Google
Câu hỏi: Trong không gian với hệ tọa độ Oxyz, khoảng cách giữa mặt phẳng $\left( \alpha \right):2x+4y+4z+1=0$ và mặt phẳng $\left( \beta \right):x+2y+2z+2=0$ bằng
A. $\dfrac{3}{2}$.
B. $\dfrac{1}{3}$.
C. $\dfrac{1}{2}$.
D. 1.
A. $\dfrac{3}{2}$.
B. $\dfrac{1}{3}$.
C. $\dfrac{1}{2}$.
D. 1.
Cách 1: Chọn $M\left( -2;0;0 \right)\in \left( \beta \right)$
Do $\left( \alpha \right)\text{//}\left( \beta \right)$ ta có: $d\left( \left( \alpha \right),\left( \beta \right) \right)=d\left( M,\left( \alpha \right) \right)=\dfrac{\left| -4+0+0+1 \right|}{\sqrt{{{2}^{2}}+{{4}^{2}}+{{4}^{2}}}}=\dfrac{1}{2}$.
Cách 2:
$\left( \alpha \right):2x+4y+4z+1=0\Leftrightarrow x+2y+2x+\dfrac{1}{2}=0$.
Khi đó: $d\left( \left( \alpha \right),\left( \beta \right) \right)=\dfrac{\left| \dfrac{1}{2}-2 \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}}=\dfrac{1}{2}$.
Tổng quát: Khoảng cách giữa hai mặt phẳng song song $\left( \alpha \right):ax+by+cz+{{d}_{1}}=0$, $\left( \beta \right):ax+by+cz+{{d}_{2}}=0$ là:
$d\left( \left( \alpha \right),\left( \beta \right) \right)=\dfrac{\left| {{d}_{1}}-{{d}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
Do $\left( \alpha \right)\text{//}\left( \beta \right)$ ta có: $d\left( \left( \alpha \right),\left( \beta \right) \right)=d\left( M,\left( \alpha \right) \right)=\dfrac{\left| -4+0+0+1 \right|}{\sqrt{{{2}^{2}}+{{4}^{2}}+{{4}^{2}}}}=\dfrac{1}{2}$.
Cách 2:
$\left( \alpha \right):2x+4y+4z+1=0\Leftrightarrow x+2y+2x+\dfrac{1}{2}=0$.
Khi đó: $d\left( \left( \alpha \right),\left( \beta \right) \right)=\dfrac{\left| \dfrac{1}{2}-2 \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}}=\dfrac{1}{2}$.
Tổng quát: Khoảng cách giữa hai mặt phẳng song song $\left( \alpha \right):ax+by+cz+{{d}_{1}}=0$, $\left( \beta \right):ax+by+cz+{{d}_{2}}=0$ là:
$d\left( \left( \alpha \right),\left( \beta \right) \right)=\dfrac{\left| {{d}_{1}}-{{d}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
Đáp án C.