Compute area of triangle
$begingroup$
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
asked Dec 10 '18 at 19:03
TukiTuki
1,019416
$endgroup$
add a comment |
$begingroup$
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
asked Dec 10 '18 at 19:03
TukiTuki
1,019416
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 10 '18 at 19:03
TukiTuki
1,019416
$endgroup$
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
asked Dec 10 '18 at 19:03
TukiTuki
1,019416
asked Dec 10 '18 at 19:03
TukiTuki
1,019416
asked Dec 10 '18 at 19:03
TukiTuki
1,019416
asked Dec 10 '18 at 19:03
TukiTuki
1,019416
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
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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.
answered Dec 10 '18 at 19:07
J.G.J.G.
27.6k22843
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 10 '18 at 19:07
J.G.J.G.
27.6k22843
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 10 '18 at 19:07
J.G.J.G.
27.6k22843
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 10 '18 at 19:07
J.G.J.G.
27.6k22843
$endgroup$
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.
answered Dec 10 '18 at 19:07
J.G.J.G.
27.6k22843
edited Dec 10 '18 at 19:35
answered Dec 10 '18 at 19:07
J.G.J.G.
27.6k22843
answered Dec 10 '18 at 19:07
J.G.J.G.
27.6k22843
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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