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.

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

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

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

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

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

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 ?

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.