Area of a triangle using vectors 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 $= \frac{1}{2}ab\sin{C}$
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} = (5\sqrt{3})(15)\sin{\theta} $
$\therefore \sin{\theta} = -\dfrac{1}{\sqrt{3}}$
If I substitute these values into the general formula:
area $= \frac{1}{2}ab\sin{C}$
I get:
area $= \frac{1}{2}(5\sqrt{3})(15)(-\dfrac{1}{\sqrt{3}})$
$\therefore$ 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. 

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:
$Area=(1/2) | a \times b |$.
