Reflection of a line in a plane












-1












$begingroup$



The line $l_1$ has the equation $r=(6i+2-2k)+lambda(4i+5j-k)$ and the plane $pi_1$: $2x-y+4z=4$, the line $l_2$ is the reflection of $l_1$ in the plane $pi_1$. Find the exact vector equation of line $l_2$




So the line intersects the plane when $lambda=-2$, giving the point $(-2,-8,0)$ which will be common on $l_1$ and $l_2$. But I am unsure on how to find the direction vector for $l_2$. Any help would be appreciated.






geometry vector-spaces vectors







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  • $begingroup$
    Please do not reask questions; instead edit or comment.
    $endgroup$
    – quid♦
    Dec 26 '18 at 14:19
















-1












$begingroup$



The line $l_1$ has the equation $r=(6i+2-2k)+lambda(4i+5j-k)$ and the plane $pi_1$: $2x-y+4z=4$, the line $l_2$ is the reflection of $l_1$ in the plane $pi_1$. Find the exact vector equation of line $l_2$




So the line intersects the plane when $lambda=-2$, giving the point $(-2,-8,0)$ which will be common on $l_1$ and $l_2$. But I am unsure on how to find the direction vector for $l_2$. Any help would be appreciated.






geometry vector-spaces vectors







share|cite|improve this question









$endgroup$












  • $begingroup$
    Please do not reask questions; instead edit or comment.
    $endgroup$
    – quid♦
    Dec 26 '18 at 14:19














-1












-1








-1


0



$begingroup$



The line $l_1$ has the equation $r=(6i+2-2k)+lambda(4i+5j-k)$ and the plane $pi_1$: $2x-y+4z=4$, the line $l_2$ is the reflection of $l_1$ in the plane $pi_1$. Find the exact vector equation of line $l_2$




So the line intersects the plane when $lambda=-2$, giving the point $(-2,-8,0)$ which will be common on $l_1$ and $l_2$. But I am unsure on how to find the direction vector for $l_2$. Any help would be appreciated.






geometry vector-spaces vectors







share|cite|improve this question









$endgroup$





The line $l_1$ has the equation $r=(6i+2-2k)+lambda(4i+5j-k)$ and the plane $pi_1$: $2x-y+4z=4$, the line $l_2$ is the reflection of $l_1$ in the plane $pi_1$. Find the exact vector equation of line $l_2$




So the line intersects the plane when $lambda=-2$, giving the point $(-2,-8,0)$ which will be common on $l_1$ and $l_2$. But I am unsure on how to find the direction vector for $l_2$. Any help would be appreciated.






geometry vector-spaces vectors




geometry vector-spaces vectors






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asked Dec 20 '18 at 10:17









H.LinkhornH.Linkhorn

485213




485213












  • $begingroup$
    Please do not reask questions; instead edit or comment.
    $endgroup$
    – quid♦
    Dec 26 '18 at 14:19


















  • $begingroup$
    Please do not reask questions; instead edit or comment.
    $endgroup$
    – quid♦
    Dec 26 '18 at 14:19
















$begingroup$
Please do not reask questions; instead edit or comment.
$endgroup$
– quid♦
Dec 26 '18 at 14:19




$begingroup$
Please do not reask questions; instead edit or comment.
$endgroup$
– quid♦
Dec 26 '18 at 14:19










3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint...Construct the line through $(6,2,-2)$ with direction the same as the normal to the plane, and find where this line meets the plane. This point will be the midpoint of the line joining $(6,2,-2)$ and its reflection in the plane. Once you have this reflection point you can form the line of reflection because you now have two points.



the answer I get is $r=-2i-8j+t(88i+103j-13k)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    doing this I got an answer of $(-2i-8j+0k)+mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer?
    $endgroup$
    – H.Linkhorn
    Dec 20 '18 at 12:03












  • $begingroup$
    Looks like you need to check your calculations
    $endgroup$
    – David Quinn
    Dec 20 '18 at 22:49










  • $begingroup$
    Would it be possible for someone to solve the problem and give me a work solution for me to compare against.
    $endgroup$
    – H.Linkhorn
    Dec 21 '18 at 16:28










