Question

[doshu tmato]

giup mình voi ạ

Answer 1

[Hoàng Tử Hà]

a/ \(\left(P\right):3\left(x-0\right)+0\left(y-2\right)+1\left(z+5\right)=0\Rightarrow\left(P\right):3x+z+5=0\)

b/\(\overrightarrow{AB}\left(2;4;-9\right);\overrightarrow{AC}\left(4;0;-7\right)\)

 \(\overrightarrow{n_{\left(P\right)}}=\left[\overrightarrow{AB},\overrightarrow{AC}\right]=\left(4.\left(-7\right)-0.\left(-9\right);\left(-9\right).4-\left(-7\right).2;2.0-4.4\right)=\left(-28;-22;-16\right)\)

\(\Rightarrow\left(P\right):-28\left(x-0\right)-22\left(y-1\right)-16\left(z-7\right)=0\Rightarrow\left(P\right):28x+22y+16z-134=0\)

c/ Truc Oy di qua O(0;0;0) va co vtcp \(\overrightarrow{j}\left(0;1;0\right)\)

\(\overrightarrow{OD}\left(3;-6;2\right)\)

\(\Rightarrow\overrightarrow{n_{\left(P\right)}}=\left[\overrightarrow{j};\overrightarrow{OD}\right]=\left(2;0;-3\right)\)

\(\Rightarrow\left(P\right):2\left(x-0\right)+0\left(y-1\right)-3\left(z-0\right)=0\Rightarrow\left(P\right):2x-3z=0\)

d/ \(\overrightarrow{Oz}\left(0;0;1\right)\)

\(\overrightarrow{DE}\left(5;-2;-7\right)\)

\(\Rightarrow\overrightarrow{n_{\left(P\right)}}=\left[\overrightarrow{Oz};\overrightarrow{DE}\right]=\left(2;5;0\right)\)

\(\Rightarrow\left(P\right):2\left(x-0\right)+5\left(y-0\right)+0\left(z-1\right)=0\Rightarrow\left(P\right):2x+5y=0\)

e/ \(\overrightarrow{n_{Oyz}}=\overrightarrow{n_{\left(P\right)}}=\left(1;0;0\right)\)

\(\Rightarrow\left(P\right):1\left(x-3\right)=0\Rightarrow\left(P\right):x-3=0\)

f/ Cách làm giống câu b

g/ \(\overrightarrow{HI}=\overrightarrow{IK}\Rightarrow\left\{{}\begin{matrix}x_I=\dfrac{3-1}{2}=1\\y_I=\dfrac{-1+5}{2}=2\\z_1=\dfrac{2-4}{2}=-1\end{matrix}\right.\Rightarrow I\left(1;2;-1\right)\)

\(\overrightarrow{n_{\left(P\right)}}=\overrightarrow{HK}\left(-4;6;-6\right)\)

\(\Rightarrow\left(P\right):-4\left(x-1\right)+6\left(y-2\right)-6\left(z+1\right)=0\Rightarrow\left(P\right):-4x+6y-6z+2=0\)

P/s: Bạn tính toán lại kết quả hộ mình nhé !