3D Transformation in Computer Graphics
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3D transformations in computer graphics include translation, scaling, and rotation of 3D objects. Translation moves an object by adding translation offsets to the x, y, and z coordinates. Scaling enlarges or shrinks an object by multiplying the coordinates by scaling factors. Rotation rotates an object by applying rotation matrices to change the orientation. Reflection mirrors an object across planes by flipping the sign of coordinates on one axis. These transformations are used to manipulate 3D objects in computer graphics and animation.
![This translation is achieved by adding the translation coordinates to the old coordinates of the
object as-
• Xnew = Xold + Tx (This denotes translation towards X axis)
• Ynew = Yold + Ty (This denotes translation towards Y axis)
• Znew = Zold + Tz (This denotes translation towards Z axis)
In Matrix form, the above translation equations may be represented as-
Problem-
Given a 3D object with coordinate points A(-6, 22, -7), B(-19, 17, 13). Apply the translation with
the distance 2 towards X axis, 2 towards Y axis and 2 towards Z axis and obtain the new
coordinates of the object.
Solution
We know that
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 𝑇𝑥
0 1 0 𝑇𝑦
0 0 1 𝑇𝑧
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| (1)
For point A(-6, 22, -7) equation (1) becomes
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 2
0 1 0 2
0 0 1 2
0 0 0 1
| |
−6
22
−7
1
| = |
−6 + 0 − 0 + 2
−0 + 22 − 0 + 2
−0 + 0 − 7 + 2
−0 + 0 − 0 + 1
| = |
−4
24
−5
1
|
Therefore the new coordinates of A(-6, 22, -7) will be A’(-4, 24, -5)](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-2-320.jpg)
![For point B(-19, 17, 13) equation (1) becomes
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 2
0 1 0 2
0 0 1 2
0 0 0 1
| |
−19
17
13
1
| = |
−19 + 0 + 0 + 2
−0 + 17 + 0 + 2
−0 + 0 + 13 + 2
−0 + 0 + 0 + 1
| = |
−17
19
15
1
|
Therefore the new coordinates of B(-19, 17, 13) will be B’(-17, 19, 15)
3D Scaling in Computer Graphics-
In computer graphics, scaling is a process of modifying or altering the size of objects.
• Scaling may be used to increase or reduce the size of object.
• Scaling factor determines whether the object size is to be increased or reduced.
• If scaling factor > 1, then the object size is increased.
• If scaling factor < 1, then the object size is reduced.
Consider a point object O has to be scaled in a 3D plane.
Let-
• Initial coordinates of the object O = (Xold, Yold,Zold)
• Scaling factor for X-axis = Sx
• Scaling factor for Y-axis = Sy
• Scaling factor for Z-axis = Sz
• New coordinates of the object O after scaling = (Xnew, Ynew, Znew)
This scaling is achieved by using the following scaling equations-
• Xnew = Xold x Sx
• Ynew = Yold x Sy
• Znew = Zold x Sz
In Matrix form, the above scaling equations may be represented as-](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-3-320.jpg)
![Problem
Given a 3D object with coordinate points A(0, 3, 3), B(3, 3, 6). Apply the scaling parameter 2
towards X axis, 3 towards Y axis and 3 towards Z axis and obtain the new coordinates of the
object.
Solution
We know that
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
𝑆𝑥 0 0 0
0 𝑆𝑦 0 0
0 0 𝑆𝑧 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| (1)
For point A(0, 3, 3) equation (1) becomes
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
2 0 0 0
0 3 0 0
0 0 3 0
0 0 0 1
| |
0
3
3
1
| = |
0 + 0 + 0 + 0
0 + 9 + 0 + 0
0 + 0 + 9 + 0
0 + 0 + 0 + 1
| = |
0
9
9
1
|
Therefore, the new coordinates of A(0, 3, 3) will be A’(0, 9, 9)
For point B(3, 3, 6) equation (1) becomes
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
2 0 0 0
0 3 0 0
0 0 3 0
0 0 0 1
| |
3
3
6
1
| = |
6 + 0 + 0 + 0
0 + 9 + 0 + 0
0 + 0 + 18 + 0
0 + 0 + 0 + 1
| = |
6
9
18
1
|
Therefore, the new coordinates of B(3, 3, 6) will be B’(6, 9, 18)
3D Rotation in Computer Graphics-
In Computer graphics,
3D Rotation is a process of rotating an object with respect to an angle in a three-dimensional plane.
