Area of a triangle using vectors












3












$begingroup$


I have to find the area of a triangle whose vertices have coordinates



O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$



I thought that perhaps I should use the dot product to find the angle between the lines $vec{OA}$ and $vec{OB}$ and use this angle in the formula:



area $= frac{1}{2}absin{C}$



These are my steps for doing this:



$mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} sin{theta} $



Let $mathbf{a} = begin{pmatrix} 1 \ -5 \ -7 end{pmatrix}$ and let $mathbf{b} = begin{pmatrix} 10 \ 10 \ 5 end{pmatrix}$



$therefore begin{pmatrix} 1 \ -5 \ -7 end{pmatrix} cdot begin{pmatrix} 10 \ 10 \ 5 end{pmatrix} = (5sqrt{3})(15)sin{theta} $



$therefore sin{theta} = -dfrac{1}{sqrt{3}}$



If I substitute these values into the general formula:



area $= frac{1}{2}absin{C}$



I get:



area $= frac{1}{2}(5sqrt{3})(15)(-dfrac{1}{sqrt{3}})$



$therefore$ area $= -dfrac{75}{2}$



However this isn't right, the area should be $dfrac{75}{sqrt{2}}$



I feel I'm missing something really obvious but I can't spot it, can anyone help?



Thank you.






trigonometry triangle area







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  • 1




    $begingroup$
    $mathbf{a}cdot mathbf{b}=|mathbf{a}||mathbf{b}| cos theta$.
    $endgroup$
    – Meow
    Feb 7 '14 at 16:54










  • $begingroup$
    Haha, thank you, can't believe I didn't check that.
    $endgroup$
    – Elise
    Feb 7 '14 at 16:55






  • 2




    $begingroup$
    Use the cross product ...
    $endgroup$
    – Mark Bennet
    Feb 7 '14 at 17:31










  • $begingroup$
    @MarkBennet that would make it a lot more efficient, thank you :)
    $endgroup$
    – Elise
    Feb 7 '14 at 18:05
















3












$begingroup$


I have to find the area of a triangle whose vertices have coordinates



O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$



I thought that perhaps I should use the dot product to find the angle between the lines $vec{OA}$ and $vec{OB}$ and use this angle in the formula:



area $= frac{1}{2}absin{C}$



These are my steps for doing this:



$mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} sin{theta} $



Let $mathbf{a} = begin{pmatrix} 1 \ -5 \ -7 end{pmatrix}$ and let $mathbf{b} = begin{pmatrix} 10 \ 10 \ 5 end{pmatrix}$



$therefore begin{pmatrix} 1 \ -5 \ -7 end{pmatrix} cdot begin{pmatrix} 10 \ 10 \ 5 end{pmatrix} = (5sqrt{3})(15)sin{theta} $



$therefore sin{theta} = -dfrac{1}{sqrt{3}}$



If I substitute these values into the general formula:



area $= frac{1}{2}absin{C}$



I get:



area $= frac{1}{2}(5sqrt{3})(15)(-dfrac{1}{sqrt{3}})$



$therefore$ area $= -dfrac{75}{2}$



However this isn't right, the area should be $dfrac{75}{sqrt{2}}$



I feel I'm missing something really obvious but I can't spot it, can anyone help?



Thank you.






trigonometry triangle area







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $mathbf{a}cdot mathbf{b}=|mathbf{a}||mathbf{b}| cos theta$.
    $endgroup$
    – Meow
    Feb 7 '14 at 16:54










  • $begingroup$
    Haha, thank you, can't believe I didn't check that.
    $endgroup$
    – Elise
    Feb 7 '14 at 16:55






  • 2




    $begingroup$
    Use the cross product ...
    $endgroup$
    – Mark Bennet
    Feb 7 '14 at 17:31










  • $begingroup$
    @MarkBennet that would make it a lot more efficient, thank you :)
    $endgroup$
    – Elise
    Feb 7 '14 at 18:05














