**Question:**

Let ABC be a pyramid whose base ABC is a right triangle at A. Side AC = a, \(BC = a\sqrt 5 \). The plane (SAB) is perpendicular to the base plane and the equilateral triangle SAB. Call K points on the side SC such that SC = 3SK. Calculate the distance \(d\) between the two lines AC and BK according to a.

**Reference explanation:**

**Correct answer:**

Let H be the midpoint of AB \( \Rightarrow SH \bot AB\) (due to equilateral triangle SAB)

Do \((SAB) \bot (ABC) \Rightarrow SH \bot (ABC)\)

Since triangle ABC is right-angled at A, AB=2a\( \Rightarrow SH = a\sqrt 3 .\)

\({S_{ABC}} = \frac{1}{2}AB.AC = \frac{1}{2}.2a.a = {a^2}\)

Draw KM parallel to AC cutting SA at M. Then AC//KM deduces AC//(BKM)

So d(AC,BK)=d(AC,(BKM))

We have \(AC \bot AB;AC \bot SH\) so \(AC \bot (SAB)\)

The guy \(AI \bot BM,\) is done by KM//AC so \(AI \bot KM\) infers \(AI \bot \left( {BKM} \right)\)

Infer d(AC,BK)=d(AC,(BKM))=d(A,(BKM))=AI

We have: \(\frac{{MA}}{{SA}} = \frac{{KC}}{{SC}} = \frac{2}{3} \Rightarrow {S_{AMB}} = \frac {2}{3}{S_{SAB}} = \frac{2}{3}{(2a)^2}\frac{{\sqrt 3 }}{4} = \frac{2}{3}{ a^2}\sqrt 3 .\)

We have again \(BM = \sqrt {A{B^2} + A{M^2} – AB.AM.\cos {{60}^0}} = \frac{{2a\sqrt 7 }}{ 3}\)

So \(AI = \frac{{2{S_{ABM}}}}{{BM}} = \frac{{2\sqrt {21} a}}{7}.\) So \(d(AC) ,BK) = \frac{{2\sqrt {21} a}}{7}.\)

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