Compute area of triangle












1












$begingroup$


Problem



Triangle can be formed in between three points. These three points are in $mathbb{R}^3$ and
in my case these points are: $p_1=(0,4,6),p_2=(-5,3,1),p_3=(2,1,2)$. Compute area of this triangle.



Attempt to solve



I take one point and draw two vectors $vec{u}$ and $vec{v}$ that define two sides of this triangle. Then i compute cross product of these two vectors which length gives me size of parallelogram. Dividing this parallelogram should give area of this triangle.



$$ vec{u}=begin{bmatrix} 0-5 \4+3 \ 6+1 end{bmatrix} = begin{bmatrix} -5 \ 7 \ 7 end{bmatrix}, vec{v}=begin{bmatrix} 0+2 \ 4+1 \ 6+2 end{bmatrix}= begin{bmatrix} 2\ 5 \ 8 end{bmatrix} $$



It shouldn't matter from which point i compute the other two vectors. Now length cross product vector should give us the area of parallelogram.



$$ vec{u} times vec{u} = begin{vmatrix} i & j & k \ -5 & 7 & 7 \ 2 & 5 & 8 end{vmatrix} = ibegin{vmatrix} 7 & 7 \ 5 & 8 end{vmatrix} - j begin{vmatrix} -5 & 7 \ 2 & 8 end{vmatrix} + kbegin{vmatrix} -5 & 7 \ 2 & 5 end{vmatrix} $$



$$ = i(7cdot 8(7cdot 5)- j(-5cdot 8-7cdot 2) + k(-5 cdot 5 - 7 cdot 2) $$



$$ i(56-35)-j(40-14)+k(-25-14) $$



$$ 21i+54j-39k $$



$$ text{Area} = frac{|vec{u} times vec{v}|}{2} = frac{sqrt{21^2+52^2+(-39)^2}}{2} approx 34.154 $$



However this solution seems to be incorrect. WolframAlpha gives solution to this. Did i compute something simply wrong or is there more fundamental probelm on how i understand the problem ?






vectors area







share|cite|improve this question









$endgroup$












  • $begingroup$
    Notice that $|utimes v|^2=|u|^2|v|^2-(ucdot v)^2$, so there's no need to calculate the cross product explicitly.
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 20:00
















1












$begingroup$


Problem



Triangle can be formed in between three points. These three points are in $mathbb{R}^3$ and
in my case these points are: $p_1=(0,4,6),p_2=(-5,3,1),p_3=(2,1,2)$. Compute area of this triangle.



Attempt to solve



I take one point and draw two vectors $vec{u}$ and $vec{v}$ that define two sides of this triangle. Then i compute cross product of these two vectors which length gives me size of parallelogram. Dividing this parallelogram should give area of this triangle.



$$ vec{u}=begin{bmatrix} 0-5 \4+3 \ 6+1 end{bmatrix} = begin{bmatrix} -5 \ 7 \ 7 end{bmatrix}, vec{v}=begin{bmatrix} 0+2 \ 4+1 \ 6+2 end{bmatrix}= begin{bmatrix} 2\ 5 \ 8 end{bmatrix} $$



It shouldn't matter from which point i compute the other two vectors. Now length cross product vector should give us the area of parallelogram.



$$ vec{u} times vec{u} = begin{vmatrix} i & j & k \ -5 & 7 & 7 \ 2 & 5 & 8 end{vmatrix} = ibegin{vmatrix} 7 & 7 \ 5 & 8 end{vmatrix} - j begin{vmatrix} -5 & 7 \ 2 & 8 end{vmatrix} + kbegin{vmatrix} -5 & 7 \ 2 & 5 end{vmatrix} $$



$$ = i(7cdot 8(7cdot 5)- j(-5cdot 8-7cdot 2) + k(-5 cdot 5 - 7 cdot 2) $$



$$ i(56-35)-j(40-14)+k(-25-14) $$



$$ 21i+54j-39k $$



$$ text{Area} = frac{|vec{u} times vec{v}|}{2} = frac{sqrt{21^2+52^2+(-39)^2}}{2} approx 34.154 $$



However this solution seems to be incorrect. WolframAlpha gives solution to this. Did i compute something simply wrong or is there more fundamental probelm on how i understand the problem ?






vectors area







share|cite|improve this question









$endgroup$












  • $begingroup$
    Notice that $|utimes v|^2=|u|^2|v|^2-(ucdot v)^2$, so there's no need to calculate the cross product explicitly.
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 20:00














1












1








1





$begingroup$


Problem



Triangle can be formed in between three points. These three points are in $mathbb{R}^3$ and
in my case these points are: $p_1=(0,4,6),p_2=(-5,3,1),p_3=(2,1,2)$. Compute area of this triangle.