Consider a point object O has to be rotated from one angle to another in a 3D plane.
Let-
• Initial coordinates of the object O = (Xold, Yold, Zold)
• Initial angle of the object O with respect to origin = Φ
• Rotation angle = θ
• New coordinates of the object O after rotation = (Xnew, Ynew, Znew)](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-4-320.jpg)
![For Z-Axis Rotation-
This rotation is achieved by using the following rotation equations-
• Xnew = Xold x cosθ – Yold x sinθ
• Ynew = Xold x sinθ + Yold x cosθ
• Znew = Zold
In Matrix form, the above rotation equations may be represented as-
Problem
Given a homogeneous point (1, 2, 3). Apply rotation 90 degree towards X, Y and Z axis and find
out the new coordinate points.
Solution
Rotation along X-axis
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
0 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0
0 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
|
= |
1 0 0 0
0 𝑐𝑜𝑠900
−𝑠𝑖𝑛900
0
0 𝑠𝑖𝑛900
𝑐𝑜𝑠900
0
0 0 0 1
| |
1
2
3
1
| = |
1 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 1
| |
1
2
3
1
| = |
1 + 0 + 0 + 0
0 + 0 − 3 + 0
0 + 2 + 0 + 0
0 + 0 + 0 + 1
| = |
1
−3
2
1
|
Therefore, the new coordinates for the homogenous point will be (1, -3, 2)
Rotation Y-axis
[
𝒙𝒏𝒆𝒘
𝒚𝒏𝒆𝒘
𝒛𝒏𝒆𝒘
𝟏
] = |
𝒄𝒐𝒔𝜽 𝟎 𝒔𝒊𝒏𝜽 𝟎
𝟎 𝟏 𝟎 𝟎
−𝒔𝒊𝒏𝜽 𝟎 𝒄𝒐𝒔𝜽 𝟎
𝟎 𝟎 𝟎 𝟏
| |
𝒙𝒐𝒍𝒅
𝒚𝒐𝒍𝒅
𝒛𝒐𝒍𝒅
𝟏
|](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-6-320.jpg)
![= |
𝒄𝒐𝒔𝟗𝟎𝟎
𝟎 𝒔𝒊𝒏𝟗𝟎𝟎
𝟎
𝟎 𝟏 𝟎 𝟎
−𝒔𝒊𝒏𝟗𝟎𝟎
𝟎 𝒄𝒐𝒔𝟗𝟎𝟎
𝟎
𝟎 𝟎 𝟎 𝟏
| |
𝟏
𝟐
𝟑
𝟏
| = |
𝟎 𝟎 𝟏 𝟎
𝟎 𝟏 𝟎 𝟎
−𝟏 𝟎 𝟎 𝟎
𝟎 𝟎 𝟎 𝟏
| |
𝟏
𝟐
𝟑
𝟏
| = |
𝟎 + 𝟎 + 𝟑 + 𝟎
𝟎 + 𝟐 + 𝟎 + 𝟎
−𝟏 + 𝟎 + 𝟎 + 𝟎
𝟎 + 𝟎 + 𝟎 + 𝟏
| = |
𝟑
𝟐
−𝟏
𝟏
|
Therefore, the new coordinates for the homogenous point will be (3, 2, -1)
Rotation Z-axis
[
𝒙𝒏𝒆𝒘
𝒚𝒏𝒆𝒘
𝒛𝒏𝒆𝒘
𝟏
] = |
𝒄𝒐𝒔𝜽 −𝒔𝒊𝒏𝜽 𝟎 𝟎
𝒔𝒊𝒏𝜽 𝒄𝒐𝒔𝜽 𝟎 𝟎
𝟎 𝟎 𝟏 𝟎
𝟎 𝟎 𝟎 𝟏
| |
𝒙𝒐𝒍𝒅
𝒚𝒐𝒍𝒅
𝒛𝒐𝒍𝒅
𝟏
|
= |
𝒄𝒐𝒔𝟗𝟎𝟎
−𝒔𝒊𝒏𝟗𝟎𝟎
𝟎 𝟎
𝒔𝒊𝒏𝟗𝟎𝟎
𝒄𝒐𝒔𝟗𝟎𝟎
𝟎 𝟎
𝟎 𝟎 𝟏 𝟎
𝟎 𝟎 𝟎 𝟏
| |
𝟏
𝟐
𝟑
𝟏
| = |
𝟎 −𝟏 𝟎 𝟎
𝟏 𝟎 𝟎 𝟎
𝟎 𝟎 𝟏 𝟎
𝟎 𝟎 𝟎 𝟏
| |
𝟏
𝟐
𝟑
𝟏
| = |
𝟎 − 𝟐 + 𝟎 + 𝟎
𝟏 + 𝟎 + 𝟎 + 𝟎
𝟎 + 𝟎 + 𝟑 + 𝟎
𝟎 + 𝟎 + 𝟎 + 𝟏
| = |
−𝟐
𝟏
𝟑
𝟏
|
Therefore, the new coordinates for the homogenous point will be (-2, 1, 3)
3D Reflection in Computer Graphics-
• Reflection is a kind of rotation where the angle of rotation is 180 degree.
