Intersection of ray with plane given in homogeneous coordinates












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$begingroup$


Given




  • a plane $pinmathbb{P}^3$ expressed in homogeneous coordinates ($p_x$, $p_y$, $p_z$, $p_w$)

  • a ray expressed as


    1. a source point $sinmathbb{P^3}$ expressed in homogeneous coordinates ($s_x$, $s_y$, $s_z$, $s_w$)

    2. a direction $din S^2$ expressed in coordinates ($d_x$, $d_y$, $d_z$)




how can I find the point at which the ray intersects the plane?



I can sometimes work out efficient algorithms with cross products in $mathbb{P^2}$ but I battle with $mathbb{P^3}$ due to having to visualize 4 dimensions.










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$endgroup$

















    0












    $begingroup$


    Given




    • a plane $pinmathbb{P}^3$ expressed in homogeneous coordinates ($p_x$, $p_y$, $p_z$, $p_w$)

    • a ray expressed as


      1. a source point $sinmathbb{P^3}$ expressed in homogeneous coordinates ($s_x$, $s_y$, $s_z$, $s_w$)

      2. a direction $din S^2$ expressed in coordinates ($d_x$, $d_y$, $d_z$)




    how can I find the point at which the ray intersects the plane?



    I can sometimes work out efficient algorithms with cross products in $mathbb{P^2}$ but I battle with $mathbb{P^3}$ due to having to visualize 4 dimensions.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Given




      • a plane $pinmathbb{P}^3$ expressed in homogeneous coordinates ($p_x$, $p_y$, $p_z$, $p_w$)

      • a ray expressed as


        1. a source point $sinmathbb{P^3}$ expressed in homogeneous coordinates ($s_x$, $s_y$, $s_z$, $s_w$)

        2. a direction $din S^2$ expressed in coordinates ($d_x$, $d_y$, $d_z$)




      how can I find the point at which the ray intersects the plane?



      I can sometimes work out efficient algorithms with cross products in $mathbb{P^2}$ but I battle with $mathbb{P^3}$ due to having to visualize 4 dimensions.










      share|cite|improve this question









      $endgroup$




      Given




      • a plane $pinmathbb{P}^3$ expressed in homogeneous coordinates ($p_x$, $p_y$, $p_z$, $p_w$)

      • a ray expressed as


        1. a source point $sinmathbb{P^3}$ expressed in homogeneous coordinates ($s_x$, $s_y$, $s_z$, $s_w$)

        2. a direction $din S^2$ expressed in coordinates ($d_x$, $d_y$, $d_z$)




      how can I find the point at which the ray intersects the plane?



      I can sometimes work out efficient algorithms with cross products in $mathbb{P^2}$ but I battle with $mathbb{P^3}$ due to having to visualize 4 dimensions.







      projective-geometry






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      asked Sep 26 '18 at 21:44









      MusefulMuseful

      1839




      1839






















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          One way to represent a line in $mathbb P^3$ is via its Plücker matrix: if $mathbf p$ and $mathbf q$ are two distinct points on the line, then its Plücker matrix is $L=mathbf pmathbf q^T-mathbf qmathbf p^T$. It’s an easy exercise to verify that $L$ is, up to a scale factor, independent of the choice of the two points. A handy feature of this representation is that if $mathbfpi$ is a homogeneous vector that represents a plane, its intersection with the line is simply $Lmathbfpi$.



          Applying this to your problem, you have the two points $mathbf s = (s_x,s_y,s_z,s_w)^T$ and $mathbf d = (d_x,d_y,d_z,0)^T$ that define the line, and the plane $mathbfpi = (p_x,p_y,p_z,p_w)^T$, so $$Lmathbfpi = (mathbf smathbf d^T-mathbf dmathbf s^T)mathbfpi = (mathbfpi^Tmathbf d)mathbf s - (mathbfpi^Tmathbf s)mathbf d,tag{*}$$ i.e., two dot products, two multiplications of a vector by a scalar, a vector addition, and three divisions to dehomogenize. If this expression vanishes, then the line lies on the plane. If you’re specifically interested in the intersection of a ray with the plane, you’ll also need to test that this point lies on the correct side of the line from $s$, or equivalently, that $d$ points toward the plane from $s$.



          The expression (*) can also be derived directly: The line through $mathbf s$ and $mathbf d$ consists of all nontrivial linear combinations $lambdamathbf s+mumathbf d$ of the two points, and the intersection of the line with $mathbfpi$ can therefore be found by solving $$mathbfpi^T(lambdamathbf s+mumathbf d) = begin{bmatrix}mathbfpi^Tmathbf s&mathbfpi^Tmathbf dend{bmatrix} begin{bmatrix}lambda\muend{bmatrix}=0$$ for $lambda$ and $mu$, which is obviously satisfied by $lambda=mathbfpi^Tmathbf d$, $mu=-mathbfpi^Tmathbf s$. It can also be viewed as a special case of a principle known as Plücker’s mu.



