. If we take your line $\mathbf l^T\mathbf x = 0 \Rightarrow ax + by + c = 0$, then we have that, after taking the partial derivatives of this line, we have the (non-unit) tangent vector $(a, b)$. $\endgroup$ Commented Nov 14, 2019 at 18:56

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