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- 2075 Asoj · 7 marks
Shear cube w.r.t. ZX plane, constant 3
Short answer
Apply x' = x+3y, z' = z+3y, y unchanged. Final coordinates: (0,0,0), (0,0,2), (6,2,6), (6,2,8), (2,0,0), (2,0,2), (8,2,6), (8,2,8).
A unit length cube with diagonal passing through (0, 0, 0) and (2, 2, 2) is sheared with respect to ZX-plane with shear constants = 3 in both directions. Obtain the final coordinates of the cube after shearing.
Assumptions: the question says "unit length", but a diagonal from (0,0,0) to (2,2,2) means the cube has side 2, with corners at 0 and 2 on each axis. Shear "with respect to the ZX plane" keeps y fixed (the ZX plane, y = 0, does not move). x and z are displaced in proportion to y with shear constants sh_x = sh_z = 3.
Equations: x' = x + 3y, y' = y, z' = z + 3y
| x' | | 1 3 0 0 | | x |
| y' | = | 0 1 0 0 | | y |
| z' | | 0 3 1 0 | | z |
| 1 | | 0 0 0 1 | | 1 |
| Corner | (x, y, z) | x + 3y | z + 3y | New corner |
|---|---|---|---|---|
| A | (0,0,0) | 0 | 0 | (0,0,0) |
| B | (0,0,2) | 0 | 2 | (0,0,2) |
| C | (0,2,0) | 6 | 6 | (6,2,6) |
| D | (0,2,2) | 6 | 8 | (6,2,8) |
| E | (2,0,0) | 2 | 0 | (2,0,0) |
| F | (2,0,2) | 2 | 2 | (2,0,2) |
| G | (2,2,0) | 8 | 6 | (8,2,6) |
| H | (2,2,2) | 8 | 8 | (8,2,8) |
Answer: (0,0,0), (0,0,2), (6,2,6), (6,2,8), (2,0,0), (2,0,2), (8,2,6), (8,2,8). The bottom face (y = 0) stays in place, and the top face (y = 2) is moved by 6 in both x and z.
✓ Answer checked by Claude Opus 5.5 (high effort)
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