The line $frac{x+6}{5}=frac{y+10}{3}=frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle...












4












$begingroup$


The line $dfrac{x+6}{5}=dfrac{y+10}{3}=dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.




My Attempt:

Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.

Let the equation of $AB$ be $dfrac{x-7}{a_1}=dfrac{y-2}{b_1}=dfrac{z-4}{c_1}$

and the equation of $BC$ be $dfrac{x-7}{a_2}=dfrac{y-2}{b_2}=dfrac{z-4}{c_2}$

As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$

Also as the triangle is a isosceles right triangle, so angle between $AC$ and
$BC$ will be same as the angle between $CA$ and $BA$. Let it be $theta$.


So $costheta=dfrac{5a_1+3b_1+8c_1}{sqrt{5^2+3^2+8^2}sqrt{a_1^2+b_1^2+c_1^2}}=dfrac{5a_2+3b_2+8c_2}{sqrt{5^2+3^2+8^2}sqrt{a_2^2+b_2^2+c_2^2}}$

Squaring both sides we get

$dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$



But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.






geometry analytic-geometry 3d







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    4












    $begingroup$


    The line $dfrac{x+6}{5}=dfrac{y+10}{3}=dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.




    My Attempt:

    Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.

    Let the equation of $AB$ be $dfrac{x-7}{a_1}=dfrac{y-2}{b_1}=dfrac{z-4}{c_1}$

    and the equation of $BC$ be $dfrac{x-7}{a_2}=dfrac{y-2}{b_2}=dfrac{z-4}{c_2}$

    As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$

    Also as the triangle is a isosceles right triangle, so angle between $AC$ and
    $BC$ will be same as the angle between $CA$ and $BA$. Let it be $theta$.


    So $costheta=dfrac{5a_1+3b_1+8c_1}{sqrt{5^2+3^2+8^2}sqrt{a_1^2+b_1^2+c_1^2}}=dfrac{5a_2+3b_2+8c_2}{sqrt{5^2+3^2+8^2}sqrt{a_2^2+b_2^2+c_2^2}}$

    Squaring both sides we get

    $dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$



    But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.






    geometry analytic-geometry 3d







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    $endgroup$















      4












      4








      4


      2



      $begingroup$


      The line $dfrac{x+6}{5}=dfrac{y+10}{3}=dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.




      My Attempt:

      Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.

      Let the equation of $AB$ be $dfrac{x-7}{a_1}=dfrac{y-2}{b_1}=dfrac{z-4}{c_1}$

      and the equation of $BC$ be $dfrac{x-7}{a_2}=dfrac{y-2}{b_2}=dfrac{z-4}{c_2}$

      As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$

      Also as the triangle is a isosceles right triangle, so angle between $AC$ and
      $BC$ will be same as the angle between $CA$ and $BA$. Let it be $theta$.


      So $costheta=dfrac{5a_1+3b_1+8c_1}{sqrt{5^2+3^2+8^2}sqrt{a_1^2+b_1^2+c_1^2}}=dfrac{5a_2+3b_2+8c_2}{sqrt{5^2+3^2+8^2}sqrt{a_2^2+b_2^2+c_2^2}}$

      Squaring both sides we get

      $dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$



      But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.






      geometry analytic-geometry 3d







      share|cite|improve this question











      $endgroup$




      The line $dfrac{x+6}{5}=dfrac{y+10}{3}=dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.




      My Attempt:

      Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.

      Let the equation of $AB$ be $dfrac{x-7}{a_1}=dfrac{y-2}{b_1}=dfrac{z-4}{c_1}$

      and the equation of $BC$ be $dfrac{x-7}{a_2}=dfrac{y-2}{b_2}=dfrac{z-4}{c_2}$

      As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$

      Also as the triangle is a isosceles right triangle, so angle between $AC$ and
      $BC$ will be same as the angle between $CA$ and $BA$. Let it be $theta$.


