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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.

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1  
$\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos \theta$. –  Alyosha Feb 7 at 16:54
    
Haha, thank you, can't believe I didn't check that. –  Elise Feb 7 at 16:55
1  
Use the cross product ... –  Mark Bennet Feb 7 at 17:31
    
@MarkBennet that would make it a lot more efficient, thank you :) –  Elise Feb 7 at 18:05

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