          The formula (*) is the three-dimensional analogue of using cross products to find the intersection of the line $mathbf l$ with the line defined by $mathbf s$ and $mathbf d$ in $mathbb P^2$, since we have the identity $$mathbf ltimes(mathbf stimesmathbf d) = (mathbf l^Tmathbf d)mathbf s - (mathbf l^Tmathbf s)mathbf d.$$






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          $endgroup$













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            1 Answer
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            0












            $begingroup$

            One way to represent a line in $mathbb P^3$ is via its Plücker matrix: if $mathbf p$ and $mathbf q$ are two distinct points on the line, then its Plücker matrix is $L=mathbf pmathbf q^T-mathbf qmathbf p^T$. It’s an easy exercise to verify that $L$ is, up to a scale factor, independent of the choice of the two points. A handy feature of this representation is that if $mathbfpi$ is a homogeneous vector that represents a plane, its intersection with the line is simply $Lmathbfpi$.



            Applying this to your problem, you have the two points $mathbf s = (s_x,s_y,s_z,s_w)^T$ and $mathbf d = (d_x,d_y,d_z,0)^T$ that define the line, and the plane $mathbfpi = (p_x,p_y,p_z,p_w)^T$, so $$Lmathbfpi = (mathbf smathbf d^T-mathbf dmathbf s^T)mathbfpi = (mathbfpi^Tmathbf d)mathbf s - (mathbfpi^Tmathbf s)mathbf d,tag{*}$$ i.e., two dot products, two multiplications of a vector by a scalar, a vector addition, and three divisions to dehomogenize. If this expression vanishes, then the line lies on the plane. If you’re specifically interested in the intersection of a ray with the plane, you’ll also need to test that this point lies on the correct side of the line from $s$, or equivalently, that $d$ points toward the plane from $s$.



            The expression (*) can also be derived directly: The line through $mathbf s$ and $mathbf d$ consists of all nontrivial linear combinations $lambdamathbf s+mumathbf d$ of the two points, and the intersection of the line with $mathbfpi$ can therefore be found by solving $$mathbfpi^T(lambdamathbf s+mumathbf d) = begin{bmatrix}mathbfpi^Tmathbf s&mathbfpi^Tmathbf dend{bmatrix} begin{bmatrix}lambda\muend{bmatrix}=0$$ for $lambda$ and $mu$, which is obviously satisfied by $lambda=mathbfpi^Tmathbf d$, $mu=-mathbfpi^Tmathbf s$. It can also be viewed as a special case of a principle known as Plücker’s mu.



            The formula (*) is the three-dimensional analogue of using cross products to find the intersection of the line $mathbf l$ with the line defined by $mathbf s$ and $mathbf d$ in $mathbb P^2$, since we have the identity $$mathbf ltimes(mathbf stimesmathbf d) = (mathbf l^Tmathbf d)mathbf s - (mathbf l^Tmathbf s)mathbf d.$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              One way to represent a line in $mathbb P^3$ is via its Plücker matrix: if $mathbf p$ and $mathbf q$ are two distinct points on the line, then its Plücker matrix is $L=mathbf pmathbf q^T-mathbf qmathbf p^T$. It’s an easy exercise to verify that $L$ is, up to a scale factor, independent of the choice of the two points. A handy feature of this representation is that if $mathbfpi$ is a homogeneous vector that represents a plane, its intersection with the line is simply $Lmathbfpi$.



              Applying this to your problem, you have the two points $mathbf s = (s_x,s_y,s_z,s_w)^T$ and $mathbf d = (d_x,d_y,d_z,0)^T$ that define the line, and the plane $mathbfpi = (p_x,p_y,p_z,p_w)^T$, so $$Lmathbfpi = (mathbf smathbf d^T-mathbf dmathbf s^T)mathbfpi = (mathbfpi^Tmathbf d)mathbf s - (mathbfpi^Tmathbf s)mathbf d,tag{*}$$ i.e., two dot products, two multiplications of a vector by a scalar, a vector addition, and three divisions to dehomogenize. If this expression vanishes, then the line lies on the plane. If you’re specifically interested in the intersection of a ray with the plane, you’ll also need to test that this point lies on the correct side of the line from $s$, or equivalently, that $d$ points toward the plane from $s$.



              The expression (*) can also be derived directly: The line through $mathbf s$ and $mathbf d$ consists of all nontrivial linear combinations $lambdamathbf s+mumathbf d$ of the two points, and the intersection of the line with $mathbfpi$ can therefore be found by solving $$mathbfpi^T(lambdamathbf s+mumathbf d) = begin{bmatrix}mathbfpi^Tmathbf s&mathbfpi^Tmathbf dend{bmatrix} begin{bmatrix}lambda\muend{bmatrix}=0$$ for $lambda$ and $mu$, which is obviously satisfied by $lambda=mathbfpi^Tmathbf d$, $mu=-mathbfpi^Tmathbf s$. It can also be viewed as a special case of a principle known as Plücker’s mu.



