Area of a triangle using vectors












3














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


















  • 1




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










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






  • 2




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










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
















3














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


















  • 1




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










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






  • 2




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










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














3












3








3


0





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













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











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share|cite|improve this question










asked Feb 7 '14 at 16:50









Elise

395314




395314








  • 1




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










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






  • 2




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










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














  • 1




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










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






  • 2




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










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








1




1




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




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












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




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




2




2




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




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












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




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










2 Answers
2






active

oldest

votes


















0














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





























    0














    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





















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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      0














      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


























        0














        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
























          0












          0








          0






          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












          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 at 3:18









          HugoTeixeira

          3281313




          3281313























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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 at 3:40









                  NicNic8

                  3,8103922




                  3,8103922






























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