Question

Sejam os pontos A(1, 2, 1), B(−2, −1, 2) e C(2, 3, 1). Determine a altura do triângulo ABC em relação a base AC e determine sua área.
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Answer 1

$A(1,2,1) , B(-2,-1,2) , C(2,3,1)$

$Vetores:$

$\overrightarrow{CB}=-4\overrightarrow{i}+ -4\overrightarrow{j}+1\overrightarrow{k}$

$\overrightarrow{CA}=-1\overrightarrow{i}-1\overrightarrow{j}+0\overrightarrow{k}$

$| \overrightarrow{CA}| = \sqrt{(-1)^2+(-1)^2+(0)^2} \\| \overrightarrow{CA}| = \sqrt{2}$

$| \overrightarrow{CB}| = \sqrt{(-4)^2+(-4)^2+(1)^2}\\ | \overrightarrow{CB}| = \sqrt{33}$

$Nao \ estou \ conseguindo \ usar \ o \ LaTeX \ direito . \ Entao\ CB\ representa\ o \\ vetor\ \overrightarrow{CB} \ e\ assim\ por\ diante.$

$Proj_{CA}^{CB} = \frac{CB.CA}{(|CA|)^2}\ . \ \overrightarrow{CA}$

$Proj_{CA}^{CB} = \frac{(-4).(-1)+(-4).(-1)+(1).(0)}{(\sqrt{2})^2 } \ . \ (-1,-1,0) \\\\Proj_{CA}^{CB} = \frac{-8}{2}\ . \ (-1,-1,0) \\\\Proj_{CA}^{BC} = (4,4,0)\\\\\\|Proj_{CA}^{CB}| = \sqrt{4^2 + 4^2 + 0^2}\\\\|Proj_{CA}^{CB}| = 4\sqrt{2}$

$(| \overrightarrow{CB}|)^2= (|Proj_{CA}^{BC}| )^2 + (| \overrightarrow{h}|)^2\\\\(\sqrt{33})^2 = (4\sqrt{2})^2 + (| \overrightarrow{h}|)^2\\\\| \overrightarrow{h}| = 1\ (u.m.)$

$A\ altura\ relativa\ a\ base\ CA \ ( sendo\ CA\ que\ agora\ nao\ e\ um\ vetor)\\\\e\ igual\ | \overrightarrow{h}| = 1.$

$Area:\\\\S = \frac{1}{2}\ . \ | \overrightarrow{CA}| \ . \ | \overrightarrow{h}|\\\\S = \frac{1}{2}\ . \sqrt{2} \ . \ 1 \\\\S = \frac{\sqrt{2} }{2}\ (u.m.)^2$

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