3












3








3


0



$begingroup$


I have to find the area of a triangle whose vertices have coordinates



O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$



I thought that perhaps I should use the dot product to find the angle between the lines $vec{OA}$ and $vec{OB}$ and use this angle in the formula:



area $= frac{1}{2}absin{C}$



These are my steps for doing this:



$mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} sin{theta} $



Let $mathbf{a} = begin{pmatrix} 1 \ -5 \ -7 end{pmatrix}$ and let $mathbf{b} = begin{pmatrix} 10 \ 10 \ 5 end{pmatrix}$



$therefore begin{pmatrix} 1 \ -5 \ -7 end{pmatrix} cdot begin{pmatrix} 10 \ 10 \ 5 end{pmatrix} = (5sqrt{3})(15)sin{theta} $



$therefore sin{theta} = -dfrac{1}{sqrt{3}}$



If I substitute these values into the general formula:



area $= frac{1}{2}absin{C}$



I get:



area $= frac{1}{2}(5sqrt{3})(15)(-dfrac{1}{sqrt{3}})$



$therefore$ area $= -dfrac{75}{2}$



However this isn't right, the area should be $dfrac{75}{sqrt{2}}$



I feel I'm missing something really obvious but I can't spot it, can anyone help?



Thank you.






trigonometry triangle area







share|cite|improve this question









$endgroup$




I have to find the area of a triangle whose vertices have coordinates



O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$



I thought that perhaps I should use the dot product to find the angle between the lines $vec{OA}$ and $vec{OB}$ and use this angle in the formula:



area $= frac{1}{2}absin{C}$



These are my steps for doing this:



$mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} sin{theta} $



Let $mathbf{a} = begin{pmatrix} 1 \ -5 \ -7 end{pmatrix}$ and let $mathbf{b} = begin{pmatrix} 10 \ 10 \ 5 end{pmatrix}$



$therefore begin{pmatrix} 1 \ -5 \ -7 end{pmatrix} cdot begin{pmatrix} 10 \ 10 \ 5 end{pmatrix} = (5sqrt{3})(15)sin{theta} $



$therefore sin{theta} = -dfrac{1}{sqrt{3}}$



If I substitute these values into the general formula:



area $= frac{1}{2}absin{C}$



I get:



area $= frac{1}{2}(5sqrt{3})(15)(-dfrac{1}{sqrt{3}})$



$therefore$ area $= -dfrac{75}{2}$



However this isn't right, the area should be $dfrac{75}{sqrt{2}}$



I feel I'm missing something really obvious but I can't spot it, can anyone help?



Thank you.






trigonometry triangle area




trigonometry triangle area






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 7 '14 at 16:50









EliseElise

395314




395314








  • 1




    $begingroup$
    $mathbf{a}cdot mathbf{b}=|mathbf{a}||mathbf{b}| cos theta$.
    $endgroup$
    – Meow
    Feb 7 '14 at 16:54










  • $begingroup$
    Haha, thank you, can't believe I didn't check that.
    $endgroup$
    – Elise
    Feb 7 '14 at 16:55






  • 2




    $begingroup$
    Use the cross product ...
    $endgroup$
    – Mark Bennet
    Feb 7 '14 at 17:31










  • $begingroup$
    @MarkBennet that would make it a lot more efficient, thank you :)
    $endgroup$
    – Elise
    Feb 7 '14 at 18:05














  • 1




    $begingroup$
    $mathbf{a}cdot mathbf{b}=|mathbf{a}||mathbf{b}| cos theta$.
    $endgroup$
    – Meow
    Feb 7 '14 at 16:54










  • $begingroup$
    Haha, thank you, can't believe I didn't check that.
    $endgroup$
    – Elise
    Feb 7 '14 at 16:55






  • 2




    $begingroup$
    Use the cross product ...
    $endgroup$
    – Mark Bennet
    Feb 7 '14 at 17:31