• The reflected object is always formed on the other side of mirror.
• The size of reflected object is same as the size of original object.
Consider a point object O has to be reflected in a 3D plane.
Let-
• Initial coordinates of the object O = (Xold, Yold, Zold)
• New coordinates of the reflected object O after reflection = (Xnew, Ynew,Znew)
In 3 dimensions, there are 3 possible types of reflection-
• Reflection relative to XY plane
• Reflection relative to YZ plane
• Reflection relative to XZ plane
Reflection Relative to XY Plane
This reflection is achieved by using the following reflection equations-
• Xnew = Xold
• Ynew = Yold
• Znew = -Zold](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-7-320.jpg)
![In Matrix form, the above reflection equations may be represented as-
Problem
Given a 3D triangle with coordinate points A(3, 4, 1), B(6, 4, 2). Apply the reflection and find out
the new coordinates of the object.
Solution
Refection relative to XY Plane
For point A(3, 4, 1)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 1
| |
3
4
1
1
| = |
3 + 0 + 0 + 0
0 + 4 + 0 + 0
0 + 0 − 1 + 0
0 + 0 + 0 + 1
| = |
3
4
−1
1
|
Therefore the new coordinates for point A(3, 4, 1) will be A’(3, 4, -1)
For point B(6, 4, 2)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 1
| |
6
4
2
1
| = |
6 + 0 + 0 + 0
0 + 4 + 0 + 0
0 + 0 − 2 + 0
0 + 0 + 0 + 1
| = |
6
4
−2
1
|
Therefore the new coordinates for point B(6, 4, 2) will be B’(6, 4, -2)
Refection relative to YZ Plane
For point A(3, 4, 1)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
−1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
−1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
| |
6
4
2
1
| = |
−3 + 0 + 0 + 0
0 + 4 + 0 + 0
0 + 0 + 1 + 0
0 + 0 + 0 + 1
| = |
−3
4
1
1
|
Therefore the new coordinates for point A(3, 4, 1) will be A’(-3, 4, 1)](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-9-320.jpg)
![For point B(6, 4, 2)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
−1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
−1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
| |
3
4
1
1
| = |
−6 + 0 + 0 + 0
0 + 4 + 0 + 0
0 + 0 + 2 + 0
0 + 0 + 0 + 1
| = |
−6
4
2
1
|
Therefore the new coordinates for point B(6, 4, 2) will be B’(-6, 4, 2)
Refection relative to XZ Plane
For point A(3, 4, 1)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 1
| |
3
4
1
1
| = |
3 + 0 + 0 + 0
0 − 4 + 0 + 0
0 + 0 + 1 + 0
0 + 0 + 0 + 1
| = |
3
−4
1
1
|
Therefore the new coordinates for point A(3, 4, 1) will be A’(3, -4, 1)
For point B(6, 4, 2)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 1
| |
6
4
2
1
| = |
6 + 0 + 0 + 0
0 − 4 + 0 + 0
0 + 0 + 2 + 0
0 + 0 + 0 + 1
| = |
6
−4
2
1
|
Therefore the new coordinates for point B(6, 4, 2) will be B’(6, -4, 2)
3D Shearing in Computer Graphics-
In Computer graphics,
3D Shearing is an ideal technique to change the shape of an existing object in a three-dimensional plane.
