Question

[Thái Thùy Linh]

Cho điểm A(1,-3,1) B(2,5,1) vectơ OC = -3i+2j+5k a) Chứng minh ABC tạo thành một tam giác b) Tìm toạ độ điểm E sao cho OAEB là hình thang với hai đáy OA=2BE c) Tìm toạ độ điểm M sao cho 3vectoAB + 2vectoAM = 3vectoCM

Answer 1

[Nguyễn Việt Lâm]

\(\overrightarrow{OC}=-3i+2j+5k\Rightarrow C\left(-3;2;5\right)\)

\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(1;8;0\right)\\\overrightarrow{AC}=\left(-4;5;4\right)\end{matrix}\right.\)

Hai vecto \(\overrightarrow{AB};\overrightarrow{AC}\) không cùng phương nên A;B;C tạo thành 1 tam giác

b. Gọi \(E\left(x;y;z\right)\Rightarrow\overrightarrow{BE}=\left(x-2;y-5;z-1\right)\)

\(\overrightarrow{OA}=\left(1;-3;1\right)\) , đồng thời OA=2BE

\(\Rightarrow\left[{}\begin{matrix}\overrightarrow{OA}=2\overrightarrow{BE}\\\overrightarrow{OA}=-2\overrightarrow{BE}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(1;-3;1\right)=\left(2x-4;2y-10;2z-2\right)\\\left(1;-3;1\right)=\left(4-2x;10-2y;2-2z\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}E\left(\dfrac{5}{2};\dfrac{7}{2};\dfrac{3}{2}\right)\\E\left(\dfrac{3}{2};\dfrac{13}{2};\dfrac{1}{2}\right)\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

c.

Gọi \(M\left(x;y;z\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(1;10;0\right)\\\overrightarrow{AM}=\left(x-1;y+3;z-1\right)\\\overrightarrow{CM}=\left(x+3;y-2;z-5\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3\overrightarrow{AB}=\left(3;30;0\right)\\2\overrightarrow{AM}=\left(2x-2;2y+6;2z-2\right)\\3\overrightarrow{CM}=\left(3x+9;3y-6;3z-15\right)\end{matrix}\right.\)

\(3\overrightarrow{AB}+2\overrightarrow{AM}=3\overrightarrow{CM}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3+2x-2=3x+9\\30+2y+6=3y-6\\0+2z-2=3z-15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=42\\z=13\end{matrix}\right.\)

\(\Rightarrow M\left(-8;42;13\right)\)