Calcular a área do triangulo de vértices:
b) A (1, 0, 1) B (4, 2, 1) C ( 1, 2, 0)
c) A (2, 3, -1) B (3, 1, -2) C ( -1, 0, 2)
d) A (-1, 2, -2) B (2, 3, -1) C (0, 1, 1)
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Pelo modo mais fácil... Determinante dividido por 2
A)![\left[\begin{array}{ccc}1&0&1\\4&2&1\\1&2&0\end{array}\right] = 8 - 4 = \frac{4}{2} = 2](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B0%26amp%3B1%5C%5C4%26amp%3B2%26amp%3B1%5C%5C1%26amp%3B2%26amp%3B0%5Cend%7Barray%7D%5Cright%5D+%3D+8+-+4+%3D++%5Cfrac%7B4%7D%7B2%7D++%3D+2 "\left[\begin{array}{ccc}1&0&1\\4&2&1\\1&2&0\end{array}\right] = 8 - 4 = \frac{4}{2} = 2")
B)![\left[\begin{array}{ccc}2&3&-1\\3&1&-2\\-1&0&2\end{array}\right] = 6+4-18-1 = \frac{-9}{2} =|-4,5| = 4,5](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B3%26amp%3B-1%5C%5C3%26amp%3B1%26amp%3B-2%5C%5C-1%26amp%3B0%26amp%3B2%5Cend%7Barray%7D%5Cright%5D+%3D+6%2B4-18-1+%3D++%5Cfrac%7B-9%7D%7B2%7D+%3D%7C-4%2C5%7C+%3D+4%2C5 "\left[\begin{array}{ccc}2&3&-1\\3&1&-2\\-1&0&2\end{array}\right] = 6+4-18-1 = \frac{-9}{2} =|-4,5| = 4,5")
C)![\left[\begin{array}{ccc}-1&2&-2\\2&3&-1\\0&1&1\end{array}\right] =-3-4-4-1 = \frac{-12}{2} =|-6 |= +6](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26amp%3B2%26amp%3B-2%5C%5C2%26amp%3B3%26amp%3B-1%5C%5C0%26amp%3B1%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%3D-3-4-4-1+%3D++%5Cfrac%7B-12%7D%7B2%7D+%3D%7C-6+%7C%3D+%2B6 "\left[\begin{array}{ccc}-1&2&-2\\2&3&-1\\0&1&1\end{array}\right] =-3-4-4-1 = \frac{-12}{2} =|-6 |= +6")
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