Dados os ponto A, B e C

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 176 Ex. 4

by AMMA · Published 8 de Novembro de 2010 · Updated 12 de Janeiro de 2022

Enunciado

Considere um referencial ortonormado $(O,\vec{i},\vec{j},\vec{k})$.

Dados os pontos $A\,(-2,1,5)$, $B\,(1,-3,0)$ e $C\,(2,2,-1)$

Determine as coordenadas do ponto M, tal que $\overrightarrow{BM}=3\,\overrightarrow{AB}+2\,\overrightarrow{AC}$.
Determine as coordenadas do ponto N, tal que $2\,\overrightarrow{NA}=3\,\overrightarrow{NB}$.
Calcule as coordenadas do ponto médio I de [MN].
Calcule as coordenadas do simétrico de C em relação a I.

Resolução

Ora,
$\begin{array}{*{35}{l}}
\overrightarrow{BM} & = & 3\,\overrightarrow{AB}+2\,\overrightarrow{AC} \\
{} & = & 3(1+2,-3-1,0-5)+2(2+2,2-1,-1-5) \\
{} & = & (9,-12,-15)+(8,2,-12) \\
{} & = & (17,-10,-27) \\
\end{array}$
logo:
$\begin{array}{*{35}{l}}
\overrightarrow{BM}=(17,-10,-27) & \Leftrightarrow & M-B=(17,-10,-27) \\
{} & \Leftrightarrow & M=B+(17,-10,-27) \\
{} & \Leftrightarrow & M=(1,-3,0)+(17,-10,-27) \\
{} & \Leftrightarrow & M=(18,-13,-27) \\
\end{array}$
Considerando $N\,(x,y,z)$, vem:
$\begin{array}{*{35}{l}}
2\,\overrightarrow{NA}=3\,\overrightarrow{NB} & \Leftrightarrow & 2(-2-x,1-y,5-z)=3(1-x,-3-y,0-z) \\
{} & \Leftrightarrow & (-4-2x,2-2y,10-2z)=(3-3x,-9-3y,-3z) \\
{} & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}}
-4-2x=3-3x \\
2-2y=-9-3y \\
10-2z=-3z \\
\end{array} \right. \\
{} & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}}
x=7 \\
y=-11 \\
z=-10 \\
\end{array} \right. \\
\end{array}$
Logo, $N\,(7,-11,-10)$.

ALTERNATIVA:

$\begin{array}{*{35}{l}}
2\,\overrightarrow{NA}=3\,\overrightarrow{NB} & \Leftrightarrow & 2(A-N)=3(B-N) \\
{} & \Leftrightarrow & 3N-2N=3B-2A \\
{} & \Leftrightarrow & N=3(1,-3,0)-2(-2,1,5) \\
{} & \Leftrightarrow & N=(7,-11,-10) \\
\end{array}$
As coordenadas de I são $(\frac{18+7}{2},\frac{-13+(-11)}{2},\frac{-27+(-10)}{2})=(\frac{25}{2},-12,-\frac{37}{2})$.
Seja C’ o simétrico de C em relação a I.
Ora, C’ é tal que $C’=I+\overrightarrow{CI}$. (Verifique graficamente)
Logo,
$\begin{array}{*{35}{l}}
C’ & = & (\frac{25}{2},-12,-\frac{37}{2})+(\frac{25}{2}-2,-12-2,-\frac{37}{2}+1) \\
{} & = & (23,-26,-36) \\
\end{array}$

ALTERNATIVA:

O ponto médio do segmento [CC’] será o ponto I (Verifique graficamente). Logo, sendo C’ (x,y.z), vem:
\[\begin{array}{*{35}{l}}
(\frac{x+2}{2},\frac{y+2}{2},\frac{z-1}{2})=(\frac{25}{2},-12,-\frac{37}{2}) & \Leftrightarrow & (x+2,y+2,z-1)=(25,-24,-37) \\
{} & \Leftrightarrow & (x,y,z)=(23,-26,-36) \\
\end{array}\]