  • $begingroup$
    I have added an answer
    $endgroup$
    – David Quinn
    Dec 21 '18 at 23:10










  • $begingroup$
    Would you be able to include a method for me to see where ive gone wrong?
    $endgroup$
    – H.Linkhorn
    Dec 22 '18 at 14:13



















0












$begingroup$

HINT



In the first place, I advise you to obtain the normal component of $(4,5,-1)$ related to the place $pi_{1}$. After so, all you have to do is to substract twice it from the original direction of $l_{1}$ in order to obtain the direction of $l_{2}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$"
    $endgroup$
    – H.Linkhorn
    Dec 24 '18 at 17:54










  • $begingroup$
    I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$langle(4,5,-1),(2,-1,4)rangle (2,-1,4)/lVert(2,-1,4)rVert^{2}$$
    $endgroup$
    – APC89
    Dec 24 '18 at 17:58





















0












$begingroup$

Given a plane $Pi_1$ and a line $L$



$$
Pi_1to (p-p_1)cdot vec n_1 = 0\
Lto p = p_0 +lambda vec n_2
$$



first we determine the intersection point



$$
p^* = Pi_1cap L
$$



by making



$$
(p_0-p_1+lambdavec n_2)cdot vec n_1 = 0Rightarrow lambda = -frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}
$$



then



$$
p^* = p_0-frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}vec n_2
$$



After that the reflection for $vec n_2$ regarding $Pi_1$ is obtained as follows:



$$
vec n_2 = alpha vec n_1+vec mRightarrow vec n_1cdotvec n_2 = alpha||vec n_1||^2
$$



then



$$
alpha = frac{vec n_1cdotvec n_2}{||vec n_1||^2}Rightarrow vec m = vec n_2- frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
$$



and then the reflected line is



$$
L_Rto p = p^* +lambda vec n_R
$$



with



$$
vec n_R = vec m - alpha vec n_1 = vec n_2- 2frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
$$






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint...Construct the line through $(6,2,-2)$ with direction the same as the normal to the plane, and find where this line meets the plane. This point will be the midpoint of the line joining $(6,2,-2)$ and its reflection in the plane. Once you have this reflection point you can form the line of reflection because you now have two points.



    the answer I get is $r=-2i-8j+t(88i+103j-13k)$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      doing this I got an answer of $(-2i-8j+0k)+mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer?
      $endgroup$
      – H.Linkhorn
      Dec 20 '18 at 12:03












    • $begingroup$
      Looks like you need to check your calculations
      $endgroup$
      – David Quinn
      Dec 20 '18 at 22:49










    • $begingroup$
      Would it be possible for someone to solve the problem and give me a work solution for me to compare against.
      $endgroup$
      – H.Linkhorn
      Dec 21 '18 at 16:28










    • $begingroup$
      I have added an answer
      $endgroup$
      – David Quinn
      Dec 21 '18 at 23:10










    • $begingroup$
      Would you be able to include a method for me to see where ive gone wrong?
      $endgroup$
      – H.Linkhorn
      Dec 22 '18 at 14:13
















    1












    $begingroup$

    Hint...Construct the line through $(6,2,-2)$ with direction the same as the normal to the plane, and find where this line meets the plane. This point will be the midpoint of the line joining $(6,2,-2)$ and its reflection in the plane. Once you have this reflection point you can form the line of reflection because you now have two points.



    the answer I get is $r=-2i-8j+t(88i+103j-13k)$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      doing this I got an answer of $(-2i-8j+0k)+mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer?
      $endgroup$
      – H.Linkhorn
      Dec 20 '18 at 12:03












    • $begingroup$
      Looks like you need to check your calculations
      $endgroup$
      – David Quinn
      Dec 20 '18 at 22:49










    • $begingroup$
      Would it be possible for someone to solve the problem and give me a work solution for me to compare against.
      $endgroup$
      – H.Linkhorn
      Dec 21 '18 at 16:28










    • $begingroup$
      I have added an answer
      $endgroup$
      – David Quinn
      Dec 21 '18 at 23:10










