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.
Best Answer
The correct formula is $\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \cos{\theta} $
So what you really have is $\cos{\theta} = \cfrac{-1}{\sqrt{3}}$
Therefore $$\sin{\theta} = \sqrt{1 - \cos^2{\theta}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$$
Finally, the area of the triangle is:
$$ Area = \frac{1}{2} (5 \sqrt{3}) (15) \frac{\sqrt{2}}{\sqrt{3}} = \frac{75 \sqrt{2}}{2} $$
We can just multiply $\frac{\sqrt{2}}{\sqrt{2}}$ to the area, and then we get the answer you posted:
$$ Area = \frac{75 \sqrt{2}}{2} \left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{75}{\sqrt{2}} $$