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Shear unit cube w.r.t. yz plane
Short answer
Shearing with respect to the yz plane fixes x and displaces y and z by 2x. Original unit cube corners (0,0,0) to (1,1,1) become (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,2,2), (1,2,3), (1,3,2), (1,3,3).
A unit length cube with diagonal passing through (0,0,0) and (1,1,1) is sheared with respect to yz plane with the shear constants = 2 in both directions. Obtain the coordinates of all the corners of the cube after shearing.
Assumption: shear "with respect to the yz plane" keeps the x-coordinate fixed (the yz plane, x = 0, does not move). The y and z coordinates are displaced in proportion to x with shear constants sh_y = sh_z = 2. The unit cube has corners from (0,0,0) to (1,1,1).
Equations: x' = x, y' = y + 2x, z' = z + 2x
| x' | | 1 0 0 0 | | x |
| y' | = | 2 1 0 0 | | y |
| z' | | 2 0 1 0 | | z |
| 1 | | 0 0 0 1 | | 1 |
| Corner | (x, y, z) | y + 2x | z + 2x | New corner |
|---|---|---|---|---|
| A | (0,0,0) | 0 | 0 | (0,0,0) |
| B | (0,0,1) | 0 | 1 | (0,0,1) |
| C | (0,1,0) | 1 | 0 | (0,1,0) |
| D | (0,1,1) | 1 | 1 | (0,1,1) |
| E | (1,0,0) | 2 | 2 | (1,2,2) |
| F | (1,0,1) | 2 | 3 | (1,2,3) |
| G | (1,1,0) | 3 | 2 | (1,3,2) |
| H | (1,1,1) | 3 | 3 | (1,3,3) |
The face x = 0 stays in place and the face x = 1 slides by 2 units in both y and z.
Answer: (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,2,2), (1,2,3), (1,3,2), (1,3,3).
✓ Answer checked by Claude Opus 5.5 (high effort)
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