    • $begingroup$
      Would you be able to include a method for me to see where ive gone wrong?
      $endgroup$
      – H.Linkhorn
      Dec 22 '18 at 14:13














    1












    1








    1





    $begingroup$

    Hint...Construct the line through $(6,2,-2)$ with direction the same as the normal to the plane, and find where this line meets the plane. This point will be the midpoint of the line joining $(6,2,-2)$ and its reflection in the plane. Once you have this reflection point you can form the line of reflection because you now have two points.



    the answer I get is $r=-2i-8j+t(88i+103j-13k)$






    share|cite|improve this answer











    $endgroup$



    Hint...Construct the line through $(6,2,-2)$ with direction the same as the normal to the plane, and find where this line meets the plane. This point will be the midpoint of the line joining $(6,2,-2)$ and its reflection in the plane. Once you have this reflection point you can form the line of reflection because you now have two points.



    the answer I get is $r=-2i-8j+t(88i+103j-13k)$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 21 '18 at 23:09

























    answered Dec 20 '18 at 10:30









    David QuinnDavid Quinn

    24.1k21141




    24.1k21141












    • $begingroup$
      doing this I got an answer of $(-2i-8j+0k)+mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer?
      $endgroup$
      – H.Linkhorn
      Dec 20 '18 at 12:03












    • $begingroup$
      Looks like you need to check your calculations
      $endgroup$
      – David Quinn
      Dec 20 '18 at 22:49










    • $begingroup$
      Would it be possible for someone to solve the problem and give me a work solution for me to compare against.
      $endgroup$
      – H.Linkhorn
      Dec 21 '18 at 16:28










    • $begingroup$
      I have added an answer
      $endgroup$
      – David Quinn
      Dec 21 '18 at 23:10










    • $begingroup$
      Would you be able to include a method for me to see where ive gone wrong?
      $endgroup$
      – H.Linkhorn
      Dec 22 '18 at 14:13


















    • $begingroup$
      doing this I got an answer of $(-2i-8j+0k)+mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer?
      $endgroup$
      – H.Linkhorn
      Dec 20 '18 at 12:03












    • $begingroup$
      Looks like you need to check your calculations
      $endgroup$
      – David Quinn
      Dec 20 '18 at 22:49










    • $begingroup$
      Would it be possible for someone to solve the problem and give me a work solution for me to compare against.
      $endgroup$
      – H.Linkhorn
      Dec 21 '18 at 16:28










    • $begingroup$
      I have added an answer
      $endgroup$
      – David Quinn
      Dec 21 '18 at 23:10










    • $begingroup$
      Would you be able to include a method for me to see where ive gone wrong?
      $endgroup$
      – H.Linkhorn
      Dec 22 '18 at 14:13
















    $begingroup$
    doing this I got an answer of $(-2i-8j+0k)+mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer?
    $endgroup$
    – H.Linkhorn
    Dec 20 '18 at 12:03






    $begingroup$
    doing this I got an answer of $(-2i-8j+0k)+mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer?
    $endgroup$
    – H.Linkhorn
    Dec 20 '18 at 12:03














    $begingroup$
    Looks like you need to check your calculations
    $endgroup$
    – David Quinn
    Dec 20 '18 at 22:49




    $begingroup$
    Looks like you need to check your calculations
    $endgroup$
    – David Quinn
    Dec 20 '18 at 22:49












    $begingroup$
    Would it be possible for someone to solve the problem and give me a work solution for me to compare against.
    $endgroup$
    – H.Linkhorn
    Dec 21 '18 at 16:28




    $begingroup$
    Would it be possible for someone to solve the problem and give me a work solution for me to compare against.
    $endgroup$
    – H.Linkhorn
    Dec 21 '18 at 16:28












    $begingroup$
    I have added an answer
    $endgroup$
    – David Quinn
    Dec 21 '18 at 23:10




    $begingroup$
    I have added an answer
    $endgroup$
    – David Quinn
    Dec 21 '18 at 23:10












    $begingroup$
    Would you be able to include a method for me to see where ive gone wrong?
    $endgroup$
    – H.Linkhorn
    Dec 22 '18 at 14:13