  • $begingroup$
    @MarkBennet that would make it a lot more efficient, thank you :)
    $endgroup$
    – Elise
    Feb 7 '14 at 18:05








1




1




$begingroup$
$mathbf{a}cdot mathbf{b}=|mathbf{a}||mathbf{b}| cos theta$.
$endgroup$
– Meow
Feb 7 '14 at 16:54




$begingroup$
$mathbf{a}cdot mathbf{b}=|mathbf{a}||mathbf{b}| cos theta$.
$endgroup$
– Meow
Feb 7 '14 at 16:54












$begingroup$
Haha, thank you, can't believe I didn't check that.
$endgroup$
– Elise
Feb 7 '14 at 16:55




$begingroup$
Haha, thank you, can't believe I didn't check that.
$endgroup$
– Elise
Feb 7 '14 at 16:55




2




2




$begingroup$
Use the cross product ...
$endgroup$
– Mark Bennet
Feb 7 '14 at 17:31




$begingroup$
Use the cross product ...
$endgroup$
– Mark Bennet
Feb 7 '14 at 17:31












$begingroup$
@MarkBennet that would make it a lot more efficient, thank you :)
$endgroup$
– Elise
Feb 7 '14 at 18:05




$begingroup$
@MarkBennet that would make it a lot more efficient, thank you :)
$endgroup$
– Elise
Feb 7 '14 at 18:05










2 Answers
2






active

oldest

votes


















0












$begingroup$

The correct formula is $mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} cos{theta} $



So what you really have is $cos{theta} = cfrac{-1}{sqrt{3}}$



Therefore $$sin{theta} = sqrt{1 - cos^2{theta}} = sqrt{1 - frac{1}{3}} = sqrt{frac{2}{3}} = frac{sqrt{2}}{sqrt{3}}$$



Finally, the area of the triangle is:



$$
Area = frac{1}{2} (5 sqrt{3}) (15) frac{sqrt{2}}{sqrt{3}} = frac{75 sqrt{2}}{2}
$$



We can just multiply $frac{sqrt{2}}{sqrt{2}}$ to the area, and then we get the answer you posted:



$$
Area = frac{75 sqrt{2}}{2} left(frac{sqrt{2}}{sqrt{2}}right) = frac{75}{sqrt{2}}
$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since your vectors are in $mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value:
    $Area=(1/2) | a times b |$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Alternative approach is en.wikipedia.org/wiki/Heron%27s_formula
      $endgroup$
      – user2661923
      Dec 31 '18 at 14:23











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The correct formula is $mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} cos{theta} $



    So what you really have is $cos{theta} = cfrac{-1}{sqrt{3}}$



    Therefore $$sin{theta} = sqrt{1 - cos^2{theta}} = sqrt{1 - frac{1}{3}} = sqrt{frac{2}{3}} = frac{sqrt{2}}{sqrt{3}}$$



    Finally, the area of the triangle is:



    $$
    Area = frac{1}{2} (5 sqrt{3}) (15) frac{sqrt{2}}{sqrt{3}} = frac{75 sqrt{2}}{2}
    $$



    We can just multiply $frac{sqrt{2}}{sqrt{2}}$ to the area, and then we get the answer you posted:



    $$
    Area = frac{75 sqrt{2}}{2} left(frac{sqrt{2}}{sqrt{2}}right) = frac{75}{sqrt{2}}
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The correct formula is $mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} cos{theta} $



      So what you really have is $cos{theta} = cfrac{-1}{sqrt{3}}$



      Therefore $$sin{theta} = sqrt{1 - cos^2{theta}} = sqrt{1 - frac{1}{3}} = sqrt{frac{2}{3}} = frac{sqrt{2}}{sqrt{3}}$$



      Finally, the area of the triangle is:



      $$
      Area = frac{1}{2} (5 sqrt{3}) (15) frac{sqrt{2}}{sqrt{3}} = frac{75 sqrt{2}}{2}
      $$