In a three-dimensional plane, the object size can be changed along X direction, Y direction as well
as Z direction.
So, there are three versions of shearing-
• Shearing in X direction
• Shearing in Y direction
• Shearing in Z direction
Consider a point object O has to be sheared in a 3D plane.
Let-
• Initial coordinates of the object O = (Xold, Yold, Zold)
• Shearing parameter towards X direction = Shx](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-10-320.jpg)
![• Ynew = Yold + Shy x Zold
• Znew = Zold
In Matrix form, the above shearing equations may be represented as-
Problem
Given a 3D triangle with points (0, 0, 0), (1, 1, 2) and (1, 1, 3). Apply shear parameter 2 on X axis,
2 on Y axis and 3 on Z axis and find out the new coordinates of the object.
Solution
Shearing along X-axis
For point (0, 0, 0)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
𝑆ℎ𝑦 1 0 0
𝑆ℎ𝑧 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 0 0
2 1 0 0
3 0 1 0
0 0 0 1
| |
0
0
0
1
| = |
0 + 0 + 0 + 0
0 + 0 + 0 + 0
0 + 0 + 0 + 0
0 + 0 + 0 + 1
| = |
0
0
0
1
|
Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0)
For point (1, 1, 2)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
𝑆ℎ𝑦 1 0 0
𝑆ℎ𝑧 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 0 0
2 1 0 0
3 0 1 0
0 0 0 1
| |
1
1
2
1
| = |
1 + 0 + 0 + 0
2 + 1 + 0 + 0
3 + 0 + 2 + 0
0 + 0 + 0 + 1
| = |
1
3
5
1
|
Therefore, the new coordinates for point (1, 1, 2) will be (1, 3, 5)
For point (1, 1, 3)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 0 0
𝑆ℎ𝑦 1 0 0
𝑆ℎ𝑧 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 0 0
2 1 0 0
3 0 1 0
0 0 0 1
| |
1
1
3
1
| = |
1 + 0 + 0 + 0
2 + 1 + 0 + 0
3 + 0 + 3 + 0
0 + 0 + 0 + 1
| = |
1
3
6
1
|](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-12-320.jpg)
![Therefore, the new coordinates for point (1, 1, 3) will be (1, 3, 6)
Shearing along Y-axis
For point (0, 0, 0)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 𝑆ℎ𝑥 0 0
0 1 0 0
0 𝑆ℎ𝑧 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 2 0 0
0 1 0 0
0 3 1 0
0 0 0 1
| |
0
0
0
1
| = |
0 + 0 + 0 + 0
0 + 0 + 0 + 0
0 + 0 + 0 + 0
0 + 0 + 0 + 1
| = |
0
0
0
1
|
Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0)
For point (1, 1, 2)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 𝑆ℎ𝑥 0 0
0 1 0 0
0 𝑆ℎ𝑧 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 2 0 0
0 1 0 0
0 3 1 0
0 0 0 1
| |
1
1
2
1
| = |
1 + 2 + 0 + 0
0 + 1 + 0 + 0
0 + 3 + 2 + 0
0 + 0 + 0 + 1
| = |
3
1
5
1
|
Therefore, the new coordinates for point (1, 1, 2) will be (3, 1, 5)
For point (1, 1, 3)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 𝑆ℎ𝑥 0 0
0 1 0 0
0 𝑆ℎ𝑧 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 2 0 0
0 1 0 0
0 3 1 0
0 0 0 1
| |
1
1
3
1
| = |
1 + 2 + 0 + 0
0 + 1 + 0 + 0
0 + 3 + 3 + 0
0 + 0 + 0 + 1
| = |
3
1
6
1
|
Therefore, the new coordinates for point (1, 1, 3) will be (3, 1, 6)
Shearing along Z-axis
For point (0, 0, 0)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 𝑆ℎ𝑥 0
0 1 𝑆ℎ𝑦 0
0 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 2 0
0 1 2 0
0 0 1 0
0 0 0 1
| |
0
0
0
1
| = |
0 + 0 + 0 + 0
0 + 0 + 0 + 0
0 + 0 + 0 + 0
0 + 0 + 0 + 1
| = |
0
0
0
1
|
Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0)
