Calculate area of triangle in space given points












1












$begingroup$


Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:



$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$



My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.



Is there a better solution, either in terms of elegance or ease of mental calculation?






vectors area







share|cite|improve this question











$endgroup$












  • $begingroup$
    The linked question answers in 2D, you could do the same for 3D.
    $endgroup$
    – StubbornAtom
    Aug 17 '16 at 16:49






  • 1




    $begingroup$
    Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
    $endgroup$
    – ITA
    Aug 17 '16 at 16:51










  • $begingroup$
    @IvanAbraham: Whoops. Indeed it is. Edited.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:30










  • $begingroup$
    @StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:32










  • $begingroup$
    $$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:34


















1












$begingroup$


Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:



$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$



My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.



Is there a better solution, either in terms of elegance or ease of mental calculation?






vectors area







share|cite|improve this question











$endgroup$












  • $begingroup$
    The linked question answers in 2D, you could do the same for 3D.
    $endgroup$
    – StubbornAtom
    Aug 17 '16 at 16:49






  • 1




    $begingroup$
    Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
    $endgroup$
    – ITA
    Aug 17 '16 at 16:51










  • $begingroup$
    @IvanAbraham: Whoops. Indeed it is. Edited.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:30










  • $begingroup$
    @StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:32










  • $begingroup$
    $$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:34
















1












1








1


1



$begingroup$


Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:



$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$



My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.



Is there a better solution, either in terms of elegance or ease of mental calculation?






vectors area







share|cite|improve this question











$endgroup$




Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:



$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$



My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.



Is there a better solution, either in terms of elegance or ease of mental calculation?






vectors area




vectors area






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:20









Community♦

1




1










asked Aug 17 '16 at 16:46









ZazZaz

5481826




5481826












  • $begingroup$
    The linked question answers in 2D, you could do the same for 3D.
    $endgroup$
    – StubbornAtom
    Aug 17 '16 at 16:49






  • 1




    $begingroup$
    Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
    $endgroup$
    – ITA
    Aug 17 '16 at 16:51










  • $begingroup$
    @IvanAbraham: Whoops. Indeed it is. Edited.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:30










  • $begingroup$
    @StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:32










  • $begingroup$
    $$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:34




















  • $begingroup$
    The linked question answers in 2D, you could do the same for 3D.
    $endgroup$
    – StubbornAtom
    Aug 17 '16 at 16:49






  • 1




    $begingroup$
    Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
    $endgroup$
    – ITA
    Aug 17 '16 at 16:51










  • $begingroup$
    @IvanAbraham: Whoops. Indeed it is. Edited.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:30










  • $begingroup$
    @StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:32










  • $begingroup$
    $$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
    $endgroup$
    – Zaz
    Aug 17 '16 at 17:34


















$begingroup$
The linked question answers in 2D, you could do the same for 3D.
$endgroup$
– StubbornAtom
Aug 17 '16 at 16:49




$begingroup$
The linked question answers in 2D, you could do the same for 3D.
$endgroup$
– StubbornAtom
Aug 17 '16 at 16:49




1




1




$begingroup$
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
$endgroup$
– ITA
Aug 17 '16 at 16:51




$begingroup$
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
$endgroup$
– ITA
Aug 17 '16 at 16:51












$begingroup$
@IvanAbraham: Whoops. Indeed it is. Edited.
$endgroup$
– Zaz
Aug 17 '16 at 17:30




$begingroup$
@IvanAbraham: Whoops. Indeed it is. Edited.
$endgroup$
– Zaz
Aug 17 '16 at 17:30












$begingroup$
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
$endgroup$
– Zaz
Aug 17 '16 at 17:32




$begingroup$
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
$endgroup$
– Zaz
Aug 17 '16 at 17:32












$begingroup$
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
$endgroup$
– Zaz
Aug 17 '16 at 17:34






$begingroup$
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
$endgroup$
– Zaz
Aug 17 '16 at 17:34












2 Answers
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$begingroup$

I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.



http://www.mathopenref.com/heronsformula.html






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Even if we stick with cross products, we can make the result considerably more elegant in the sense of making the symmetry in indices manifest, viz.$$(P_1-P_0)times (P_2-P_0)=P_1times P_2-P_1times P_0-P_0times P_2+P_0times P_0\=P_0times P_1+P_1times P_2+P_2times P_0.$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      0












      $begingroup$

      I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.



      http://www.mathopenref.com/heronsformula.html






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.



        http://www.mathopenref.com/heronsformula.html






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.



          http://www.mathopenref.com/heronsformula.html






          share|cite|improve this answer









          $endgroup$



          I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.



          http://www.mathopenref.com/heronsformula.html







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 17 '16 at 16:49









          Alfred YergerAlfred Yerger

          10.5k2249




          10.5k2249























              0












              $begingroup$

              Even if we stick with cross products, we can make the result considerably more elegant in the sense of making the symmetry in indices manifest, viz.$$(P_1-P_0)times (P_2-P_0)=P_1times P_2-P_1times P_0-P_0times P_2+P_0times P_0\=P_0times P_1+P_1times P_2+P_2times P_0.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Even if we stick with cross products, we can make the result considerably more elegant in the sense of making the symmetry in indices manifest, viz.$$(P_1-P_0)times (P_2-P_0)=P_1times P_2-P_1times P_0-P_0times P_2+P_0times P_0\=P_0times P_1+P_1times P_2+P_2times P_0.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Even if we stick with cross products, we can make the result considerably more elegant in the sense of making the symmetry in indices manifest, viz.$$(P_1-P_0)times (P_2-P_0)=P_1times P_2-P_1times P_0-P_0times P_2+P_0times P_0\=P_0times P_1+P_1times P_2+P_2times P_0.$$






                  share|cite|improve this answer









                  $endgroup$



                  Even if we stick with cross products, we can make the result considerably more elegant in the sense of making the symmetry in indices manifest, viz.$$(P_1-P_0)times (P_2-P_0)=P_1times P_2-P_1times P_0-P_0times P_2+P_0times P_0\=P_0times P_1+P_1times P_2+P_2times P_0.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 9:04









                  J.G.J.G.

                  30.5k23149




                  30.5k23149






























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