      We can just multiply $frac{sqrt{2}}{sqrt{2}}$ to the area, and then we get the answer you posted:



      $$
      Area = frac{75 sqrt{2}}{2} left(frac{sqrt{2}}{sqrt{2}}right) = frac{75}{sqrt{2}}
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The correct formula is $mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} cos{theta} $



        So what you really have is $cos{theta} = cfrac{-1}{sqrt{3}}$



        Therefore $$sin{theta} = sqrt{1 - cos^2{theta}} = sqrt{1 - frac{1}{3}} = sqrt{frac{2}{3}} = frac{sqrt{2}}{sqrt{3}}$$



        Finally, the area of the triangle is:



        $$
        Area = frac{1}{2} (5 sqrt{3}) (15) frac{sqrt{2}}{sqrt{3}} = frac{75 sqrt{2}}{2}
        $$



        We can just multiply $frac{sqrt{2}}{sqrt{2}}$ to the area, and then we get the answer you posted:



        $$
        Area = frac{75 sqrt{2}}{2} left(frac{sqrt{2}}{sqrt{2}}right) = frac{75}{sqrt{2}}
        $$






        share|cite|improve this answer









        $endgroup$



        The correct formula is $mathbf{a} cdot mathbf{b} = begin{vmatrix} {mathbf{a}} end{vmatrix}begin{vmatrix} {mathbf{b}} end{vmatrix} cos{theta} $



        So what you really have is $cos{theta} = cfrac{-1}{sqrt{3}}$



        Therefore $$sin{theta} = sqrt{1 - cos^2{theta}} = sqrt{1 - frac{1}{3}} = sqrt{frac{2}{3}} = frac{sqrt{2}}{sqrt{3}}$$



        Finally, the area of the triangle is:



        $$
        Area = frac{1}{2} (5 sqrt{3}) (15) frac{sqrt{2}}{sqrt{3}} = frac{75 sqrt{2}}{2}
        $$



        We can just multiply $frac{sqrt{2}}{sqrt{2}}$ to the area, and then we get the answer you posted:



        $$
        Area = frac{75 sqrt{2}}{2} left(frac{sqrt{2}}{sqrt{2}}right) = frac{75}{sqrt{2}}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 14 '18 at 3:18









        HugoTeixeiraHugoTeixeira

        3281313




        3281313























            0












            $begingroup$

            Since your vectors are in $mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value:
            $Area=(1/2) | a times b |$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Alternative approach is en.wikipedia.org/wiki/Heron%27s_formula
              $endgroup$
              – user2661923
              Dec 31 '18 at 14:23
















            0












            $begingroup$

            Since your vectors are in $mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value:
            $Area=(1/2) | a times b |$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Alternative approach is en.wikipedia.org/wiki/Heron%27s_formula
              $endgroup$
              – user2661923
              Dec 31 '18 at 14:23














            0












            0








            0





            $begingroup$

            Since your vectors are in $mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value:
            $Area=(1/2) | a times b |$.






            share|cite|improve this answer









            $endgroup$



            Since your vectors are in $mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value:
            $Area=(1/2) | a times b |$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 14 '18 at 3:40









            NicNic8NicNic8

            4,14531023




            4,14531023












            • $begingroup$
              Alternative approach is en.wikipedia.org/wiki/Heron%27s_formula
              $endgroup$
              – user2661923
              Dec 31 '18 at 14:23


















            • $begingroup$
              Alternative approach is en.wikipedia.org/wiki/Heron%27s_formula
              $endgroup$
              – user2661923
              Dec 31 '18 at 14:23
















            $begingroup$
            Alternative approach is en.wikipedia.org/wiki/Heron%27s_formula
            $endgroup$
            – user2661923
            Dec 31 '18 at 14:23




            $begingroup$
            Alternative approach is en.wikipedia.org/wiki/Heron%27s_formula
            $endgroup$
            – user2661923
            Dec 31 '18 at 14:23


















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