Question

Answer 1

A(2;1;5); B(0;-1;-2)

=>\(\overrightarrow{AB}=\left(0-2;-1-1;-2-5\right)\)

=>\(\overrightarrow{AB}=\left(-2;-2;-7\right)\)

=>\(\overrightarrow{BA}_{}=\left(2;2;7\right)\)

A(2;1;5); C(-4;1;-2)

\(\overrightarrow{AC}=\left(-4-2;1-1;-2-5\right)\)

=>\(\overrightarrow{AC}=\left(-6;0;-7\right)\)

=>\(\overrightarrow{CA}=\left(6;0;7\right)\)

B(0;-1;-2); C(-4;1;-2)

=>\(\overrightarrow{BC}=\left(-4-0;1+1;-2+2\right)\)

=>\(\overrightarrow{BC}=\left(-4;2;0\right)\)

=>\(\overrightarrow{CB}=\left(4;-2;0\right)\)

Tọa độ vecto \(\left\lbrack\overrightarrow{AB};\overrightarrow{AC}\right\rbrack\) là:

\(\begin{cases}x=\left(-2\right)\cdot\left(-7\right)-\left(-7\right)\cdot0=14-0=14\\ y=-7\cdot\left(-6\right)-\left(-2\right)\cdot\left(-7\right)=42-14=28\\ z=\left(-2\right)\cdot0-\left(-2\right)\cdot\left(-6\right)=-12\end{cases}\)

Tọa độ vecto \(\left\lbrack\overrightarrow{AB};\overrightarrow{BC}\right\rbrack\) là:

\(\begin{cases}x=-2\cdot0-\left(-7\right)\cdot2=14\\ y=-7\cdot\left(-4\right)-\left(-2\right)\cdot0=28\\ z=-2\cdot2-\left(-2\right)\cdot\left(-4\right)=-4-8=-12\end{cases}\)

Tọa độ của vecto \(\left\lbrack\overrightarrow{AC};\overrightarrow{CB}\right\rbrack\) là:

\(\begin{cases}x=0\cdot0-\left(-7\right)\cdot\left(-2\right)=0-14=-14\\ y=-7\cdot4-\left(-6\right)\cdot0=-28\\ z=-6\cdot\left(-2\right)-0\cdot4=12\end{cases}\)