Trong không gian tọa độ $Oxyz$, cho hai mặt phẳng $(P): x+y+z-3=0$ và $(Q): x-y+z-1=0$. Viết phương trình mặt phẳng $(R)$ vuông góc với$ (P), (Q)$ sao cho khoảng cách từ $O$ đến $(R)$ bằng $2$.
Do $1:1:1\neq 1:(-1):1$, nên $(P)\cap (Q)=d$, trong đó có vectơ chỉ phương :
  $\overrightarrow{u_d}=(\left| {\begin{array}{*{20}{c}}
{{1}}&{{1}}\\
{{-1}}&{{1}}
\end{array}} \right|;\left| {\begin{array}{*{20}{c}}
{{1}}&{{1}}\\
{{1}}&{{1}}
\end{array}} \right|;\left| {\begin{array}{*{20}{c}}
{{1}}&{{1}}\\
{{1}}&{{-1}}
\end{array}} \right|)=(2;0;-2)//(1;0;-1)$
Do $d\bot (R)$ nên có thể lấy $\overrightarrow{u_d}$ làm vectơ pháp tuyến của $(R)$.
Vì thế $(R)$ có dạng: $x-z+d=0$
từ $d(O;(R))=2\Leftrightarrow \frac{|D|}{\sqrt{2}}=2\Leftrightarrow D=\pm 2\sqrt{2}$.
Vậy có hai mặt phẳng $(R)$ thỏa mãn yêu cầu đầu bài là: $(R_1):x-z+2\sqrt{2}=0; (R_2): x-z-2\sqrt{2}=0$.


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