For point (1, 1, 2)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 𝑆ℎ𝑥 0
0 1 𝑆ℎ𝑦 0
0 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 2 0
0 1 2 0
0 0 1 0
0 0 0 1
| |
1
1
2
1
| = |
1 + 0 + 4 + 0
0 + 1 + 4 + 0
0 + 0 + 2 + 0
0 + 0 + 0 + 1
| = |
5
5
2
1
|
Therefore, the new coordinates for point (1, 1, 2) will be (5, 5, 2)](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-13-320.jpg)
![For point (1, 1, 3)
[
𝑥𝑛𝑒𝑤
𝑦𝑛𝑒𝑤
𝑧𝑛𝑒𝑤
1
] = |
1 0 𝑆ℎ𝑥 0
0 1 𝑆ℎ𝑦 0
0 0 1 0
0 0 0 1
| |
𝑥𝑜𝑙𝑑
𝑦𝑜𝑙𝑑
𝑧𝑜𝑙𝑑
1
| = |
1 0 2 0
0 1 2 0
0 0 1 0
0 0 0 1
| |
1
1
3
1
| = |
1 + 0 + 6 + 0
0 + 1 + 6 + 0
0 + 0 + 3 + 0
0 + 0 + 0 + 1
| = |
7
7
3
1
|
Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0)](https://image.slidesharecdn.com/3dtransformationincomputergraphics-211026180224/85/3D-Transformation-in-Computer-Graphics-14-320.jpg)
3D Transformation in Computer Graphics
- 1. 3D Transformation in Computer Graphics- In Computer graphics, Transformation is a process of modifying and re-positioning the existing graphics. • 3D Transformations take place in a three dimensional plane. • 3D Transformations are important and a bit more complex than 2D Transformations. • Transformations are helpful in changing the position, size, orientation, shape etc of the object. 3D Translation in Computer Graphics- In Computer graphics, 3D Translation is a process of moving an object from one position to another in a three dimensional plane. Consider a point object O has to be moved from one position to another in a 3D plane. Let- • Initial coordinates of the object O = (Xold, Yold, Zold) • New coordinates of the object O after translation = (Xnew, Ynew, Zold) • Translation vector or Shift vector = (Tx, Ty, Tz) Given a Translation vector (Tx, Ty, Tz)- • Tx defines the distance the Xold coordinate has to be moved. • Ty defines the distance the Yold coordinate has to be moved. • Tz defines the distance the Zold coordinate has to be moved.
- 2. This translation is achieved by adding the translation coordinates to the old coordinates of the object as- • Xnew = Xold + Tx (This denotes translation towards X axis) • Ynew = Yold + Ty (This denotes translation towards Y axis) • Znew = Zold + Tz (This denotes translation towards Z axis) In Matrix form, the above translation equations may be represented as- Problem- Given a 3D object with coordinate points A(-6, 22, -7), B(-19, 17, 13). Apply the translation with the distance 2 towards X axis, 2 towards Y axis and 2 towards Z axis and obtain the new coordinates of the object. Solution We know that [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 𝑇𝑥 0 1 0 𝑇𝑦 0 0 1 𝑇𝑧 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | (1) For point A(-6, 22, -7) equation (1) becomes [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 2 0 1 0 2 0 0 1 2 0 0 0 1 | | −6 22 −7 1 | = | −6 + 0 − 0 + 2 −0 + 22 − 0 + 2 −0 + 0 − 7 + 2 −0 + 0 − 0 + 1 | = | −4 24 −5 1 | Therefore the new coordinates of A(-6, 22, -7) will be A’(-4, 24, -5)
