Chọn hệ trục tọa độ \(Oxyz\)z sao cho :
\(A\left(0;0;0\right);B\left(a;0;0\right);C\left(a;a;0\right);D\left(0;a;0\right);S\left(0;0;2a\right);M\left(\dfrac{a}{2};a;0\right)\)
\(\overrightarrow{SB}=\left(a;0;-2a\right)\)
\(\overrightarrow{SD}=\left(0;a;-2a\right)\)
\(\overrightarrow{n\left(SBD\right)}=\left[\overrightarrow{SB}.\overrightarrow{SD}\right]=\left(2a^2;2a^2;a^2\right)=\left(2;2;1\right)\)
\(\Rightarrow\left(SBD\right):2\left(x-a\right)+2\left(y-0\right)+\left(z-0\right)=0\) hay \(2x+2y+z-2a=0\)
\(d\left(M;\left(SBD\right)\right)=\dfrac{\left|2.\dfrac{a}{2}+2.a+0-2a\right|}{\sqrt{2^2+2^2+1^2}}=\dfrac{a}{3}\)



