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Question 1: In 3D projective space $\Bbb P^3$(i.e. points and planes are represented by homogeneous 4-D vectors), there is a line $l$ definded by two distinct points $A,B$ and two distinct planes $P,Q$(i.e. this line passes through $A,B$ and $P,Q$ intersect at $l$). Now this line can be represented by a Plücker matrix: $$L=AB^T-BA^T= \begin{bmatrix}0 & l_{12} & l_{13} & l_{14} \\ -l_{12} & 0 & l_{23} & -l_{42} \\ -l_{13} & -l_{23} & 0 & l_{34} \\ - l_{14} & l_{42} & -l_{34} & 0 \end{bmatrix}$$ Similarily,define dual form of $L$ as $L^*$: $$L^*=PQ^T-QP^T$$ Now how to show $l_{12}:l_{13}:l_{14}:l_{23}:l_{42}:l_{34}=l^*_{34}:l^*_{42}:l^*_{23}:l^*_{14}:l^*_{13}:l^*_{12}$?

Question 2: Represent two lines $\mathcal L,\hat {\mathcal L}$ with Plücker line coordinates: $$\mathcal L=\{l_{12},l_{13},l_{14},l_{23},l_{42},l_{34}\}$$ $$\hat {\mathcal L}=\{\hat {l_{12}},\hat {l_{13}},\hat {l_{14}},\hat {l_{23}},\hat {l_{42}},\hat {l_{34}}\}$$ $\mathcal L \text{ and } \hat {\mathcal L}$ pass through $A,B$ and $\hat A, \hat B$ separately, now how to show $$det[A,B,\hat A, \hat B]=l_{12}\hat {l_{34}}+\hat {l_{12}}l_{34}+l_{13}\hat {l_{42}}+\hat {l_{13}}l_{42}+l_{14}\hat {l_{23}}+l_{23}\hat {l_{14}}$$ Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision from page 71 to 72.

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  • $\begingroup$ Do you have any thoughts whatsoever about how to approach these exercises? $\endgroup$
    – amd
    Nov 2, 2017 at 18:04
  • $\begingroup$ I have tried to compute determinant of L:$det(L)=(l_{12}l_{34}+l_{13}l_{42}+l_{14}l_{23})^2$, while it relates little with my question. One more known constraint is $Rank(L)=Rank(L^*)=2$ $\endgroup$
    – Finley
    Nov 3, 2017 at 2:10

1 Answer 1

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Proof to Question2:

From definitions of $L \text{ and } \hat L$, I can derive $$l_{ij}=A_iB_j-B_iA_j,\hat {l_{ij}}=\hat{A_i}\hat{B_j}-\hat{B_i}\hat{A_j}$$ so $$det(A,B,\hat A,\hat B)=\begin{vmatrix} A_1 & B_1 & \hat{A_1} & \hat{B_1} \\ A_2 & B_2 & \hat{A_2} & \hat{B_2}\\ A_3 & B_3 & \hat{A_3} & \hat{B_3}\\ A_4 & B_4 & \hat{A_4} & \hat{B_4} \end{vmatrix} $$$$=A_1\begin{vmatrix} B_2 & \hat{A_2} & \hat{B_2}\\ B_3 & \hat{A_3} & \hat{B_3}\\ B_4 & \hat{A_4} & \hat{B_4} \end{vmatrix}-A_2\begin{vmatrix} B_1 & \hat{A_1} & \hat{B_1} \\ B_3 & \hat{A_3} & \hat{B_3}\\ B_4 & \hat{A_4} & \hat{B_4} \end{vmatrix}+A_3\begin{vmatrix} B_1 & \hat{A_1} & \hat{B_1} \\ B_2 & \hat{A_2} & \hat{B_2}\\ B_4 & \hat{A_4} & \hat{B_4} \end{vmatrix}-A_4\begin{vmatrix} B_1 & \hat{A_1} & \hat{B_1} \\ B_2 & \hat{A_2} & \hat{B_2}\\ B_3 & \hat{A_3} & \hat{B_3}\\ \end{vmatrix} $$ $$=A_1(B_2 \hat{l_{34}}+B_3\hat{l_{42}}+B_4\hat{l_{23}})-A_2(B_1\hat{l_{34}}-B_3\hat{l_{14}}+B_4\hat{l_{13}})+A_3(-B_1\hat{l_{42}}-B_2\hat{l_{14}}+B_4\hat{l_{12}})-A_4(B_1\hat{l_{23}}-B_2\hat{l_{13}}+B_3\hat{l_{12}})$$ $$=l_{12}\hat{l_{34}}+\hat{l_{12}}l_{34}+l_{13}\hat{l_{42}}+\hat{l_{13}}l_{42}+l_{14}\hat{l_{23}}+\hat{l_{14}}l_{23}$$

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