              The formula (*) is the three-dimensional analogue of using cross products to find the intersection of the line $mathbf l$ with the line defined by $mathbf s$ and $mathbf d$ in $mathbb P^2$, since we have the identity $$mathbf ltimes(mathbf stimesmathbf d) = (mathbf l^Tmathbf d)mathbf s - (mathbf l^Tmathbf s)mathbf d.$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                One way to represent a line in $mathbb P^3$ is via its Plücker matrix: if $mathbf p$ and $mathbf q$ are two distinct points on the line, then its Plücker matrix is $L=mathbf pmathbf q^T-mathbf qmathbf p^T$. It’s an easy exercise to verify that $L$ is, up to a scale factor, independent of the choice of the two points. A handy feature of this representation is that if $mathbfpi$ is a homogeneous vector that represents a plane, its intersection with the line is simply $Lmathbfpi$.



                Applying this to your problem, you have the two points $mathbf s = (s_x,s_y,s_z,s_w)^T$ and $mathbf d = (d_x,d_y,d_z,0)^T$ that define the line, and the plane $mathbfpi = (p_x,p_y,p_z,p_w)^T$, so $$Lmathbfpi = (mathbf smathbf d^T-mathbf dmathbf s^T)mathbfpi = (mathbfpi^Tmathbf d)mathbf s - (mathbfpi^Tmathbf s)mathbf d,tag{*}$$ i.e., two dot products, two multiplications of a vector by a scalar, a vector addition, and three divisions to dehomogenize. If this expression vanishes, then the line lies on the plane. If you’re specifically interested in the intersection of a ray with the plane, you’ll also need to test that this point lies on the correct side of the line from $s$, or equivalently, that $d$ points toward the plane from $s$.



                The expression (*) can also be derived directly: The line through $mathbf s$ and $mathbf d$ consists of all nontrivial linear combinations $lambdamathbf s+mumathbf d$ of the two points, and the intersection of the line with $mathbfpi$ can therefore be found by solving $$mathbfpi^T(lambdamathbf s+mumathbf d) = begin{bmatrix}mathbfpi^Tmathbf s&mathbfpi^Tmathbf dend{bmatrix} begin{bmatrix}lambda\muend{bmatrix}=0$$ for $lambda$ and $mu$, which is obviously satisfied by $lambda=mathbfpi^Tmathbf d$, $mu=-mathbfpi^Tmathbf s$. It can also be viewed as a special case of a principle known as Plücker’s mu.



                The formula (*) is the three-dimensional analogue of using cross products to find the intersection of the line $mathbf l$ with the line defined by $mathbf s$ and $mathbf d$ in $mathbb P^2$, since we have the identity $$mathbf ltimes(mathbf stimesmathbf d) = (mathbf l^Tmathbf d)mathbf s - (mathbf l^Tmathbf s)mathbf d.$$






                share|cite|improve this answer











                $endgroup$



                One way to represent a line in $mathbb P^3$ is via its Plücker matrix: if $mathbf p$ and $mathbf q$ are two distinct points on the line, then its Plücker matrix is $L=mathbf pmathbf q^T-mathbf qmathbf p^T$. It’s an easy exercise to verify that $L$ is, up to a scale factor, independent of the choice of the two points. A handy feature of this representation is that if $mathbfpi$ is a homogeneous vector that represents a plane, its intersection with the line is simply $Lmathbfpi$.



                Applying this to your problem, you have the two points $mathbf s = (s_x,s_y,s_z,s_w)^T$ and $mathbf d = (d_x,d_y,d_z,0)^T$ that define the line, and the plane $mathbfpi = (p_x,p_y,p_z,p_w)^T$, so $$Lmathbfpi = (mathbf smathbf d^T-mathbf dmathbf s^T)mathbfpi = (mathbfpi^Tmathbf d)mathbf s - (mathbfpi^Tmathbf s)mathbf d,tag{*}$$ i.e., two dot products, two multiplications of a vector by a scalar, a vector addition, and three divisions to dehomogenize. If this expression vanishes, then the line lies on the plane. If you’re specifically interested in the intersection of a ray with the plane, you’ll also need to test that this point lies on the correct side of the line from $s$, or equivalently, that $d$ points toward the plane from $s$.



                The expression (*) can also be derived directly: The line through $mathbf s$ and $mathbf d$ consists of all nontrivial linear combinations $lambdamathbf s+mumathbf d$ of the two points, and the intersection of the line with $mathbfpi$ can therefore be found by solving $$mathbfpi^T(lambdamathbf s+mumathbf d) = begin{bmatrix}mathbfpi^Tmathbf s&mathbfpi^Tmathbf dend{bmatrix} begin{bmatrix}lambda\muend{bmatrix}=0$$ for $lambda$ and $mu$, which is obviously satisfied by $lambda=mathbfpi^Tmathbf d$, $mu=-mathbfpi^Tmathbf s$. It can also be viewed as a special case of a principle known as Plücker’s mu.



                The formula (*) is the three-dimensional analogue of using cross products to find the intersection of the line $mathbf l$ with the line defined by $mathbf s$ and $mathbf d$ in $mathbb P^2$, since we have the identity $$mathbf ltimes(mathbf stimesmathbf d) = (mathbf l^Tmathbf d)mathbf s - (mathbf l^Tmathbf s)mathbf d.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 6 '18 at 0:53

























                answered Sep 27 '18 at 0:08









                amdamd

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                29.8k21050






























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