Finding an equation of a plane through the origin that is parallel to a given plane and parallel to a line. ...

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Finding an equation of a plane through the origin that is parallel to a given plane and parallel to a line.



The Next CEO of Stack OverflowFind the equation of a plane which is perpendicular to another planeFind the equation of the plane that goes through the origin and is parallel to the linesEquation of plane parallel to $x$-axis.Line equation through point, parallel to plane and intersecting lineEquation of plane passing through intersection of line and planeForm the equation of a plane, given 2 points and a parallel straight lineFind the equation of the plane that passes through a point & is parallel/orthogonal to a plane/lineFinding a plane parallel to a lineWhy line passing through origin in the intersection of three planes is parallel to the planes?Given a plane and a line, find the equation of another plane that has an angle 30 of degree to the given plane and contains the given line.












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A plane through the origin is perpendicular to the plane $2x-y-z=5$ and parallel to the line joining the points $(1,2,3)$ and $(4,-1,2)$. Find the equation of the plane.




Analyzing this problem I found that the normal vector of the required plane is perpendicular to the line parallel to the plane. And, the given plane's normal vector is parallel to the line. I am stuck after this. Can I do a cross product between the vector formed by the two points and the normal vector of the given plane?






vectors analytic-geometry 3d







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    2












    $begingroup$



    A plane through the origin is perpendicular to the plane $2x-y-z=5$ and parallel to the line joining the points $(1,2,3)$ and $(4,-1,2)$. Find the equation of the plane.




    Analyzing this problem I found that the normal vector of the required plane is perpendicular to the line parallel to the plane. And, the given plane's normal vector is parallel to the line. I am stuck after this. Can I do a cross product between the vector formed by the two points and the normal vector of the given plane?






    vectors analytic-geometry 3d







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      A plane through the origin is perpendicular to the plane $2x-y-z=5$ and parallel to the line joining the points $(1,2,3)$ and $(4,-1,2)$. Find the equation of the plane.




      Analyzing this problem I found that the normal vector of the required plane is perpendicular to the line parallel to the plane. And, the given plane's normal vector is parallel to the line. I am stuck after this. Can I do a cross product between the vector formed by the two points and the normal vector of the given plane?






      vectors analytic-geometry 3d







      share|cite|improve this question











      $endgroup$





      A plane through the origin is perpendicular to the plane $2x-y-z=5$ and parallel to the line joining the points $(1,2,3)$ and $(4,-1,2)$. Find the equation of the plane.




      Analyzing this problem I found that the normal vector of the required plane is perpendicular to the line parallel to the plane. And, the given plane's normal vector is parallel to the line. I am stuck after this. Can I do a cross product between the vector formed by the two points and the normal vector of the given plane?






      vectors analytic-geometry 3d




      vectors analytic-geometry 3d






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 17 at 7:06









      Parcly Taxel

      44.7k1376109




      44.7k1376109










      asked Mar 17 at 7:00









      Joshua Roberto GrutaJoshua Roberto Gruta

      111




      111






















          2 Answers
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          1












          $begingroup$

          We know that the required plane's normal is perpendicular to both the given plane's normal $(2,-1,-1)$ and the line's direction vector $(3,-3,-1)$. We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs:
          $$(2,-1,-1)times(3,-3,1)=dots$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something?
            $endgroup$
            – Joshua Roberto Gruta
            Mar 17 at 7:33






          • 1




            $begingroup$
            @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal.
            $endgroup$
            – Parcly Taxel
            Mar 17 at 8:23



















          0












          $begingroup$

          You could as well find the parametric form of the plane:
          $$x=lambdabegin{pmatrix}2\-1\-1end{pmatrix}+mubegin{pmatrix}3\-3\1end{pmatrix},lambda,muinmathbb{R}$$
          where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by
          $$begin{pmatrix}2\-1\-1end{pmatrix}timesbegin{pmatrix}3\-3\1end{pmatrix}=begin{pmatrix}-4\-5\-3end{pmatrix}$$
          from what You get the coordinate equation of the plane:
          $$4x+5y+3z=0.$$






          share|cite|improve this answer









          $endgroup$














            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            We know that the required plane's normal is perpendicular to both the given plane's normal $(2,-1,-1)$ and the line's direction vector $(3,-3,-1)$. We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs:
            $$(2,-1,-1)times(3,-3,1)=dots$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something?
              $endgroup$
              – Joshua Roberto Gruta
              Mar 17 at 7:33






            • 1




              $begingroup$
              @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal.
              $endgroup$
              – Parcly Taxel
              Mar 17 at 8:23
















