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2014-05-23T16:35:06-03:00
A= (-4,1,1)\\\\B=(1,0,1)\\\\C=(0,-1,3)

os lados desse triangulo são
AB ; AC ; BC

AB = B-A
determiinando as coordenadas de AB
(1,0,1) - (-4,1,1)\\\\AB=(1-(-4));(0-1);(1-1)\\\\\boxed{AB=(5,-1,0)}

AC = C-A
AC = (0,-1,3)-(-4,1,1)\\\\\boxed{AC=(4,-2,2)}

BC=(0,-1,3)-(1,0,1)\\\\BC=(-1,-1,2)

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para calcualr a area desse triangulo é só fazer o produto vetorial entre
AB E AC

  \left[\begin{array}{ccc}i&j&k\\5&-1&0\\4&-2&2\end{array}\right] \\\\\\i=-2\\j=-10\\k=-6\\\\(-2,-10,-6)

a area vai ser o modulo disso dividido por 2
A= \sqrt{-2^2+ -10^2 + -6^2} \\\\A= \sqrt{4+100+36} \\\\A= \sqrt{140} \\\\Area=  \frac{ \sqrt{140} }{2}

fatorando a raíz de 140
140| 2
70 | 2
35 | 5
7   | 7
1
 \sqrt{140} =2 \sqrt{5*7} =2 \sqrt{35}
então a area sera
AREA =  \frac{2 \sqrt{35} }{2} = \sqrt{35}
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a altura relativa ao lado BC

|BC|= \sqrt{1^2+1^1+2^2} \\\\|BC|= \sqrt{6}

area de um triangulo = (base*H)/2
H é a altura
base = |BC|
area = √35

 \sqrt{35}=  \frac{1}{2} *\sqrt{6} *H

isolando H temos
\frac{2* \sqrt{35} }{ \sqrt{6} } =H

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