Question

Answer 1

Vd1:

a: A(-3;0;2); B(0;-1;4); C(1;5;0)

\(\overrightarrow{AB}=\left(0+3;-1-0;4-2\right)=\left(3;-1;2\right)\)

\(\overrightarrow{AC}=\left(1+3;5-0;0-2\right)=\left(4;5;-2\right)\)

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

\(\begin{cases}x=\left(-1\right)\cdot\left(-2\right)-2\cdot5=2-10=-8\\ y=2\cdot4-3\cdot\left(-2\right)=8+6=14\\ z=3\cdot5-\left(-1\right)\cdot4=15+4=19\end{cases}\)

b: Phương trình mp(ABC) là:

-8(x+3)+14(y-0)+19(z-2)=0

=>-8x-24+14y+19z-38=0

=>-8x+14y+19z=62