            1












            $begingroup$

            We know that the required plane's normal is perpendicular to both the given plane's normal $(2,-1,-1)$ and the line's direction vector $(3,-3,-1)$. We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs:
            $$(2,-1,-1)times(3,-3,1)=dots$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something?
              $endgroup$
              – Joshua Roberto Gruta
              Mar 17 at 7:33






            • 1




              $begingroup$
              @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal.
              $endgroup$
              – Parcly Taxel
              Mar 17 at 8:23














            1












            1








            1





            $begingroup$

            We know that the required plane's normal is perpendicular to both the given plane's normal $(2,-1,-1)$ and the line's direction vector $(3,-3,-1)$. We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs:
            $$(2,-1,-1)times(3,-3,1)=dots$$






            share|cite|improve this answer









            $endgroup$



            We know that the required plane's normal is perpendicular to both the given plane's normal $(2,-1,-1)$ and the line's direction vector $(3,-3,-1)$. We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs:
            $$(2,-1,-1)times(3,-3,1)=dots$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 17 at 7:10









            Parcly TaxelParcly Taxel

            44.7k1376109




            44.7k1376109












            • $begingroup$
              Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something?
              $endgroup$
              – Joshua Roberto Gruta
              Mar 17 at 7:33






            • 1




              $begingroup$
              @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal.
              $endgroup$
              – Parcly Taxel
              Mar 17 at 8:23


















            • $begingroup$
              Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something?
              $endgroup$
              – Joshua Roberto Gruta
              Mar 17 at 7:33






            • 1




              $begingroup$
              @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal.
              $endgroup$
              – Parcly Taxel
              Mar 17 at 8:23
















            $begingroup$
            Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something?
            $endgroup$
            – Joshua Roberto Gruta
            Mar 17 at 7:33




            $begingroup$
            Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something?
            $endgroup$
            – Joshua Roberto Gruta
            Mar 17 at 7:33




            1




            1




            $begingroup$
            @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal.
            $endgroup$
            – Parcly Taxel
            Mar 17 at 8:23




            $begingroup$
            @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal.
            $endgroup$
            – Parcly Taxel
            Mar 17 at 8:23











            0












            $begingroup$

            You could as well find the parametric form of the plane:
            $$x=lambdabegin{pmatrix}2\-1\-1end{pmatrix}+mubegin{pmatrix}3\-3\1end{pmatrix},lambda,muinmathbb{R}$$
            where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by
            $$begin{pmatrix}2\-1\-1end{pmatrix}timesbegin{pmatrix}3\-3\1end{pmatrix}=begin{pmatrix}-4\-5\-3end{pmatrix}$$
            from what You get the coordinate equation of the plane:
            $$4x+5y+3z=0.$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You could as well find the parametric form of the plane:
              $$x=lambdabegin{pmatrix}2\-1\-1end{pmatrix}+mubegin{pmatrix}3\-3\1end{pmatrix},lambda,muinmathbb{R}$$
              where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by
              $$begin{pmatrix}2\-1\-1end{pmatrix}timesbegin{pmatrix}3\-3\1end{pmatrix}=begin{pmatrix}-4\-5\-3end{pmatrix}$$
              from what You get the coordinate equation of the plane:
              $$4x+5y+3z=0.$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You could as well find the parametric form of the plane:
                $$x=lambdabegin{pmatrix}2\-1\-1end{pmatrix}+mubegin{pmatrix}3\-3\1end{pmatrix},lambda,muinmathbb{R}$$
                where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by
                $$begin{pmatrix}2\-1\-1end{pmatrix}timesbegin{pmatrix}3\-3\1end{pmatrix}=begin{pmatrix}-4\-5\-3end{pmatrix}$$
                from what You get the coordinate equation of the plane:
                $$4x+5y+3z=0.$$






                share|cite|improve this answer









                $endgroup$



                You could as well find the parametric form of the plane:
                $$x=lambdabegin{pmatrix}2\-1\-1end{pmatrix}+mubegin{pmatrix}3\-3\1end{pmatrix},lambda,muinmathbb{R}$$
                where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by
                $$begin{pmatrix}2\-1\-1end{pmatrix}timesbegin{pmatrix}3\-3\1end{pmatrix}=begin{pmatrix}-4\-5\-3end{pmatrix}$$
                from what You get the coordinate equation of the plane:
                $$4x+5y+3z=0.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 17 at 9:35









                Peter MelechPeter Melech

                2,702813




                2,702813






























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