Cross Products vs using geometry












1












$begingroup$


Say I wanted to calculate the area of a figure comprised of the coordinates (0,1,1) (1,0,1) and (0,0,0). When I used cross products I get the area as the square root of 3/2, however, when I use standard geometry I get the answer as 1/6. Can someone please point out my mistake when using cross products( I got the vectors as <0,1,1> and <-1,1,0> and I got the cross product as <-1,1,1>






euclidean-geometry







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$endgroup$












  • $begingroup$
    What is this "standard geometry"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:06










  • $begingroup$
    Since the pyramid has base area 1/2 and height 1 the area of the pyramid would be 0.5* 1/3*1 =1/6
    $endgroup$
    – user501887
    Dec 31 '18 at 19:11






  • 1




    $begingroup$
    Pyramid? You have three vertices; those define a triangle for me, not a pyramid.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:13






  • 1




    $begingroup$
    @user501887 to have a pyramid in 3d, you have to have 4 vertices. You only have three. There is no shape with non-zero area in the euclidean geometry that has three vertices, other than a tringle, no matter what dimension you talk about
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:45






  • 1




    $begingroup$
    If the figure were described as in your above comment —"[a] pyramid [having] base area 1/2 and height 1"— then $0.5cdotfrac13cdot 1=frac16$ wouldn't be its area, but its volume. That said, as others have commented, three points make a triangle, not a pyramid. To your comment about our (in)ability to form triangles "in the 3d plane": every triangle you have ever encountered in your life exists in 3d. :)
    $endgroup$
    – Blue
    Dec 31 '18 at 22:35
















1












$begingroup$


Say I wanted to calculate the area of a figure comprised of the coordinates (0,1,1) (1,0,1) and (0,0,0). When I used cross products I get the area as the square root of 3/2, however, when I use standard geometry I get the answer as 1/6. Can someone please point out my mistake when using cross products( I got the vectors as <0,1,1> and <-1,1,0> and I got the cross product as <-1,1,1>






euclidean-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is this "standard geometry"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:06










  • $begingroup$
    Since the pyramid has base area 1/2 and height 1 the area of the pyramid would be 0.5* 1/3*1 =1/6
    $endgroup$
    – user501887
    Dec 31 '18 at 19:11






  • 1




    $begingroup$
    Pyramid? You have three vertices; those define a triangle for me, not a pyramid.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:13






  • 1




    $begingroup$
    @user501887 to have a pyramid in 3d, you have to have 4 vertices. You only have three. There is no shape with non-zero area in the euclidean geometry that has three vertices, other than a tringle, no matter what dimension you talk about
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:45






  • 1




    $begingroup$
    If the figure were described as in your above comment —"[a] pyramid [having] base area 1/2 and height 1"— then $0.5cdotfrac13cdot 1=frac16$ wouldn't be its area, but its volume. That said, as others have commented, three points make a triangle, not a pyramid. To your comment about our (in)ability to form triangles "in the 3d plane": every triangle you have ever encountered in your life exists in 3d. :)
    $endgroup$
    – Blue
    Dec 31 '18 at 22:35














1












1








1





$begingroup$


Say I wanted to calculate the area of a figure comprised of the coordinates (0,1,1) (1,0,1) and (0,0,0). When I used cross products I get the area as the square root of 3/2, however, when I use standard geometry I get the answer as 1/6. Can someone please point out my mistake when using cross products( I got the vectors as <0,1,1> and <-1,1,0> and I got the cross product as <-1,1,1>






euclidean-geometry







share|cite|improve this question









$endgroup$




Say I wanted to calculate the area of a figure comprised of the coordinates (0,1,1) (1,0,1) and (0,0,0). When I used cross products I get the area as the square root of 3/2, however, when I use standard geometry I get the answer as 1/6. Can someone please point out my mistake when using cross products( I got the vectors as <0,1,1> and <-1,1,0> and I got the cross product as <-1,1,1>






euclidean-geometry




euclidean-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 19:04









user501887user501887

183




183












  • $begingroup$
    What is this "standard geometry"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:06










  • $begingroup$
    Since the pyramid has base area 1/2 and height 1 the area of the pyramid would be 0.5* 1/3*1 =1/6
    $endgroup$
    – user501887
    Dec 31 '18 at 19:11






