a.
C là trung điểm của AD nên tọa độ D thỏa mãn:
\(\left\{{}\begin{matrix}x_D=2x_C-x_A=-3\\y_D=2y_C-y_A=3\\z_D=2z_C-z_A=4\end{matrix}\right.\) \(\Rightarrow D\left(-3;3;4\right)\)
b.
Gọi \(E\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(3;0;-3\right)\\\overrightarrow{EC}=\left(-2-x;2-y;3-z\right)\end{matrix}\right.\)
ABCE là hbh \(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{EC}\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2-x=3\\2-y=0\\3-z=-3\end{matrix}\right.\) \(\Leftrightarrow E\left(-5;2;6\right)\)
c.
Gọi \(F\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{FA}=\left(-1-x;1-y;2-z\right)\\\overrightarrow{FB}=\left(2-x;1-y;-1-z\right)\\\overrightarrow{FC}=\left(-2-x;2-y;3-z\right)\end{matrix}\right.\)
\(2\overrightarrow{FA}+3\overrightarrow{FB}=\overrightarrow{FC}\Leftrightarrow\left\{{}\begin{matrix}2\left(-1-x\right)+3\left(2-x\right)=-2-x\\2\left(1-y\right)+3\left(1-y\right)=2-y\\2\left(2-z\right)+3\left(-1-z\right)=3-z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{3}{4}\\z=-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow F\left(\dfrac{3}{2};\dfrac{3}{4};-\dfrac{1}{2}\right)\)
d.
Gọi G có tọa độ dạng: \(G\left(x;y;0\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AG}=\left(x+1;y-1;-2\right)\\\overrightarrow{BG}=\left(x-2;y-1;1\right)\end{matrix}\right.\)
Ba điểm A;B;G thẳng hàng khi:
\(\dfrac{x-2}{x+1}=\dfrac{y-1}{y-1}=\dfrac{1}{-2}\)
\(\Rightarrow\) Không tồn tại G thỏa mãn yêu cầu đề bài
e.
Gọi \(H\left(0;y;0\right)\) và H' là trọng tâm tam giác HBC
\(\Rightarrow H'\left(0;\dfrac{y+3}{3};\dfrac{2}{3}\right)\)
H' thuộc Oz khi và chỉ khi \(\dfrac{y+3}{3}=0\Leftrightarrow y=-3\)
\(\Rightarrow H\left(0;-3;0\right)\)
f.
\(\left\{{}\begin{matrix}S_{ABC}=\dfrac{1}{2}AB.d\left(C;AB\right)\\S_{ABI}=\dfrac{1}{2}AB.d\left(I;AB\right)\end{matrix}\right.\)
Mà \(S_{ABC}=3S_{ABI}\Rightarrow d\left(C;AB\right)=3d\left(I;AB\right)\)
\(\Rightarrow\overrightarrow{CB}=3\overrightarrow{IB}\)
Gọi \(I\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{CB}=\left(4;-1;-4\right)\\\overrightarrow{IB}=\left(2-x;1-y;-1-z\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3\left(2-x\right)=4\\3\left(1-y\right)=-1\\3\left(-1-z\right)=-4\end{matrix}\right.\) \(\Leftrightarrow...\) (bạn tự giải ra kết quả)
g.
Gọi \(X\left(x_X;y_X;z_X\right)\) ; \(Y\left(x_Y;y_Y;z_Y\right)\); \(Z\left(x_Z;y_Z;z_Z\right)\)
Do AB là đường trung bình tam giác XYZ \(\Rightarrow AB=\dfrac{1}{2}XZ=XC=CZ\)
\(\Rightarrow\overrightarrow{AB}=\overrightarrow{XC}=\overrightarrow{CZ}\)
Ta có: \(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(3;0;-3\right)\\\overrightarrow{XC}=\left(-2-x_X;2-y_X;3-z_X\right)\\\overrightarrow{CZ}=\left(x_Z+2;y_Z-2;z_Z-3\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-2-x_X=3\\2-y_X=0\\3-z_X=-3\end{matrix}\right.\) đồng thời \(\left\{{}\begin{matrix}x_Z+2=3\\y_z-2=0\\z_Z-3=-3\end{matrix}\right.\)
\(\Rightarrow X\left(-5;2;6\right)\) và \(Z\left(1;2;0\right)\)
Do A là trung điểm XY \(\Rightarrow\left\{{}\begin{matrix}x_Y=2x_A-x_X=3\\y_Y=2y_A-y_X=0\\z_Y=2z_A-z_X=-2\end{matrix}\right.\)
\(\Rightarrow Y\left(3;0;-2\right)\)
h.