Attempt to solve



I take one point and draw two vectors $vec{u}$ and $vec{v}$ that define two sides of this triangle. Then i compute cross product of these two vectors which length gives me size of parallelogram. Dividing this parallelogram should give area of this triangle.



$$ vec{u}=begin{bmatrix} 0-5 \4+3 \ 6+1 end{bmatrix} = begin{bmatrix} -5 \ 7 \ 7 end{bmatrix}, vec{v}=begin{bmatrix} 0+2 \ 4+1 \ 6+2 end{bmatrix}= begin{bmatrix} 2\ 5 \ 8 end{bmatrix} $$



It shouldn't matter from which point i compute the other two vectors. Now length cross product vector should give us the area of parallelogram.



$$ vec{u} times vec{u} = begin{vmatrix} i & j & k \ -5 & 7 & 7 \ 2 & 5 & 8 end{vmatrix} = ibegin{vmatrix} 7 & 7 \ 5 & 8 end{vmatrix} - j begin{vmatrix} -5 & 7 \ 2 & 8 end{vmatrix} + kbegin{vmatrix} -5 & 7 \ 2 & 5 end{vmatrix} $$



$$ = i(7cdot 8(7cdot 5)- j(-5cdot 8-7cdot 2) + k(-5 cdot 5 - 7 cdot 2) $$



$$ i(56-35)-j(40-14)+k(-25-14) $$



$$ 21i+54j-39k $$



$$ text{Area} = frac{|vec{u} times vec{v}|}{2} = frac{sqrt{21^2+52^2+(-39)^2}}{2} approx 34.154 $$



However this solution seems to be incorrect. WolframAlpha gives solution to this. Did i compute something simply wrong or is there more fundamental probelm on how i understand the problem ?






vectors area







share|cite|improve this question









$endgroup$




Problem



Triangle can be formed in between three points. These three points are in $mathbb{R}^3$ and
in my case these points are: $p_1=(0,4,6),p_2=(-5,3,1),p_3=(2,1,2)$. Compute area of this triangle.



Attempt to solve



I take one point and draw two vectors $vec{u}$ and $vec{v}$ that define two sides of this triangle. Then i compute cross product of these two vectors which length gives me size of parallelogram. Dividing this parallelogram should give area of this triangle.



$$ vec{u}=begin{bmatrix} 0-5 \4+3 \ 6+1 end{bmatrix} = begin{bmatrix} -5 \ 7 \ 7 end{bmatrix}, vec{v}=begin{bmatrix} 0+2 \ 4+1 \ 6+2 end{bmatrix}= begin{bmatrix} 2\ 5 \ 8 end{bmatrix} $$



It shouldn't matter from which point i compute the other two vectors. Now length cross product vector should give us the area of parallelogram.



$$ vec{u} times vec{u} = begin{vmatrix} i & j & k \ -5 & 7 & 7 \ 2 & 5 & 8 end{vmatrix} = ibegin{vmatrix} 7 & 7 \ 5 & 8 end{vmatrix} - j begin{vmatrix} -5 & 7 \ 2 & 8 end{vmatrix} + kbegin{vmatrix} -5 & 7 \ 2 & 5 end{vmatrix} $$



$$ = i(7cdot 8(7cdot 5)- j(-5cdot 8-7cdot 2) + k(-5 cdot 5 - 7 cdot 2) $$



$$ i(56-35)-j(40-14)+k(-25-14) $$



$$ 21i+54j-39k $$



$$ text{Area} = frac{|vec{u} times vec{v}|}{2} = frac{sqrt{21^2+52^2+(-39)^2}}{2} approx 34.154 $$



However this solution seems to be incorrect. WolframAlpha gives solution to this. Did i compute something simply wrong or is there more fundamental probelm on how i understand the problem ?