    $begingroup$
    Would you be able to include a method for me to see where ive gone wrong?
    $endgroup$
    – H.Linkhorn
    Dec 22 '18 at 14:13











    0












    $begingroup$

    HINT



    In the first place, I advise you to obtain the normal component of $(4,5,-1)$ related to the place $pi_{1}$. After so, all you have to do is to substract twice it from the original direction of $l_{1}$ in order to obtain the direction of $l_{2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$"
      $endgroup$
      – H.Linkhorn
      Dec 24 '18 at 17:54










    • $begingroup$
      I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$langle(4,5,-1),(2,-1,4)rangle (2,-1,4)/lVert(2,-1,4)rVert^{2}$$
      $endgroup$
      – APC89
      Dec 24 '18 at 17:58


















    0












    $begingroup$

    HINT



    In the first place, I advise you to obtain the normal component of $(4,5,-1)$ related to the place $pi_{1}$. After so, all you have to do is to substract twice it from the original direction of $l_{1}$ in order to obtain the direction of $l_{2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$"
      $endgroup$
      – H.Linkhorn
      Dec 24 '18 at 17:54










    • $begingroup$
      I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$langle(4,5,-1),(2,-1,4)rangle (2,-1,4)/lVert(2,-1,4)rVert^{2}$$
      $endgroup$
      – APC89
      Dec 24 '18 at 17:58
















    0












    0








    0





    $begingroup$

    HINT



    In the first place, I advise you to obtain the normal component of $(4,5,-1)$ related to the place $pi_{1}$. After so, all you have to do is to substract twice it from the original direction of $l_{1}$ in order to obtain the direction of $l_{2}$.






    share|cite|improve this answer









    $endgroup$



    HINT



    In the first place, I advise you to obtain the normal component of $(4,5,-1)$ related to the place $pi_{1}$. After so, all you have to do is to substract twice it from the original direction of $l_{1}$ in order to obtain the direction of $l_{2}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 24 '18 at 17:33









    APC89APC89

    2,371720




    2,371720












    • $begingroup$
      what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$"
      $endgroup$
      – H.Linkhorn
      Dec 24 '18 at 17:54










    • $begingroup$
      I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$langle(4,5,-1),(2,-1,4)rangle (2,-1,4)/lVert(2,-1,4)rVert^{2}$$
      $endgroup$
      – APC89
      Dec 24 '18 at 17:58




















    • $begingroup$
      what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$"
      $endgroup$
      – H.Linkhorn
      Dec 24 '18 at 17:54










    • $begingroup$
      I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$langle(4,5,-1),(2,-1,4)rangle (2,-1,4)/lVert(2,-1,4)rVert^{2}$$
      $endgroup$
      – APC89
      Dec 24 '18 at 17:58


















    $begingroup$
    what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$"
    $endgroup$
    – H.Linkhorn
    Dec 24 '18 at 17:54




    $begingroup$
    what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$"
    $endgroup$
    – H.Linkhorn
    Dec 24 '18 at 17:54












    $begingroup$
    I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$langle(4,5,-1),(2,-1,4)rangle (2,-1,4)/lVert(2,-1,4)rVert^{2}$$
    $endgroup$
    – APC89
    Dec 24 '18 at 17:58






    $begingroup$
    I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$langle(4,5,-1),(2,-1,4)rangle (2,-1,4)/lVert(2,-1,4)rVert^{2}$$
    $endgroup$
    – APC89
    Dec 24 '18 at 17:58













    0












    $begingroup$

    Given a plane $Pi_1$ and a line $L$



    $$
    Pi_1to (p-p_1)cdot vec n_1 = 0\
    Lto p = p_0 +lambda vec n_2
    $$



    first we determine the intersection point



    $$
    p^* = Pi_1cap L
    $$



    by making



    $$
    (p_0-p_1+lambdavec n_2)cdot vec n_1 = 0Rightarrow lambda = -frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}
    $$



    then



    $$
    p^* = p_0-frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}vec n_2
    $$