- 3. For point B(-19, 17, 13) equation (1) becomes [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 2 0 1 0 2 0 0 1 2 0 0 0 1 | | −19 17 13 1 | = | −19 + 0 + 0 + 2 −0 + 17 + 0 + 2 −0 + 0 + 13 + 2 −0 + 0 + 0 + 1 | = | −17 19 15 1 | Therefore the new coordinates of B(-19, 17, 13) will be B’(-17, 19, 15) 3D Scaling in Computer Graphics- In computer graphics, scaling is a process of modifying or altering the size of objects. • Scaling may be used to increase or reduce the size of object. • Scaling factor determines whether the object size is to be increased or reduced. • If scaling factor > 1, then the object size is increased. • If scaling factor < 1, then the object size is reduced. Consider a point object O has to be scaled in a 3D plane. Let- • Initial coordinates of the object O = (Xold, Yold,Zold) • Scaling factor for X-axis = Sx • Scaling factor for Y-axis = Sy • Scaling factor for Z-axis = Sz • New coordinates of the object O after scaling = (Xnew, Ynew, Znew) This scaling is achieved by using the following scaling equations- • Xnew = Xold x Sx • Ynew = Yold x Sy • Znew = Zold x Sz In Matrix form, the above scaling equations may be represented as-
- 4. Problem Given a 3D object with coordinate points A(0, 3, 3), B(3, 3, 6). Apply the scaling parameter 2 towards X axis, 3 towards Y axis and 3 towards Z axis and obtain the new coordinates of the object. Solution We know that [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 𝑆𝑥 0 0 0 0 𝑆𝑦 0 0 0 0 𝑆𝑧 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | (1) For point A(0, 3, 3) equation (1) becomes [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 2 0 0 0 0 3 0 0 0 0 3 0 0 0 0 1 | | 0 3 3 1 | = | 0 + 0 + 0 + 0 0 + 9 + 0 + 0 0 + 0 + 9 + 0 0 + 0 + 0 + 1 | = | 0 9 9 1 | Therefore, the new coordinates of A(0, 3, 3) will be A’(0, 9, 9) For point B(3, 3, 6) equation (1) becomes [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 2 0 0 0 0 3 0 0 0 0 3 0 0 0 0 1 | | 3 3 6 1 | = | 6 + 0 + 0 + 0 0 + 9 + 0 + 0 0 + 0 + 18 + 0 0 + 0 + 0 + 1 | = | 6 9 18 1 | Therefore, the new coordinates of B(3, 3, 6) will be B’(6, 9, 18) 3D Rotation in Computer Graphics- In Computer graphics, 3D Rotation is a process of rotating an object with respect to an angle in a three-dimensional plane. Consider a point object O has to be rotated from one angle to another in a 3D plane. Let- • Initial coordinates of the object O = (Xold, Yold, Zold) • Initial angle of the object O with respect to origin = Φ • Rotation angle = θ • New coordinates of the object O after rotation = (Xnew, Ynew, Znew)
- 5. In 3 dimensions, there are 3 possible types of rotation- • X-axis Rotation • Y-axis Rotation • Z-axis Rotation For X-Axis Rotation- This rotation is achieved by using the following rotation equations- • Xnew = Xold • Ynew = Yold x cosθ – Zold x sinθ • Znew = Yold x sinθ + Zold x cosθ In Matrix form, the above rotation equations may be represented as- For Y-Axis Rotation- This rotation is achieved by using the following rotation equations- • Xnew = Zold x sinθ + Xold x cosθ • Ynew = Yold • Znew = Yold x cosθ – Xold x sinθ In Matrix form, the above rotation equations may be represented as-
