#### I. Using only (implicit) image relations (i.e. without an explicit 3D reconstruction) and given the images of a line $\mathtt{L}$ and point $\mathtt{X}$ (not on $\mathtt{L}$) in two views, together with $\mathtt{H_{\infty}}$ between the two views, compute the image of the line in 3-space parallel to $\mathtt{L}$ and through $\mathtt{X}$. Other examples of this implicit approach to computation are given in [Zeller-96].

The image of a line parallel to $\mathtt{L}$ will intersect the image of $\mathtt{L}$ in the vanishing point of both lines. So, if we find the vanishing point $\mathtt{v}$, we can draw the required line connecting $\mathtt{v}$ and the image of $\mathtt{X}$.

Let the vanishing points of $\mathtt{L}$ in the two images be $\mathtt{v_1}$ and $\mathtt{v_2}$. Let the images of $\mathtt{L}$ be $\mathtt{l_1}$ and $\mathtt{l_2}$. Then the two vanishing points must satisfy the following conditions.

$$\mathtt{l_1}^T\mathtt{v_1} = 0$$ $$\mathtt{l_2}^T\mathtt{v_2} = 0$$

As the infinite homography relates the images of a point on the plane at infinity, the two vanishing points being the images of a point on $\pi_{\infty}$ must satisfy.

$$\mathtt{v_2 = H_{\infty}v_1}$$

Using this, we can rewrite the second constraint above as $$\mathtt{l_2}^T\mathtt{H_{\infty}v_1} = 0$$

As $l_1$, $l_2$ and $H_{\infty}$ are known quantities, we can solve for $\mathtt{v_1}$, a homogeneous point having two degrees of freedom, using the following two linear constraints $$\mathtt{l_1}^T\mathtt{v_1} = 0$$ $$\mathtt{l_2}^T\mathtt{H_{\infty}v_1} = 0$$

$\mathtt{v_2}$ can then be determined by applying the infinite homography to $\mathtt{v_1}$.

Finally, if $\mathtt{x_1}$ and $\mathtt{x_2}$ are the images of $\mathtt{X}$, we can determine the required lines using

$$\mathtt{x_1} \times \mathtt{v_1}$$ $$\mathtt{x_2} \times \mathtt{v_2}$$

comments powered by Disqus