How to calculate the area of a triangle ABC when given three position vectors $a, b$, and $ c$ in 3D?












5














Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $










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  • We can find the lengths of the sides and apply Heron's Formula.
    – Indrayudh Roy
    Apr 3 '14 at 14:26
















5














Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $










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  • We can find the lengths of the sides and apply Heron's Formula.
    – Indrayudh Roy
    Apr 3 '14 at 14:26














5












5








5







Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $










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Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $







triangle vectors area






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asked Apr 3 '14 at 14:21









user138913user138913

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  • We can find the lengths of the sides and apply Heron's Formula.
    – Indrayudh Roy
    Apr 3 '14 at 14:26


















  • We can find the lengths of the sides and apply Heron's Formula.
    – Indrayudh Roy
    Apr 3 '14 at 14:26
















We can find the lengths of the sides and apply Heron's Formula.
– Indrayudh Roy
Apr 3 '14 at 14:26




We can find the lengths of the sides and apply Heron's Formula.
– Indrayudh Roy
Apr 3 '14 at 14:26










6 Answers
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5














Heron works of course but it would be simpler to take half the length of the cross product
$(b-a)times(c-a)$.






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    2














    Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.






    share|cite|improve this answer





























      2














      use this formula:
      $$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
      where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.






      share|cite|improve this answer





























        2














        Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.






        share|cite|improve this answer





























          0














          One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.



          AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$



          You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.






          share|cite|improve this answer





























            0














            Cross product works great as a black box, but it lacks geometric intuition.



            enter image description here



            For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
            $$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
            where
            $$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$



            Thus:
            $$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
            This is equal to:
            $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$



            This is why
            $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$






            share|cite|improve this answer























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              6 Answers
              6






              active

              oldest

              votes








              6 Answers
              6






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5














              Heron works of course but it would be simpler to take half the length of the cross product
              $(b-a)times(c-a)$.






              share|cite|improve this answer


























                5














                Heron works of course but it would be simpler to take half the length of the cross product
                $(b-a)times(c-a)$.






                share|cite|improve this answer
























                  5












                  5








                  5






                  Heron works of course but it would be simpler to take half the length of the cross product
                  $(b-a)times(c-a)$.






                  share|cite|improve this answer












                  Heron works of course but it would be simpler to take half the length of the cross product
                  $(b-a)times(c-a)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 3 '14 at 14:34









                  jenajena

                  32412




                  32412























                      2














                      Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.






                      share|cite|improve this answer


























                        2














                        Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.






                        share|cite|improve this answer
























                          2












                          2








                          2






                          Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.






                          share|cite|improve this answer












                          Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Apr 3 '14 at 14:27









                          Mr.FryMr.Fry

                          3,88521223




                          3,88521223























                              2














                              use this formula:
                              $$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
                              where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.






                              share|cite|improve this answer


























                                2














                                use this formula:
                                $$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
                                where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.






                                share|cite|improve this answer
























                                  2












                                  2








                                  2






                                  use this formula:
                                  $$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
                                  where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.






                                  share|cite|improve this answer












                                  use this formula:
                                  $$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
                                  where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Apr 3 '14 at 14:29









                                  MartialMartial

                                  9991717




                                  9991717























                                      2














                                      Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.






                                      share|cite|improve this answer


























                                        2














                                        Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.






                                        share|cite|improve this answer
























                                          2












                                          2








                                          2






                                          Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.






                                          share|cite|improve this answer












                                          Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered Apr 3 '14 at 14:32









                                          JackJack

                                          23416




                                          23416























                                              0














                                              One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.



                                              AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$



                                              You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.






                                              share|cite|improve this answer


























                                                0














                                                One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.



                                                AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$



                                                You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.






                                                share|cite|improve this answer
























                                                  0












                                                  0








                                                  0






                                                  One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.



                                                  AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$



                                                  You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.






                                                  share|cite|improve this answer












                                                  One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.



                                                  AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$



                                                  You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered Apr 3 '14 at 14:31









                                                  JangoJango

                                                  352214




                                                  352214























                                                      0














                                                      Cross product works great as a black box, but it lacks geometric intuition.



                                                      enter image description here



                                                      For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
                                                      $$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
                                                      where
                                                      $$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$



                                                      Thus:
                                                      $$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
                                                      This is equal to:
                                                      $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$



                                                      This is why
                                                      $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$






                                                      share|cite|improve this answer




























                                                        0














                                                        Cross product works great as a black box, but it lacks geometric intuition.



                                                        enter image description here



                                                        For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
                                                        $$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
                                                        where
                                                        $$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$



                                                        Thus:
                                                        $$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
                                                        This is equal to:
                                                        $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$



                                                        This is why
                                                        $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$






                                                        share|cite|improve this answer


























                                                          0












                                                          0








                                                          0






                                                          Cross product works great as a black box, but it lacks geometric intuition.



                                                          enter image description here



                                                          For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
                                                          $$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
                                                          where
                                                          $$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$



                                                          Thus:
                                                          $$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
                                                          This is equal to:
                                                          $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$



                                                          This is why
                                                          $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$






                                                          share|cite|improve this answer














                                                          Cross product works great as a black box, but it lacks geometric intuition.



                                                          enter image description here



                                                          For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
                                                          $$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
                                                          where
                                                          $$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$



                                                          Thus:
                                                          $$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
                                                          This is equal to:
                                                          $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$



                                                          This is why
                                                          $$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$







                                                          share|cite|improve this answer














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                                                          share|cite|improve this answer








                                                          edited Jan 4 at 18:25

























                                                          answered Jan 4 at 18:19









                                                          zyczyc

                                                          1114




                                                          1114






























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