    After that the reflection for $vec n_2$ regarding $Pi_1$ is obtained as follows:



    $$
    vec n_2 = alpha vec n_1+vec mRightarrow vec n_1cdotvec n_2 = alpha||vec n_1||^2
    $$



    then



    $$
    alpha = frac{vec n_1cdotvec n_2}{||vec n_1||^2}Rightarrow vec m = vec n_2- frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
    $$



    and then the reflected line is



    $$
    L_Rto p = p^* +lambda vec n_R
    $$



    with



    $$
    vec n_R = vec m - alpha vec n_1 = vec n_2- 2frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Given a plane $Pi_1$ and a line $L$



      $$
      Pi_1to (p-p_1)cdot vec n_1 = 0\
      Lto p = p_0 +lambda vec n_2
      $$



      first we determine the intersection point



      $$
      p^* = Pi_1cap L
      $$



      by making



      $$
      (p_0-p_1+lambdavec n_2)cdot vec n_1 = 0Rightarrow lambda = -frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}
      $$



      then



      $$
      p^* = p_0-frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}vec n_2
      $$



      After that the reflection for $vec n_2$ regarding $Pi_1$ is obtained as follows:



      $$
      vec n_2 = alpha vec n_1+vec mRightarrow vec n_1cdotvec n_2 = alpha||vec n_1||^2
      $$



      then



      $$
      alpha = frac{vec n_1cdotvec n_2}{||vec n_1||^2}Rightarrow vec m = vec n_2- frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
      $$



      and then the reflected line is



      $$
      L_Rto p = p^* +lambda vec n_R
      $$



      with



      $$
      vec n_R = vec m - alpha vec n_1 = vec n_2- 2frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Given a plane $Pi_1$ and a line $L$



        $$
        Pi_1to (p-p_1)cdot vec n_1 = 0\
        Lto p = p_0 +lambda vec n_2
        $$



        first we determine the intersection point



        $$
        p^* = Pi_1cap L
        $$



        by making



        $$
        (p_0-p_1+lambdavec n_2)cdot vec n_1 = 0Rightarrow lambda = -frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}
        $$



        then



        $$
        p^* = p_0-frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}vec n_2
        $$



        After that the reflection for $vec n_2$ regarding $Pi_1$ is obtained as follows:



        $$
        vec n_2 = alpha vec n_1+vec mRightarrow vec n_1cdotvec n_2 = alpha||vec n_1||^2
        $$



        then



        $$
        alpha = frac{vec n_1cdotvec n_2}{||vec n_1||^2}Rightarrow vec m = vec n_2- frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
        $$



        and then the reflected line is



        $$
        L_Rto p = p^* +lambda vec n_R
        $$



        with



        $$
        vec n_R = vec m - alpha vec n_1 = vec n_2- 2frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
        $$






        share|cite|improve this answer









        $endgroup$



        Given a plane $Pi_1$ and a line $L$



        $$
        Pi_1to (p-p_1)cdot vec n_1 = 0\
        Lto p = p_0 +lambda vec n_2
        $$



        first we determine the intersection point



        $$
        p^* = Pi_1cap L
        $$



        by making



        $$
        (p_0-p_1+lambdavec n_2)cdot vec n_1 = 0Rightarrow lambda = -frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}
        $$



        then



        $$
        p^* = p_0-frac{(p_0-p_1)cdotvec n_1}{vec n_1cdotvec n_2}vec n_2
        $$



        After that the reflection for $vec n_2$ regarding $Pi_1$ is obtained as follows:



        $$
        vec n_2 = alpha vec n_1+vec mRightarrow vec n_1cdotvec n_2 = alpha||vec n_1||^2
        $$



        then



        $$
        alpha = frac{vec n_1cdotvec n_2}{||vec n_1||^2}Rightarrow vec m = vec n_2- frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
        $$



        and then the reflected line is



        $$
        L_Rto p = p^* +lambda vec n_R
        $$



        with



        $$
        vec n_R = vec m - alpha vec n_1 = vec n_2- 2frac{vec n_1cdotvec n_2}{||vec n_1||^2}vec n_1
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 1:26









        CesareoCesareo

        9,7763517




        9,7763517






























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