      So $costheta=dfrac{5a_1+3b_1+8c_1}{sqrt{5^2+3^2+8^2}sqrt{a_1^2+b_1^2+c_1^2}}=dfrac{5a_2+3b_2+8c_2}{sqrt{5^2+3^2+8^2}sqrt{a_2^2+b_2^2+c_2^2}}$

      Squaring both sides we get

      $dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$



      But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.






      geometry analytic-geometry 3d




      geometry analytic-geometry 3d






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      edited Dec 14 '15 at 9:24









      G-man

      4,50131344




      4,50131344










      asked Nov 20 '15 at 15:48









      Vinod Kumar PuniaVinod Kumar Punia

      2,697940




      2,697940






















          3 Answers
          3






          active

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          $begingroup$

          There is a much simpler and efficient way to solve this problem.



          First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.



          Now since the triangle is isosceles and right-angled it should be obvious that the points $A$ and $C$ will be obtained by traversing a distance $d$ along the given line.



          So if the direction cosines of the line are $l,m,n$ and the point $D$ is at $(x_0,y_0,z_0)$ then clearly we have the coordinates of $A$ and $C$ as $(x_0pm ld,y_0pm md,z_0pm nd)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man
            $endgroup$
            – Vinod Kumar Punia
            Nov 20 '15 at 16:27



















          0












          $begingroup$

          Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other 2 vertex of triangle.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Welcome to MSE. Please use MathJax.
            $endgroup$
            – José Carlos Santos
            Jan 6 '18 at 19:01



















          0












          $begingroup$

          Sides AB and AC make an angle of 45∘ with the given line.



          Let B=(−6+5t,−10+3t,−14+8t)



          AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^



          cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√



          Squaring and simplifying t2–5t+6=0⟹t=2or3



          Taking 2 for B and 3 for C.



          AB→=−3i^−6j^−2k^



          AC→=2i^−3j^+6k^



          Equation of AB is x−73=y−26=z−42



          Equation of AC is x−72=y−2−3=z−46






          share|cite|improve this answer









          $endgroup$














            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            There is a much simpler and efficient way to solve this problem.



            First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.



            Now since the triangle is isosceles and right-angled it should be obvious that the points $A$ and $C$ will be obtained by traversing a distance $d$ along the given line.



            So if the direction cosines of the line are $l,m,n$ and the point $D$ is at $(x_0,y_0,z_0)$ then clearly we have the coordinates of $A$ and $C$ as $(x_0pm ld,y_0pm md,z_0pm nd)$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man
              $endgroup$
              – Vinod Kumar Punia
              Nov 20 '15 at 16:27
















            2












            $begingroup$

            There is a much simpler and efficient way to solve this problem.



            First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.



            Now since the triangle is isosceles and right-angled it should be obvious that the points $A$ and $C$ will be obtained by traversing a distance $d$ along the given line.



            So if the direction cosines of the line are $l,m,n$ and the point $D$ is at $(x_0,y_0,z_0)$ then clearly we have the coordinates of $A$ and $C$ as $(x_0pm ld,y_0pm md,z_0pm nd)$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man
              $endgroup$
              – Vinod Kumar Punia
              Nov 20 '15 at 16:27














            2












            2








            2





            $begingroup$

            There is a much simpler and efficient way to solve this problem.



            First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.



            Now since the triangle is isosceles and right-angled it should be obvious that the points $A$ and $C$ will be obtained by traversing a distance $d$ along the given line.



            So if the direction cosines of the line are $l,m,n$ and the point $D$ is at $(x_0,y_0,z_0)$ then clearly we have the coordinates of $A$ and $C$ as $(x_0pm ld,y_0pm md,z_0pm nd)$






            share|cite|improve this answer









            $endgroup$



            There is a much simpler and efficient way to solve this problem.



            First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.



            Now since the triangle is isosceles and right-angled it should be obvious that the points $A$ and $C$ will be obtained by traversing a distance $d$ along the given line.



