Mais dois vetores
Geometria Analítica: Infinito 11 A - Parte 1 Pág. 178 Ex. 10
Os vetores $\overrightarrow{u}$ e $\overrightarrow{v}$ verificam a condições, em $(O,\overrightarrow{i},\overrightarrow{j})$:
- $\begin{matrix}
\overrightarrow{u}.\overrightarrow{i}=2 & \wedge & \overrightarrow{u}.\overrightarrow{j}=-3 \\
\end{matrix}$ - $\begin{matrix}
\overrightarrow{v}.\overrightarrow{i}=-4 & \wedge & \overrightarrow{v}.\overrightarrow{j}=-5 \\
\end{matrix}$
- Quais são as coordenadas de $\overrightarrow{u}$ e de $\overrightarrow{v}$?
- Calcule $\overrightarrow{u}.\overrightarrow{v}$.
- $\overrightarrow{u}=(2,-3)$ e $\overrightarrow{v}=(-4,-5)$.
Com efeito, para $\overrightarrow{u}=({{u}_{1}},{{u}_{2}})$, obtém-se:$\begin{array}{*{35}{l}}
\overrightarrow{u}.\overrightarrow{i}=2 & \Leftrightarrow & ({{u}_{1}}\overrightarrow{i}+{{u}_{2}}\overrightarrow{j})\overrightarrow{i}=2 \\
{} & \Leftrightarrow & {{u}_{1}}\overrightarrow{i}.\overrightarrow{i}+{{u}_{2}}\overrightarrow{j}.\overrightarrow{i}=2 \\
{} & \Leftrightarrow & {{u}_{1}}\times 1+{{u}_{2}}\times 0=2 \\
{} & \Leftrightarrow & {{u}_{1}}=2 \\
\end{array}$
e$\begin{array}{*{35}{l}}
\overrightarrow{u}.\overrightarrow{j}=-3 & \Leftrightarrow & ({{u}_{1}}\overrightarrow{i}+{{u}_{2}}\overrightarrow{j})\overrightarrow{j}=-3 \\
{} & \Leftrightarrow & {{u}_{1}}\overrightarrow{i}.\overrightarrow{j}+{{u}_{2}}\overrightarrow{j}.\overrightarrow{j}=-3 \\
{} & \Leftrightarrow & {{u}_{1}}\times 0+{{u}_{2}}\times 1=-3 \\
{} & \Leftrightarrow & {{u}_{2}}=-3 \\
\end{array}$
Logo, $\overrightarrow{u}=(2,-3)$ e, de forma análoga, $\overrightarrow{v}=(-4,-5)$.
-
Como $\overrightarrow{u}=(2,-3)$ e $\overrightarrow{v}=(-4,-5)$, então:
$\begin{array}{*{35}{l}}
\overrightarrow{u}.\overrightarrow{v} & = & (2\overrightarrow{i}-3\overrightarrow{j}).(-4\overrightarrow{i}-5\overrightarrow{j}) \\
{} & = & -8\overrightarrow{i}.\overrightarrow{i}-10\overrightarrow{i}.\overrightarrow{j}+12\overrightarrow{i}.\overrightarrow{j}+15\overrightarrow{j}.\overrightarrow{j} \\
{} & = & -8-0+0+15 \\
{} & = & 7 \\
\end{array}$Ou, ainda: $\overrightarrow{u}.\overrightarrow{v}=(2,-3).(-4,-5)=2\times (-4)+(-3)\times (-5)=-8+15=7$.