vectors area




vectors area






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 19:03









TukiTuki

1,019416




1,019416












  • $begingroup$
    Notice that $|utimes v|^2=|u|^2|v|^2-(ucdot v)^2$, so there's no need to calculate the cross product explicitly.
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 20:00


















  • $begingroup$
    Notice that $|utimes v|^2=|u|^2|v|^2-(ucdot v)^2$, so there's no need to calculate the cross product explicitly.
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 20:00
















$begingroup$
Notice that $|utimes v|^2=|u|^2|v|^2-(ucdot v)^2$, so there's no need to calculate the cross product explicitly.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 20:00




$begingroup$
Notice that $|utimes v|^2=|u|^2|v|^2-(ucdot v)^2$, so there's no need to calculate the cross product explicitly.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 20:00










1 Answer
1






active

oldest

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2












$begingroup$

You've defined $vec{u}=vec{p}_1+vec{p}_2,,vec{v}=vec{p}_1+vec{p}_3$. You should have used $-$ instead of $+$ in each definition. For example, $vec{p}_1-vec{p}_2$ is the path from $vec{p}_2$ to $vec{p}_1$, forming a side.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain why ?
    $endgroup$
    – Tuki
    Dec 10 '18 at 19:08










  • $begingroup$
    @Tuki See my edit.
    $endgroup$
    – J.G.
    Dec 10 '18 at 19:09











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You've defined $vec{u}=vec{p}_1+vec{p}_2,,vec{v}=vec{p}_1+vec{p}_3$. You should have used $-$ instead of $+$ in each definition. For example, $vec{p}_1-vec{p}_2$ is the path from $vec{p}_2$ to $vec{p}_1$, forming a side.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain why ?
    $endgroup$
    – Tuki
    Dec 10 '18 at 19:08










  • $begingroup$
    @Tuki See my edit.
    $endgroup$
    – J.G.
    Dec 10 '18 at 19:09
















2












$begingroup$

You've defined $vec{u}=vec{p}_1+vec{p}_2,,vec{v}=vec{p}_1+vec{p}_3$. You should have used $-$ instead of $+$ in each definition. For example, $vec{p}_1-vec{p}_2$ is the path from $vec{p}_2$ to $vec{p}_1$, forming a side.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain why ?
    $endgroup$
    – Tuki
    Dec 10 '18 at 19:08










  • $begingroup$
    @Tuki See my edit.
    $endgroup$
    – J.G.
    Dec 10 '18 at 19:09














2












2








2





$begingroup$

You've defined $vec{u}=vec{p}_1+vec{p}_2,,vec{v}=vec{p}_1+vec{p}_3$. You should have used $-$ instead of $+$ in each definition. For example, $vec{p}_1-vec{p}_2$ is the path from $vec{p}_2$ to $vec{p}_1$, forming a side.






share|cite|improve this answer











$endgroup$



You've defined $vec{u}=vec{p}_1+vec{p}_2,,vec{v}=vec{p}_1+vec{p}_3$. You should have used $-$ instead of $+$ in each definition. For example, $vec{p}_1-vec{p}_2$ is the path from $vec{p}_2$ to $vec{p}_1$, forming a side.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 19:35

























answered Dec 10 '18 at 19:07









J.G.J.G.

27.6k22843




27.6k22843












  • $begingroup$
    Could you explain why ?
    $endgroup$
    – Tuki
    Dec 10 '18 at 19:08










  • $begingroup$
    @Tuki See my edit.
    $endgroup$
    – J.G.
    Dec 10 '18 at 19:09


















  • $begingroup$
    Could you explain why ?
    $endgroup$
    – Tuki
    Dec 10 '18 at 19:08










  • $begingroup$
    @Tuki See my edit.
    $endgroup$
    – J.G.
    Dec 10 '18 at 19:09
















$begingroup$
Could you explain why ?
$endgroup$
– Tuki
Dec 10 '18 at 19:08




$begingroup$
Could you explain why ?
$endgroup$
– Tuki
Dec 10 '18 at 19:08












$begingroup$
@Tuki See my edit.
$endgroup$
– J.G.
Dec 10 '18 at 19:09




$begingroup$
@Tuki See my edit.
$endgroup$
– J.G.
Dec 10 '18 at 19:09


















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