Cho hình chóp \(S.ABC\) có đáy \(ABC\) là tam giác đều cạnh bằng 1. Biết khoảng cách từ \(A\) đến mặt phẳng \(\left( SBC \right)\) là

Chọn B

Gọi \(O\) là chân đường cao hạ từ \(S\) xuống mặt phẳng \(\left( ABC \right)\)

Đặt \(d\left( A,BC \right)=a,d\left( B,AC \right)=b,d\left( C,AB \right)=c,SO=h\)

Ta có \({{S}_{\Delta ABC}}={{S}_{\Delta OBC}}+{{S}_{\Delta OAC}}+{{S}_{\Delta OAB}}\Rightarrow a+b+c=\frac{\sqrt{3}}{2}\left( 1 \right)\)

Mặt khác \(\frac{d\left( O,\left( SBC \right) \right)}{d\left( A,\left( SBC \right) \right)}=\frac{OM}{AM}=\frac{OI}{AK}=\frac{2a}{\sqrt{3}}\Rightarrow d\left( O,\left( SBC \right) \right)=\frac{2a}{\sqrt{3}}.\frac{\sqrt{6}}{4}=\frac{a}{\sqrt{2}}\)

\(\Rightarrow \frac{2}{{{a}^{2}}}=\frac{1}{{{h}^{2}}}+\frac{1}{{{a}^{2}}}\Rightarrow a=h\)

Tương tự\(\frac{d\left( O,\left( SAC \right) \right)}{d\left( B,\left( SAC \right) \right)}=\frac{d\left( O,AC \right)}{d\left( B,AC \right)}==\frac{2b}{\sqrt{3}}\Rightarrow d\left( O,\left( SAC \right) \right)=\frac{2b}{\sqrt{3}}.\frac{\sqrt{15}}{10}=\frac{b}{\sqrt{5}}\)

\(\Rightarrow \frac{5}{{{b}^{2}}}=\frac{1}{{{h}^{2}}}+\frac{1}{{{b}^{2}}}\Rightarrow b=2h\)

Tương tự\(\frac{d\left( O,\left( SAB \right) \right)}{d\left( C,\left( SAB \right) \right)}=\frac{d\left( O,AB \right)}{d\left( C,AB \right)}==\frac{2c}{\sqrt{3}}\Rightarrow d\left( O,\left( SAC \right) \right)=\frac{2c}{\sqrt{3}}.\frac{\sqrt{30}}{20}=\frac{c}{\sqrt{10}}\)

\(\Rightarrow \frac{10}{{{c}^{2}}}=\frac{1}{{{h}^{2}}}+\frac{1}{{{c}^{2}}}\Rightarrow c=3h\)

\(\Rightarrow a+b+c=\frac{\sqrt{3}}{2}\Leftrightarrow h=\frac{\sqrt{3}}{12}\Rightarrow V=\frac{1}{3}.SO.{{S}_{\Delta ABC}}=\frac{1}{48}\)