Finding the angle between a line and plane exercise












0












$begingroup$


The aim is to find the angle between a line and plane and if they intersect then the point where they do that.



given: $$ frac{x+1}{2}=frac{y-3}{4}=frac{z}{3}, 3x-3y+2z-5=0 $$



What I have found out:



s=(2; 4; 3)
n=(3; -3; 2)
$$|s*n|=|3*2+(-3)*4+2*3|=6-12+6=0 $$
$$|s|=sqrt{29}$$
$$|n|=sqrt{22}$$



$$ sin=frac{0}{sqrt{29}sqrt{22}} $$



angle is 0 degrees.



Solution



The angle between them is 0, this means they are parallel.



So I take the point P that is on the line



$$ P(-1,3,0) $$
and replace it in the plane equation:



$$ 3*(-1)-3*3+2*0-5=-17$$



-17 is not equal to 0



they don't intersect.






plane-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Nosrati If the line would be perpendicular to the plane, you expect the normal vector of the plane to be parallel to the line, but their dot product is 0...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:07










  • $begingroup$
    When $cos=0$ the vectors are perpendicular.
    $endgroup$
    – Nosrati
    Dec 3 '18 at 15:07












  • $begingroup$
    @Nosrati Right, but which two vectors...? A direction vector of the line and a normal vector of the plane...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:10
















0












$begingroup$


The aim is to find the angle between a line and plane and if they intersect then the point where they do that.



given: $$ frac{x+1}{2}=frac{y-3}{4}=frac{z}{3}, 3x-3y+2z-5=0 $$



What I have found out:



s=(2; 4; 3)
n=(3; -3; 2)
$$|s*n|=|3*2+(-3)*4+2*3|=6-12+6=0 $$
$$|s|=sqrt{29}$$
$$|n|=sqrt{22}$$



$$ sin=frac{0}{sqrt{29}sqrt{22}} $$



angle is 0 degrees.



Solution



The angle between them is 0, this means they are parallel.



So I take the point P that is on the line



$$ P(-1,3,0) $$
and replace it in the plane equation:



$$ 3*(-1)-3*3+2*0-5=-17$$



-17 is not equal to 0



they don't intersect.






plane-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Nosrati If the line would be perpendicular to the plane, you expect the normal vector of the plane to be parallel to the line, but their dot product is 0...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:07










  • $begingroup$
    When $cos=0$ the vectors are perpendicular.
    $endgroup$
    – Nosrati
    Dec 3 '18 at 15:07












  • $begingroup$
    @Nosrati Right, but which two vectors...? A direction vector of the line and a normal vector of the plane...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:10














0












0








0





$begingroup$


The aim is to find the angle between a line and plane and if they intersect then the point where they do that.



given: $$ frac{x+1}{2}=frac{y-3}{4}=frac{z}{3}, 3x-3y+2z-5=0 $$



What I have found out:



s=(2; 4; 3)
n=(3; -3; 2)
$$|s*n|=|3*2+(-3)*4+2*3|=6-12+6=0 $$
$$|s|=sqrt{29}$$
$$|n|=sqrt{22}$$



$$ sin=frac{0}{sqrt{29}sqrt{22}} $$



angle is 0 degrees.



Solution



The angle between them is 0, this means they are parallel.



So I take the point P that is on the line



$$ P(-1,3,0) $$
and replace it in the plane equation:



$$ 3*(-1)-3*3+2*0-5=-17$$



-17 is not equal to 0



they don't intersect.






plane-geometry







share|cite|improve this question











$endgroup$




The aim is to find the angle between a line and plane and if they intersect then the point where they do that.



given: $$ frac{x+1}{2}=frac{y-3}{4}=frac{z}{3}, 3x-3y+2z-5=0 $$



What I have found out:



s=(2; 4; 3)
n=(3; -3; 2)
$$|s*n|=|3*2+(-3)*4+2*3|=6-12+6=0 $$
$$|s|=sqrt{29}$$
$$|n|=sqrt{22}$$



$$ sin=frac{0}{sqrt{29}sqrt{22}} $$



angle is 0 degrees.



Solution



The angle between them is 0, this means they are parallel.



So I take the point P that is on the line



$$ P(-1,3,0) $$
and replace it in the plane equation:



$$ 3*(-1)-3*3+2*0-5=-17$$



-17 is not equal to 0



they don't intersect.






plane-geometry




plane-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 10:24







Student123

















asked Dec 3 '18 at 14:57









Student123Student123

536




536












  • $begingroup$
    @Nosrati If the line would be perpendicular to the plane, you expect the normal vector of the plane to be parallel to the line, but their dot product is 0...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:07










  • $begingroup$
    When $cos=0$ the vectors are perpendicular.
    $endgroup$
    – Nosrati
    Dec 3 '18 at 15:07












  • $begingroup$
    @Nosrati Right, but which two vectors...? A direction vector of the line and a normal vector of the plane...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:10


















  • $begingroup$
    @Nosrati If the line would be perpendicular to the plane, you expect the normal vector of the plane to be parallel to the line, but their dot product is 0...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:07