Do ABC.A'B'C' là lăng trụ
\(\Rightarrow\overrightarrow{AA'}=\overrightarrow{BB'}=\overrightarrow{CC'}\)
Ta có: \(\left\{{}\begin{matrix}\overrightarrow{AA'}=\left(2;0;-1\right)\\\overrightarrow{BB'}=\left(x_{B'}-2;y_{B'}-1;z_{B'}+1\right)\\\overrightarrow{CC'}=\left(x_{C'}+2;y_{C'}-2;z_{C'}-3\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_{B'}-2=2\\y_{B'}-1=0\\z_{B'}+1=-1\end{matrix}\right.\) và \(\left\{{}\begin{matrix}x_{C'}+2=2\\y_{C'}-2=0\\z_{C'}-3=-1\end{matrix}\right.\)
\(\Rightarrow B'\left(4;1;-2\right)\) và \(C'\left(0;2;2\right)\)
Gọi G' là trọng tâm tam giác AB'C' \(\Rightarrow G'\left(1;\dfrac{4}{3};\dfrac{2}{3}\right)\)
i.
\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(3;0;-3\right)\\\overrightarrow{AC}=\left(-1;1;1\right)\end{matrix}\right.\)
\(\Rightarrow cos\left(\overrightarrow{AB};\overrightarrow{AC}\right)=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=\dfrac{3.\left(-1\right)+0.1-3.1}{\sqrt{3^2+0^2+3^2}.\sqrt{1^2+1^2+1^2}}=-\dfrac{\sqrt{6}}{3}\)
\(\Rightarrow\left(\overrightarrow{AB};\overrightarrow{AC}\right)\simeq161^0\)
j.
\(\left[\overrightarrow{AB};\overrightarrow{AC}\right]=\left(3;0;3\right)=3\left(1;0;1\right)\)
\(\Rightarrow\) Mặt phẳng (ABC) nhận (1;0;1) là 1 vtpt
Phương trình (ABC): \(1\left(x+1\right)+1\left(z-2\right)=0\Leftrightarrow x+z-1=0\)
Tọa độ J là nghiệm của: \(\left\{{}\begin{matrix}x=0\\y=0\\x+z-1=0\end{matrix}\right.\) \(\Leftrightarrow J\left(0;0;1\right)\)
k.
Ở câu trên ta đã tính được \(\left[\overrightarrow{AB};\overrightarrow{AC}\right]=\left(3;0;3\right)\)
Do đó: \(S_{ABC}=\dfrac{1}{2}\left|\left[\overrightarrow{AB};\overrightarrow{AC}\right]\right|=\dfrac{1}{2}\sqrt{3^2+0+3^2}=\dfrac{3\sqrt{2}}{2}\)
\(\overrightarrow{BC}=\left(-4;1;4\right)\Rightarrow BC=\sqrt{33}\)
\(\Rightarrow h_a=\dfrac{2S}{BC}=\dfrac{3\sqrt{2}}{\sqrt{33}}\)
l.
Gọi \(L\left(x;0;0\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AL}=\left(x+1;-1;-2\right)\\\overrightarrow{BL}=\left(x-2;-1;1\right)\end{matrix}\right.\)
\(\Rightarrow\left[\overrightarrow{AL};\overrightarrow{BL}\right]=\left(-3;3-3x;-3\right)\)
\(S_{LAB}=\dfrac{1}{2}\left|\left[\overrightarrow{AL};\overrightarrow{BL}\right]\right|=\dfrac{3\sqrt{6}}{2}\)
\(\Leftrightarrow9+\left(3-3x\right)^2+9=54\)
\(\Rightarrow\left[{}\begin{matrix}3-3x=6\\3-3x=-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}L\left(-1;0;0\right)\\L\left(3;0;0\right)\end{matrix}\right.\)
m.
\(\left\{{}\begin{matrix}\overrightarrow{OA}=\left(-1;1;2\right)\\\overrightarrow{OB}=\left(2;1;-1\right)\\\overrightarrow{OC}=\left(-2;2;3\right)\end{matrix}\right.\)
\(\Rightarrow\left[\overrightarrow{OA};\overrightarrow{OB}\right]=\left(-3;3;3\right)\)
\(\Rightarrow V_{OABC}=\dfrac{1}{6}\left|\left[\overrightarrow{OA};\overrightarrow{OB}\right].\overrightarrow{OC}\right|\)
\(=\dfrac{1}{6}\left|-2.\left(-3\right)+2.3+3.3\right|=\dfrac{7}{2}\)
Gọi OH là chiều cao hạ từ O của tứ diện
\(V_{OABC}=\dfrac{1}{3}OH.S_{ABC}\Rightarrow OH=\dfrac{3V_{OABC}}{S_{ABC}}=\dfrac{7\sqrt{2}}{2}\)
n.
Gọi \(N\left(x;y;z\right)\Rightarrow\overrightarrow{ON}=\left(x;y;z\right)\)
Ta có:
\(\left\{{}\begin{matrix}\overrightarrow{ON}.\overrightarrow{AB}=0\\\overrightarrow{ON}.\overrightarrow{BC}=0\\\overrightarrow{ON}.\overrightarrow{AC}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+0y-3z=0\\-4x+y+4z=0\\-x+y+z=3\end{matrix}\right.\)
\(\Rightarrow N\left(\dfrac{1}{2};4;-\dfrac{1}{2}\right)\)
