Question

trong không gian oxyz cho hình hộp abcd.a'b'c'd' VỚI A(2;1;2) B'(1;2;1) C(-2;3;2) D'(3;0;1) tìm tọa độ điểm B

 

Answer 1

\(\overrightarrow{B'D'}=\left(2;-2;0\right)\)

Gọi \(B\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BA}=\left(2-x;1-y;2-z\right)\\\overrightarrow{BC}=\left(-2-x;3-y;2-z\right)\end{matrix}\right.\)

\(\overrightarrow{BA}+\overrightarrow{BC}=\overrightarrow{BD}=\overrightarrow{B'D'}\)

\(\Rightarrow\left\{{}\begin{matrix}2-x+\left(-2-x\right)=2\\1-y+\left(3-y\right)=-2\\2-z+\left(2-z\right)=0\end{matrix}\right.\) \(\Rightarrow B\left(-1;3;2\right)\)