Trong không gian cho hình hộp chữ nhật $ABCD.A'B'C'D'$ với tọa độ các đỉnh như sau: $A'=(0;0;0); B'=(a;0;0); D'=(0;b;0); A=(0;0;c) (a,b,c>0)$ . Gọi $P,Q,R,S$  lần lượt là trung điểm của các cạnh $AB,B'C',C'D',DD'$. Tìm mối liên hệ giữa $a,b,c$ để $PR\bot QS$.
Ta có: $C'=(a;b;0); B=(a;0;c); C=(a;b;c); D=(0;b;c)$.

Từ đó theo công thức tính tọa độ trung điểm ta có:
$P=(\frac{a}{2};0;c); Q=(a;\frac{b}{2};0); R=(\frac{a}{2};b;0); S=(0;b;\frac{c}{2})$
$\Rightarrow \overrightarrow{PR}=(0;b;-c); \overrightarrow{QS}=(-a;\frac{b}{2};\frac{c}{2})$.
Vậy $PR\bot QS\Rightarrow \overrightarrow{PR}.\overrightarrow{QS}=0$
$\Leftrightarrow \frac{b^2}{2}-\frac{c^2}{2}=0\Leftrightarrow b=c$.

Vậy nếu $b=c$ thì $PR \bot QS$ .

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