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The base of a triangular prism ABC.A'B'C' is an equilateral triangle with lengths a,


Construct a Triangle from Given Base, Obtuse Angle Adjacent to Base and Difference of Two Other SidesAcute plane triangle with two sides coinciding with a right triangleWhy can't the nth triangular number be expressed as the area of an equilateral triangle?Finding the coordinates of the vertices of an equilateral triangle.What is the maximum volume of an equilateral triangular prism inscribed in a sphere of radius 2?Prove that $PA$, $PB$ and $PC$ are the lengths of the sides of a triangle.Find the area inside of a equilateral triangleFinding the largest equilateral triangle inside a given triangleWhat are the basic properties of a toroidal prism?Find the height of an irregular triangular pyramid given the edge lengths













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The base of a triangular prism ABC.A'B'C' is an equilateral triangle with lengths a, and the lengths of its adjacent sides also equal a. Let I be the midpoint of AB and $B'I perp (ABC)$. Find the distance from the B' to the plane (ACC'A') in term of a










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    The base of a triangular prism ABC.A'B'C' is an equilateral triangle with lengths a, and the lengths of its adjacent sides also equal a. Let I be the midpoint of AB and $B'I perp (ABC)$. Find the distance from the B' to the plane (ACC'A') in term of a










    share|cite|improve this question









    $endgroup$














      1












      1








      1


      1



      $begingroup$


      The base of a triangular prism ABC.A'B'C' is an equilateral triangle with lengths a, and the lengths of its adjacent sides also equal a. Let I be the midpoint of AB and $B'I perp (ABC)$. Find the distance from the B' to the plane (ACC'A') in term of a










      share|cite|improve this question









      $endgroup$




      The base of a triangular prism ABC.A'B'C' is an equilateral triangle with lengths a, and the lengths of its adjacent sides also equal a. Let I be the midpoint of AB and $B'I perp (ABC)$. Find the distance from the B' to the plane (ACC'A') in term of a







      geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 19 '12 at 13:09









      LevanDokiteLevanDokite

      789




      789




















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
          $$
          left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
          h=fracasqrt2
          $$
          Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            but I think you are wrong read more:artofproblemsolving.com/Forum/…
            $endgroup$
            – LevanDokite
            Aug 21 '12 at 22:00











          • $begingroup$
            can anyone help me :D
            $endgroup$
            – LevanDokite
            Aug 23 '12 at 14:57


















          0












          $begingroup$

          Assume
          $$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
          Then
          $$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
          The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
          $$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
          so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
          We now have to solve the equation
          $$ncdot(B'+t n)=ncdot A$$
          for $t$ and obtain
          $$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
          The quantity we are looking for is
          $$|t n|=sqrt3over5>a .$$






          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









            0












            $begingroup$

            Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
            $$
            left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
            h=fracasqrt2
            $$
            Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              but I think you are wrong read more:artofproblemsolving.com/Forum/…
              $endgroup$
              – LevanDokite
              Aug 21 '12 at 22:00











            • $begingroup$
              can anyone help me :D
              $endgroup$
              – LevanDokite
              Aug 23 '12 at 14:57















            0












            $begingroup$

            Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
            $$
            left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
            h=fracasqrt2
            $$
            Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              but I think you are wrong read more:artofproblemsolving.com/Forum/…
              $endgroup$
              – LevanDokite
              Aug 21 '12 at 22:00











            • $begingroup$
              can anyone help me :D
              $endgroup$
              – LevanDokite
              Aug 23 '12 at 14:57













            0












            0








            0





            $begingroup$

            Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
            $$
            left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
            h=fracasqrt2
            $$
            Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.






            share|cite|improve this answer









            $endgroup$



            Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
            $$
            left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
            h=fracasqrt2
            $$
            Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 19 '12 at 17:25









            kashyap1123kashyap1123

            36625




            36625











            • $begingroup$
              but I think you are wrong read more:artofproblemsolving.com/Forum/…
              $endgroup$
              – LevanDokite
              Aug 21 '12 at 22:00











            • $begingroup$
              can anyone help me :D
              $endgroup$
              – LevanDokite
              Aug 23 '12 at 14:57
















            • $begingroup$
              but I think you are wrong read more:artofproblemsolving.com/Forum/…
              $endgroup$
              – LevanDokite
              Aug 21 '12 at 22:00











            • $begingroup$
              can anyone help me :D
              $endgroup$
              – LevanDokite
              Aug 23 '12 at 14:57















            $begingroup$
            but I think you are wrong read more:artofproblemsolving.com/Forum/…
            $endgroup$
            – LevanDokite
            Aug 21 '12 at 22:00





            $begingroup$
            but I think you are wrong read more:artofproblemsolving.com/Forum/…
            $endgroup$
            – LevanDokite
            Aug 21 '12 at 22:00













            $begingroup$
            can anyone help me :D
            $endgroup$
            – LevanDokite
            Aug 23 '12 at 14:57




            $begingroup$
            can anyone help me :D
            $endgroup$
            – LevanDokite
            Aug 23 '12 at 14:57











            0












            $begingroup$

            Assume
            $$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
            Then
            $$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
            The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
            $$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
            so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
            We now have to solve the equation
            $$ncdot(B'+t n)=ncdot A$$
            for $t$ and obtain
            $$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
            The quantity we are looking for is
            $$|t n|=sqrt3over5>a .$$






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              Assume
              $$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
              Then
              $$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
              The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
              $$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
              so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
              We now have to solve the equation
              $$ncdot(B'+t n)=ncdot A$$
              for $t$ and obtain
              $$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
              The quantity we are looking for is
              $$|t n|=sqrt3over5>a .$$






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                Assume
                $$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
                Then
                $$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
                The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
                $$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
                so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
                We now have to solve the equation
                $$ncdot(B'+t n)=ncdot A$$
                for $t$ and obtain
                $$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
                The quantity we are looking for is
                $$|t n|=sqrt3over5>a .$$






                share|cite|improve this answer









                $endgroup$



                Assume
                $$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
                Then
                $$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
                The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
                $$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
                so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
                We now have to solve the equation
                $$ncdot(B'+t n)=ncdot A$$
                for $t$ and obtain
                $$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
                The quantity we are looking for is
                $$|t n|=sqrt3over5>a .$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 17 '13 at 12:34









                Christian BlatterChristian Blatter

                176k8115328




                176k8115328



























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