  • 1




    $begingroup$
    Pyramid? You have three vertices; those define a triangle for me, not a pyramid.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:13






  • 1




    $begingroup$
    @user501887 to have a pyramid in 3d, you have to have 4 vertices. You only have three. There is no shape with non-zero area in the euclidean geometry that has three vertices, other than a tringle, no matter what dimension you talk about
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:45






  • 1




    $begingroup$
    If the figure were described as in your above comment —"[a] pyramid [having] base area 1/2 and height 1"— then $0.5cdotfrac13cdot 1=frac16$ wouldn't be its area, but its volume. That said, as others have commented, three points make a triangle, not a pyramid. To your comment about our (in)ability to form triangles "in the 3d plane": every triangle you have ever encountered in your life exists in 3d. :)
    $endgroup$
    – Blue
    Dec 31 '18 at 22:35


















  • $begingroup$
    What is this "standard geometry"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:06










  • $begingroup$
    Since the pyramid has base area 1/2 and height 1 the area of the pyramid would be 0.5* 1/3*1 =1/6
    $endgroup$
    – user501887
    Dec 31 '18 at 19:11






  • 1




    $begingroup$
    Pyramid? You have three vertices; those define a triangle for me, not a pyramid.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 19:13






  • 1




    $begingroup$
    @user501887 to have a pyramid in 3d, you have to have 4 vertices. You only have three. There is no shape with non-zero area in the euclidean geometry that has three vertices, other than a tringle, no matter what dimension you talk about
    $endgroup$
    – KKZiomek
    Dec 31 '18 at 20:45






  • 1




    $begingroup$
    If the figure were described as in your above comment —"[a] pyramid [having] base area 1/2 and height 1"— then $0.5cdotfrac13cdot 1=frac16$ wouldn't be its area, but its volume. That said, as others have commented, three points make a triangle, not a pyramid. To your comment about our (in)ability to form triangles "in the 3d plane": every triangle you have ever encountered in your life exists in 3d. :)
    $endgroup$
    – Blue
    Dec 31 '18 at 22:35
















$begingroup$
What is this "standard geometry"?
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 19:06




$begingroup$
What is this "standard geometry"?
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 19:06












$begingroup$
Since the pyramid has base area 1/2 and height 1 the area of the pyramid would be 0.5* 1/3*1 =1/6
$endgroup$
– user501887
Dec 31 '18 at 19:11




$begingroup$
Since the pyramid has base area 1/2 and height 1 the area of the pyramid would be 0.5* 1/3*1 =1/6
$endgroup$
– user501887
Dec 31 '18 at 19:11




1




1




$begingroup$
Pyramid? You have three vertices; those define a triangle for me, not a pyramid.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 19:13




$begingroup$
Pyramid? You have three vertices; those define a triangle for me, not a pyramid.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 19:13




1




1




$begingroup$
@user501887 to have a pyramid in 3d, you have to have 4 vertices. You only have three. There is no shape with non-zero area in the euclidean geometry that has three vertices, other than a tringle, no matter what dimension you talk about
$endgroup$
– KKZiomek
Dec 31 '18 at 20:45




$begingroup$
@user501887 to have a pyramid in 3d, you have to have 4 vertices. You only have three. There is no shape with non-zero area in the euclidean geometry that has three vertices, other than a tringle, no matter what dimension you talk about
$endgroup$
– KKZiomek
Dec 31 '18 at 20:45




1




1




$begingroup$
If the figure were described as in your above comment —"[a] pyramid [having] base area 1/2 and height 1"— then $0.5cdotfrac13cdot 1=frac16$ wouldn't be its area, but its volume. That said, as others have commented, three points make a triangle, not a pyramid. To your comment about our (in)ability to form triangles "in the 3d plane": every triangle you have ever encountered in your life exists in 3d. :)
$endgroup$
– Blue
Dec 31 '18 at 22:35




$begingroup$
If the figure were described as in your above comment —"[a] pyramid [having] base area 1/2 and height 1"— then $0.5cdotfrac13cdot 1=frac16$ wouldn't be its area, but its volume. That said, as others have commented, three points make a triangle, not a pyramid. To your comment about our (in)ability to form triangles "in the 3d plane": every triangle you have ever encountered in your life exists in 3d. :)
$endgroup$
– Blue
Dec 31 '18 at 22:35










