Btukfyl

Let SABC be a tetrahedron, SA = c, AB = a, AC = b, SA perpendicular to AB, AB perpendicular to AC, and AC perpendicular to SA. I am trying to find the projection H of the point A on the plane SBC. I tried with two ways.

Firt way. With some calculations, I found that H(({b^2*c^2*a/((b^2+c^2)*a^2+b^2*c^2)},{b*c^2*a^2/((b^2+c^2)*a^2+b^2*c^2)},{b^2*c*a^2/((b^2+c^2)*a^2+b^2*c^2)})).

Second way, We can prove that, H is orthocentre of triangle SBC, then I see at
Is there a command to find coordinates of projection of a point on a line in 3D?

documentclass[border=3mm,12pt,tikz]{standalone}

usepackage{fouriernc}

usepackage{tikz,tikz-3dplot} 

usepackage{tkz-euclide}

usetkzobj{all} 



usetikzlibrary{intersections,calc,backgrounds}

newcommandpgfmathsinandcos[3]{%

    pgfmathsetmacro#1{sin(#3)}%

    pgfmathsetmacro#2{cos(#3)}%

}

tikzset{projection of point/.style args={(#1,#2,#3) on line through (#4,#5,#6)

        and (#7,#8,#9)}{%

        /utils/exec=pgfmathsetmacro{myprefactor}{((#1-#4)*(#7-#4)+(#2-#5)*(#8-#5)+(#3-#6)*(#9-#6))/((#7-#4)*(#7-#4)+(#8-#5)*(#8-#5)+(#9-#6)*(#9-#6))},

        insert path={%

            ({#4+myprefactor*(#7-#4)},{#5+myprefactor*(#8-#5)},{#6+myprefactor*(#9-#6)})}

}}

begin{document}

    tdplotsetmaincoords{70}{110}

    %tdplotsetmaincoords{80}{100}

    begin{tikzpicture}[tdplot_main_coords,scale=1.5]

    pgfmathsetmacroa{4}

    pgfmathsetmacrob{3}

    pgfmathsetmacroc{4}



    % definitions

    path

    coordinate(A) at (0,0,0)

    coordinate (B) at (a,0,0)

    coordinate (C) at (0,b,0)                           

    coordinate (S) at (0,0,a)

coordinate (H) at ({b^2*c^2*a/((b^2+c^2)*a^2+b^2*c^2)},{b*c^2*a^2/((b^2+c^2)*a^2+b^2*c^2)},{b^2*c*a^2/((b^2+c^2)*a^2+b^2*c^2)});



begin{scope}

draw[dashed,thick]

(A) -- (B)  (A) -- (C)  (S)--(A) --(H) ;

draw[thick]

(S) -- (B) -- (C) -- cycle;

end{scope} 

    foreach point/position in {A/left,B/left,C/below,S/above,H/above}

    {

        fill (point) circle (1.5pt);

        node[position=3pt] at (point) {$point$};

    }

    end{tikzpicture}



    tdplotsetmaincoords{70}{110}

    %tdplotsetmaincoords{80}{100}

    begin{tikzpicture}[tdplot_main_coords,scale=1.5]

    pgfmathsetmacroa{4}

    pgfmathsetmacrob{3}

pgfmathsetmacroc{4}

    % definitions

    path

coordinate(A) at (0,0,0)

coordinate (B) at (a,0,0)

coordinate (C) at (0,b,0)                           

coordinate (S) at (0,0,a)                

;

path[projection of point={(0,0,0) on line through (a,0,0) and (0,0,a)}]

coordinate (P)

[projection of point={(0,0,0) on line through (0,b,0) and (0,0,a)}]

coordinate  (Q)

[projection of point={(0,0,0) on line through (0,b,0) and (a,0,0)}]

coordinate (R);







begin{scope}

draw [very thick] (S) -- (R);

draw [very thick, name path=B--Q] (B) -- (Q);

draw [very thick, name path=C--P] (C) -- (P);

path [name intersections={of=B--Q and C--P,by=H}];

end{scope}



begin{scope}

draw[dashed,thick]

(A) -- (B)  (A) -- (C)  (S)--(A) (A)--(H) (A)--(R) (A)--(P) (A)--(Q);

draw[thick]

(S) -- (B) -- (C) -- cycle;

end{scope}



tkzMarkRightAngle(S,R,C)

tkzMarkRightAngle(B,P,C)

tkzMarkRightAngle(B,Q,C)

tkzMarkRightAngle(A,R,B)





foreach point/position in {A/left,B/left,C/below,S/above,P/left,Q/above,R/below,H/above}

{

    fill (point) circle (1.5pt);

    node[position=3pt] at (point) {$point$};

}

end{tikzpicture}

end{document} 

Is there a command to find coordinates of projection of a point on a plane?