Question

Cho SABC, SA⊥ (ABC), SA=a, tam giác ABC đều cạnh a. Tính d(SB,AC)

Answer 1

Qua B kẻ đường thẳng d song song AC

Qua A kẻ đường thẳng d' vuông góc AC cắt d tại D

\(\Rightarrow AC//BD\Rightarrow AC//\left(SBD\right)\)

\(\Rightarrow d\left(AC;SB\right)=d\left(AC;\left(SBD\right)\right)=d\left(A;\left(SBD\right)\right)\)

\(SA\perp\left(ABC\right)\Rightarrow SA\perp BD\)

\(AD\perp AC\Rightarrow AD\perp BD\Rightarrow BD\perp\left(SAD\right)\)

Từ A kẻ \(AH\perp SD\Rightarrow AH\perp\left(SBD\right)\Rightarrow AH=d\left(A;\left(SBD\right)\right)\)

\(AD=AB.cos30^0=\frac{a\sqrt{3}}{2}\)

\(\frac{1}{AH^2}=\frac{1}{AD^2}+\frac{1}{SA^2}\Rightarrow AH=\frac{AD.SA}{\sqrt{AD^2+SA^2}}=\frac{a\sqrt{21}}{7}\)