3 d transformation reflection

3 d transformation reflection

• 1.
  3D TRANSFORMATION (Reflection &Shearing) By: Arvind Kumar Assistant Professor (Vidya College of Engg.)
• 2.
  Contents By: Arvind Kumar 3D Reflection Shearing
• 3.
  3D REFLECTION By: ArvindKumar 3D 3D Reflection about XY-Plane                             1000 0100 0010 0001 1 ' ' ' z y x MXY z y x
• 4.
  3D REFLECTION By: ArvindKumar 3D 3D Reflection about XZ-Plane z y x                             1000 0100 0010 0001 1 ' ' ' z y x MXZ
• 5.
  3D REFLECTION By: ArvindKumar 3D 3D Reflection about YZ-Plane z y x                            1000 0100 0010 0001 1 ' ' ' z y x MYZ
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  3D REFLECTION wrtANY PLANE By: Arvind Kumar 3D Steps to be Follows 1. Translate, that lies in the reflection plane to the origin. 2. Rotate the Normal Vector to the reflection plane at the origin until coincident with z axis. 3. Reflect the object through xy Plane. 4. Perform inverse Rotation transformation 5. Perform inverse Translation transformation Rxy-1 T-1 Rxy T Mxy      TRRT  xyxy xyM M11  
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  3D REFLECTION wrtANY PLANE By: Arvind Kumar 3D Step-1 Translation                 1000 100 010 001 tz ty tx T
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  3D REFLECTION wrtANY PLANE By: Arvind Kumar 3D Step-2 Rotation about xy plane let the normal vector : N= 𝒏𝟏𝒊 + 𝒏𝟐 𝒋 + 𝒏𝟑 𝒌 N= 𝒏𝟏 𝟐 + 𝒏𝟐 𝟐 + 𝒏𝟑 𝟐 ,                          1000 0 321 0 23 0 0 3121 N n N n N n nn N nn N nn N Rxy    2322 nn 
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  3D REFLECTION wrtANY PLANE By: Arvind Kumar 3D Step-3 Reflection about XY Plane               1000 0100 0010 0001 Mxy
• 10.
  3D REFLECTION wrtANY PLANE By: Arvind Kumar 3D Step-4 Inverse Rotation                          1000 0 3231 0 2321 0 1 0 1 N nn N nn N nn N nn N n N Rxy   
• 11.
  3D REFLECTION wrtANY PLANE By: Arvind Kumar 3D Step-5 Inverse Translation              1000 100 010 001 1 tz ty tx T
• 12.
  Numerical By: Arvind Kumar 3D Q1.Find the matrix for the mirror reflection wrt the plane passing through the origin and having a normal vector whose direction is N= I+J+K. Solution: Here p0 (0,0,0) and plane passes through the origin . Hence translation is not necessary. the normal vector : N= I + J + K Here, n1 = 1, n2= 1, n3=1 𝑵 = 𝟑 , λ = 𝟐 
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  Numerical By: Arvind Kumar 3D                          1000 0 3 1 3 1 3 1 0 2 1 2 1 0 0 6 1 6 1 3 2 xyR               1000 0100 0010 0001 Mxy
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  Numerical By: Arvind Kumar 3D                          1000 0 3 1 2 1 6 1 0 3 1 2 1 6 1 0 3 1 0 3 2 1 xyR Resultant Matrix:       xyxy xyM M RR 1 
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  Numerical By: Arvind Kumar 3D                 1000 03/13/23/2 03/23/13/2 03/23/23/1 M      xyxy xyM M RR 1  After Reflection points are: X’ = (1/3, -2/3, -2/3) Y’ = ( -2/3, 1/3, -2/3), Z’ = (-2/3, -2/3, 1/3)
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  3D Shearing By: ArvindKumar 3D 3D Shearing about X-Axis 3D                                      11000 010 001 0001 1 ' ' ' z y x b a z y x SHx
• 17.
  3D Shearing By: ArvindKumar 3D 3D Shearing about Y-Axis 3D                                      11000 010 0010 001 1 ' ' ' z y x b a z y x SHy
• 18.
  3D Shearing By: ArvindKumar 3D 3D Shearing about Z-Axis 3D                                      11000 0100 010 001 1 ' ' ' z y x b a z y x SHz