2 Answers
2






active

oldest

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1












$begingroup$

Here is what my standard geometry tells me. You have an isosceles triangle
with two sides of length $sqrt2$ and the angle between them is $cos^{-1}(1/2)
=pi/3$. Its area is
$$frac12(sqrt2)^2sinfracpi3=frac{sqrt3}2.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product.
    $endgroup$
    – Andrei
    Dec 31 '18 at 19:51



















1












$begingroup$

The sides are given by vectors $(1,0,1), (0,1,1), (1,-1,0).$ The triangle is equilateral with lengths of sides $a=sqrt 2$ and the area $$mathcal{A}=frac{a}{2}cdotfrac{asqrt3}{2}=frac{sqrt3}{2}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Here is what my standard geometry tells me. You have an isosceles triangle
    with two sides of length $sqrt2$ and the angle between them is $cos^{-1}(1/2)
    =pi/3$    . Its area is
    $$frac12(sqrt2)^2sinfracpi3=frac{sqrt3}2.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product.
      $endgroup$
      – Andrei
      Dec 31 '18 at 19:51
















    1












    $begingroup$

    Here is what my standard geometry tells me. You have an isosceles triangle
    with two sides of length $sqrt2$ and the angle between them is $cos^{-1}(1/2)
    =pi/3$    . Its area is
    $$frac12(sqrt2)^2sinfracpi3=frac{sqrt3}2.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product.
      $endgroup$
      – Andrei
      Dec 31 '18 at 19:51














    1












    1








    1





    $begingroup$

    Here is what my standard geometry tells me. You have an isosceles triangle
    with two sides of length $sqrt2$ and the angle between them is $cos^{-1}(1/2)
    =pi/3$    . Its area is
    $$frac12(sqrt2)^2sinfracpi3=frac{sqrt3}2.$$






    share|cite|improve this answer









    $endgroup$



    Here is what my standard geometry tells me. You have an isosceles triangle
    with two sides of length $sqrt2$ and the angle between them is $cos^{-1}(1/2)
    =pi/3$    . Its area is
    $$frac12(sqrt2)^2sinfracpi3=frac{sqrt3}2.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 19:08









    Lord Shark the UnknownLord Shark the Unknown

    105k1160133




    105k1160133












    • $begingroup$
      You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product.
      $endgroup$
      – Andrei
      Dec 31 '18 at 19:51


















    • $begingroup$
      You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product.
      $endgroup$
      – Andrei
      Dec 31 '18 at 19:51
















    $begingroup$
    You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product.
    $endgroup$
    – Andrei
    Dec 31 '18 at 19:51




    $begingroup$
    You mean you have an equilateral triangle :) And I would use Pythagoras' theorem, otherwise the formula looks too much like the cross product.
    $endgroup$
    – Andrei
    Dec 31 '18 at 19:51











    1












    $begingroup$

    The sides are given by vectors $(1,0,1), (0,1,1), (1,-1,0).$ The triangle is equilateral with lengths of sides $a=sqrt 2$ and the area $$mathcal{A}=frac{a}{2}cdotfrac{asqrt3}{2}=frac{sqrt3}{2}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The sides are given by vectors $(1,0,1), (0,1,1), (1,-1,0).$ The triangle is equilateral with lengths of sides $a=sqrt 2$ and the area $$mathcal{A}=frac{a}{2}cdotfrac{asqrt3}{2}=frac{sqrt3}{2}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The sides are given by vectors $(1,0,1), (0,1,1), (1,-1,0).$ The triangle is equilateral with lengths of sides $a=sqrt 2$ and the area $$mathcal{A}=frac{a}{2}cdotfrac{asqrt3}{2}=frac{sqrt3}{2}$$






        share|cite|improve this answer









        $endgroup$



        The sides are given by vectors $(1,0,1), (0,1,1), (1,-1,0).$ The triangle is equilateral with lengths of sides $a=sqrt 2$ and the area $$mathcal{A}=frac{a}{2}cdotfrac{asqrt3}{2}=frac{sqrt3}{2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 20:37









        user376343user376343

        3,9033829




        3,9033829






























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