What method is there to translate and transform the coordinate system of a three-dimensional graphic system?

Question

In the triangular prism A-BCD, DA=DB=DC=2, BD is perpendicular to CD, and the angle ADB equals the angle ADC, both of which are 60 degrees. E is the midpoint of BC.Vector EF equals vector DA.

I use point D as the origin of the spatial Cartesian coordinate system to establish the system. The line where DB is located is the x-axis, and the line where DC is located is the y-axis. The line perpendicular to the bottom DBC and facing upwards from point D is the z-axis.

The code is as follows:

Clear["Global`*"];
d = {0, 0, 0};
b = {2, 0, 0};
c = {0, 2, 0};
e = Mean[{b, c}];
a = {x1, y1, z1};
a = SolveValues[{VectorAngle[a - d, b - d] == 60 Degree, 
    VectorAngle[a - d, c - d] == 60 Degree, Norm[a - d] == 2, 
    z1 > 0}, {x1, y1, z1}] // First
f = {x2, y2, z2};
f = SolveValues[{a - d == f - e, z2 > 0}, {x2, y2, z2}] // First
labels = {Text[Style[A, 12, FontFamily -> "Times"], a, {-1, -1}], 
   Text[Style[B, 12, FontFamily -> "Times"], b, {1, 1}], 
   Text[Style[C, 12, FontFamily -> "Times"], c, {1, 1}], 
   Text[Style[D, 12, FontFamily -> "Times"], d, {-2, 0}], 
   Text[Style[E, 12, FontFamily -> "Times"], e, {3, 0}], 
   Text[Style[F, 12, FontFamily -> "Times"], f, {-1, -2}]};
dashLines = {Dashed, 
   AbsoluteThickness[2], {Brown, Line[{{c, a}}]}, {Red, 
    Line[{{c, b}, {d, c}}]}, {Purple, Line[{{e, f}}]}, {Black, 
    Line[{{e, a}, {d, e}}]}};
realLines = {AbsoluteThickness[2], 
   Line[{{a, d}, {a, b}, {b, d}, {a, f}, {b, f}}]};
Show[Graphics3D[{dashLines, realLines, labels}, Boxed -> False, 
  ViewPoint -> {2, -3.5, 1.28}], 
 RegionPlot3D[Polygon[{a, b, d}], MeshFunctions -> {#1 + #2 + #3 &}, 
  PlotStyle -> Opacity[0.2]], 
 RegionPlot3D[Polygon[{a, b, f}], MeshFunctions -> {#1 + #2 + #3 &}, 
  PlotStyle -> Opacity[0.3]], 
 Graphics3D[{Arrow[{{d, d + {0, 0, 2}}, {b, b + {1, 0, 0}}, {c, 
      c + {0, 1, 0}}}], 
   Text[Style["z", Red, 20, Italic, FontFamily -> "Times"], 
    d + {0, 0, 2}, {-1, -1}], 
   Text[Style["y", Red, 20, Italic, FontFamily -> "Times"], 
    c + {0, 1, 0}, {-2, -1}], 
   Text[Style["x", Red, 20, Italic, FontFamily -> "Times"], 
    b + {1, 0, 0}, {2, -1}]}]]

get the result:

My current idea is to keep the spatial geometry unchanged. Just change the origin and coordinate axis of the spatial coordinate system, and now use the E point as the origin of the spatial Cartesian coordinate system. The line ED is located on the x-axis, and the line EB is located on the y-axis to reestablish the spatial Cartesian coordinate system. Do not manually recalculate the coordinates of each point and vector in the new coordinate system. What transformation is used to transform the coordinate system so that the coordinates of the points can be automatically recalculated?

My goal is not just to rebuild the coordinate system. And it is also necessary to adapt the coordinates of each point and vector in the new coordinate system to the new system. A={1,1, Sqrt 2} This is the value in the old system. For example, the value in the new series should be a={0,0, Sqrt 2}.and so on

The final effect is as follows:

Answer 1

• Directly rebuild the Axis.

Graphics3D[{Arrow[{{a, a + Normalize[a - e]}, {b, 
     b + Normalize[b - c]}, {d, d + Normalize[d - e]}}], 
  Text[Style["z", Red, 20, Italic, FontFamily -> "Times"], 
   a + Normalize[a - e], {-1, -1}], 
  Text[Style["y", Red, 20, Italic, FontFamily -> "Times"], 
   b + Normalize[b - c], {-2, -1}], 
  Text[Style["x", Red, 20, Italic, FontFamily -> "Times"], 
   d + Normalize[d - e], {2, -1}]}]

Edit

• We can build a new frame system with original: base1,base2,base3.

original = e;
base1 = Normalize[d - e];
base2 = Normalize[b - e];
base3 = Normalize[a - e];

transframe[{x_, y_, 
   z_}] = {x1, y1, z1} /. 
    Solve[{x, y, z} == original + x1*base1 + y1*base2 + z1*base3, {x1,
       y1, z1}] // First // Simplify

The same as

transframe[{x_, y_, z_}] = ({x, y, z} - original) . 
   Inverse[{base1, base2, base3}] // Simplify

{-((-2 + x + y)/Sqrt[2]), (x - y)/Sqrt[2], z}.

• Examples

d={0,0,0};
b={2, 0, 0};
a={1,1, Sqrt[2]};
transframe[d]
transframe[b]
transframe[a]

{Sqrt[2], 0, 0}

{0, Sqrt[2], 0}.

{0, 0, Sqrt[2]}