  • $begingroup$
    When $cos=0$ the vectors are perpendicular.
    $endgroup$
    – Nosrati
    Dec 3 '18 at 15:07












  • $begingroup$
    @Nosrati Right, but which two vectors...? A direction vector of the line and a normal vector of the plane...
    $endgroup$
    – StackTD
    Dec 3 '18 at 15:10
















$begingroup$
@Nosrati If the line would be perpendicular to the plane, you expect the normal vector of the plane to be parallel to the line, but their dot product is 0...
$endgroup$
– StackTD
Dec 3 '18 at 15:07




$begingroup$
@Nosrati If the line would be perpendicular to the plane, you expect the normal vector of the plane to be parallel to the line, but their dot product is 0...
$endgroup$
– StackTD
Dec 3 '18 at 15:07












$begingroup$
When $cos=0$ the vectors are perpendicular.
$endgroup$
– Nosrati
Dec 3 '18 at 15:07






$begingroup$
When $cos=0$ the vectors are perpendicular.
$endgroup$
– Nosrati
Dec 3 '18 at 15:07














$begingroup$
@Nosrati Right, but which two vectors...? A direction vector of the line and a normal vector of the plane...
$endgroup$
– StackTD
Dec 3 '18 at 15:10




$begingroup$
@Nosrati Right, but which two vectors...? A direction vector of the line and a normal vector of the plane...
$endgroup$
– StackTD
Dec 3 '18 at 15:10










1 Answer
1






active

oldest

votes


















3












$begingroup$

What you calculated is not the angle between the line and the plane, but between (a direction vector of) the line and the normal (vector) of the plane, which is a vector perpendicular to the plane.




What I have found out:



s=(2; 4; 3)
n=(3; -3; 2)
$$mathbf{s}cdotmathbf{n}=3*2+(-3)*4+2*3=6-12+6=0 $$




This dot product is zero, so the line is perpendicular to the normal, not to the plane. But that means the line and the plane aren't perpendicular, but parallel!



That leaves two options:




  • they do not intersect, so they have no points in common;

  • the line lies within the plane, so they have all points in common.


It is easy to see that $(-1,3,0)$ is a point on the line; does it lie in the plane?






share|cite|improve this answer











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    1 Answer
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    $begingroup$

    What you calculated is not the angle between the line and the plane, but between (a direction vector of) the line and the normal (vector) of the plane, which is a vector perpendicular to the plane.




    What I have found out:



    s=(2; 4; 3)
    n=(3; -3; 2)
    $$mathbf{s}cdotmathbf{n}=3*2+(-3)*4+2*3=6-12+6=0 $$




    This dot product is zero, so the line is perpendicular to the normal, not to the plane. But that means the line and the plane aren't perpendicular, but parallel!



    That leaves two options:




    • they do not intersect, so they have no points in common;

    • the line lies within the plane, so they have all points in common.


    It is easy to see that $(-1,3,0)$ is a point on the line; does it lie in the plane?






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      What you calculated is not the angle between the line and the plane, but between (a direction vector of) the line and the normal (vector) of the plane, which is a vector perpendicular to the plane.




      What I have found out:



      s=(2; 4; 3)
      n=(3; -3; 2)
      $$mathbf{s}cdotmathbf{n}=3*2+(-3)*4+2*3=6-12+6=0 $$




      This dot product is zero, so the line is perpendicular to the normal, not to the plane. But that means the line and the plane aren't perpendicular, but parallel!



      That leaves two options:




      • they do not intersect, so they have no points in common;

      • the line lies within the plane, so they have all points in common.


      It is easy to see that $(-1,3,0)$ is a point on the line; does it lie in the plane?






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        What you calculated is not the angle between the line and the plane, but between (a direction vector of) the line and the normal (vector) of the plane, which is a vector perpendicular to the plane.




        What I have found out:



        s=(2; 4; 3)
        n=(3; -3; 2)
        $$mathbf{s}cdotmathbf{n}=3*2+(-3)*4+2*3=6-12+6=0 $$




        This dot product is zero, so the line is perpendicular to the normal, not to the plane. But that means the line and the plane aren't perpendicular, but parallel!



        That leaves two options:




        • they do not intersect, so they have no points in common;

        • the line lies within the plane, so they have all points in common.


        It is easy to see that $(-1,3,0)$ is a point on the line; does it lie in the plane?






        share|cite|improve this answer











        $endgroup$



        What you calculated is not the angle between the line and the plane, but between (a direction vector of) the line and the normal (vector) of the plane, which is a vector perpendicular to the plane.




        What I have found out:



        s=(2; 4; 3)
        n=(3; -3; 2)
        $$mathbf{s}cdotmathbf{n}=3*2+(-3)*4+2*3=6-12+6=0 $$




        This dot product is zero, so the line is perpendicular to the normal, not to the plane. But that means the line and the plane aren't perpendicular, but parallel!



        That leaves two options:




        • they do not intersect, so they have no points in common;

        • the line lies within the plane, so they have all points in common.


        It is easy to see that $(-1,3,0)$ is a point on the line; does it lie in the plane?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 15:15

























        answered Dec 3 '18 at 15:05









        StackTDStackTD

        22.6k2049




        22.6k2049






























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