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 ?