            So if the direction cosines of the line are $l,m,n$ and the point $D$ is at $(x_0,y_0,z_0)$ then clearly we have the coordinates of $A$ and $C$ as $(x_0pm ld,y_0pm md,z_0pm nd)$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 20 '15 at 15:59









            G-manG-man

            4,50131344




            4,50131344












            • $begingroup$
              This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man
              $endgroup$
              – Vinod Kumar Punia
              Nov 20 '15 at 16:27


















            • $begingroup$
              This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man
              $endgroup$
              – Vinod Kumar Punia
              Nov 20 '15 at 16:27
















            $begingroup$
            This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man
            $endgroup$
            – Vinod Kumar Punia
            Nov 20 '15 at 16:27




            $begingroup$
            This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man
            $endgroup$
            – Vinod Kumar Punia
            Nov 20 '15 at 16:27











            0












            $begingroup$

            Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other 2 vertex of triangle.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Please use MathJax.
              $endgroup$
              – José Carlos Santos
              Jan 6 '18 at 19:01
















            0












            $begingroup$

            Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other 2 vertex of triangle.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Please use MathJax.
              $endgroup$
              – José Carlos Santos
              Jan 6 '18 at 19:01














            0












            0








            0





            $begingroup$

            Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other 2 vertex of triangle.






            share|cite|improve this answer









            $endgroup$



            Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other 2 vertex of triangle.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 '18 at 18:50









            SHUBHAM SHEKHARSHUBHAM SHEKHAR

            1




            1












            • $begingroup$
              Welcome to MSE. Please use MathJax.
              $endgroup$
              – José Carlos Santos
              Jan 6 '18 at 19:01


















            • $begingroup$
              Welcome to MSE. Please use MathJax.
              $endgroup$
              – José Carlos Santos
              Jan 6 '18 at 19:01
















            $begingroup$
            Welcome to MSE. Please use MathJax.
            $endgroup$
            – José Carlos Santos
            Jan 6 '18 at 19:01




            $begingroup$
            Welcome to MSE. Please use MathJax.
            $endgroup$
            – José Carlos Santos
            Jan 6 '18 at 19:01











            0












            $begingroup$

            Sides AB and AC make an angle of 45∘ with the given line.



            Let B=(−6+5t,−10+3t,−14+8t)



            AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^



            cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√



            Squaring and simplifying t2–5t+6=0⟹t=2or3



            Taking 2 for B and 3 for C.



            AB→=−3i^−6j^−2k^



            AC→=2i^−3j^+6k^



            Equation of AB is x−73=y−26=z−42



            Equation of AC is x−72=y−2−3=z−46






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Sides AB and AC make an angle of 45∘ with the given line.



              Let B=(−6+5t,−10+3t,−14+8t)



              AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^



              cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√



              Squaring and simplifying t2–5t+6=0⟹t=2or3



              Taking 2 for B and 3 for C.



              AB→=−3i^−6j^−2k^



              AC→=2i^−3j^+6k^



              Equation of AB is x−73=y−26=z−42



              Equation of AC is x−72=y−2−3=z−46






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Sides AB and AC make an angle of 45∘ with the given line.



                Let B=(−6+5t,−10+3t,−14+8t)



                AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^



                cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√



                Squaring and simplifying t2–5t+6=0⟹t=2or3



                Taking 2 for B and 3 for C.



                AB→=−3i^−6j^−2k^



                AC→=2i^−3j^+6k^



                Equation of AB is x−73=y−26=z−42



                Equation of AC is x−72=y−2−3=z−46






                share|cite|improve this answer









                $endgroup$



                Sides AB and AC make an angle of 45∘ with the given line.



                Let B=(−6+5t,−10+3t,−14+8t)



                AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^



                cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√



                Squaring and simplifying t2–5t+6=0⟹t=2or3



                Taking 2 for B and 3 for C.



                AB→=−3i^−6j^−2k^



                AC→=2i^−3j^+6k^



                Equation of AB is x−73=y−26=z−42



                Equation of AC is x−72=y−2−3=z−46







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 21:21









                TerencePTerenceP

                338




                338






























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                    Plaza Victoria

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                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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                    Brian Clough

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                    Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
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                    Cáceres

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                    Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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