- 6. For Z-Axis Rotation- This rotation is achieved by using the following rotation equations- • Xnew = Xold x cosθ – Yold x sinθ • Ynew = Xold x sinθ + Yold x cosθ • Znew = Zold In Matrix form, the above rotation equations may be represented as- Problem Given a homogeneous point (1, 2, 3). Apply rotation 90 degree towards X, Y and Z axis and find out the new coordinate points. Solution Rotation along X-axis [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 0 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0 0 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 0 𝑐𝑜𝑠900 −𝑠𝑖𝑛900 0 0 𝑠𝑖𝑛900 𝑐𝑜𝑠900 0 0 0 0 1 | | 1 2 3 1 | = | 1 0 0 0 0 0 −1 0 0 1 0 0 0 0 0 1 | | 1 2 3 1 | = | 1 + 0 + 0 + 0 0 + 0 − 3 + 0 0 + 2 + 0 + 0 0 + 0 + 0 + 1 | = | 1 −3 2 1 | Therefore, the new coordinates for the homogenous point will be (1, -3, 2) Rotation Y-axis [ 𝒙𝒏𝒆𝒘 𝒚𝒏𝒆𝒘 𝒛𝒏𝒆𝒘 𝟏 ] = | 𝒄𝒐𝒔𝜽 𝟎 𝒔𝒊𝒏𝜽 𝟎 𝟎 𝟏 𝟎 𝟎 −𝒔𝒊𝒏𝜽 𝟎 𝒄𝒐𝒔𝜽 𝟎 𝟎 𝟎 𝟎 𝟏 | | 𝒙𝒐𝒍𝒅 𝒚𝒐𝒍𝒅 𝒛𝒐𝒍𝒅 𝟏 |
- 7. = | 𝒄𝒐𝒔𝟗𝟎𝟎 𝟎 𝒔𝒊𝒏𝟗𝟎𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝒔𝒊𝒏𝟗𝟎𝟎 𝟎 𝒄𝒐𝒔𝟗𝟎𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 | | 𝟏 𝟐 𝟑 𝟏 | = | 𝟎 𝟎 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 | | 𝟏 𝟐 𝟑 𝟏 | = | 𝟎 + 𝟎 + 𝟑 + 𝟎 𝟎 + 𝟐 + 𝟎 + 𝟎 −𝟏 + 𝟎 + 𝟎 + 𝟎 𝟎 + 𝟎 + 𝟎 + 𝟏 | = | 𝟑 𝟐 −𝟏 𝟏 | Therefore, the new coordinates for the homogenous point will be (3, 2, -1) Rotation Z-axis [ 𝒙𝒏𝒆𝒘 𝒚𝒏𝒆𝒘 𝒛𝒏𝒆𝒘 𝟏 ] = | 𝒄𝒐𝒔𝜽 −𝒔𝒊𝒏𝜽 𝟎 𝟎 𝒔𝒊𝒏𝜽 𝒄𝒐𝒔𝜽 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟏 | | 𝒙𝒐𝒍𝒅 𝒚𝒐𝒍𝒅 𝒛𝒐𝒍𝒅 𝟏 | = | 𝒄𝒐𝒔𝟗𝟎𝟎 −𝒔𝒊𝒏𝟗𝟎𝟎 𝟎 𝟎 𝒔𝒊𝒏𝟗𝟎𝟎 𝒄𝒐𝒔𝟗𝟎𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟏 | | 𝟏 𝟐 𝟑 𝟏 | = | 𝟎 −𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟏 | | 𝟏 𝟐 𝟑 𝟏 | = | 𝟎 − 𝟐 + 𝟎 + 𝟎 𝟏 + 𝟎 + 𝟎 + 𝟎 𝟎 + 𝟎 + 𝟑 + 𝟎 𝟎 + 𝟎 + 𝟎 + 𝟏 | = | −𝟐 𝟏 𝟑 𝟏 | Therefore, the new coordinates for the homogenous point will be (-2, 1, 3) 3D Reflection in Computer Graphics- • Reflection is a kind of rotation where the angle of rotation is 180 degree. • The reflected object is always formed on the other side of mirror. • The size of reflected object is same as the size of original object. Consider a point object O has to be reflected in a 3D plane. Let- • Initial coordinates of the object O = (Xold, Yold, Zold) • New coordinates of the reflected object O after reflection = (Xnew, Ynew,Znew) In 3 dimensions, there are 3 possible types of reflection- • Reflection relative to XY plane • Reflection relative to YZ plane • Reflection relative to XZ plane Reflection Relative to XY Plane This reflection is achieved by using the following reflection equations- • Xnew = Xold • Ynew = Yold • Znew = -Zold
- 8. In Matrix form, the above reflection equations may be represented as- Reflection Relative to YZ Plane This reflection is achieved by using the following reflection equations- • Xnew = -Xold • Ynew = Yold • Znew = Zold In Matrix form, the above reflection equations may be represented as- Reflection Relative to XZ Plane This reflection is achieved by using the following reflection equations- • Xnew = Xold • Ynew = -Yold • Znew = Zold
- 9. In Matrix form, the above reflection equations may be represented as- Problem Given a 3D triangle with coordinate points A(3, 4, 1), B(6, 4, 2). Apply the reflection and find out the new coordinates of the object. Solution Refection relative to XY Plane For point A(3, 4, 1) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 1 | | 3 4 1 1 | = | 3 + 0 + 0 + 0 0 + 4 + 0 + 0 0 + 0 − 1 + 0 0 + 0 + 0 + 1 | = | 3 4 −1 1 | Therefore the new coordinates for point A(3, 4, 1) will be A’(3, 4, -1) For point B(6, 4, 2) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 1 | | 6 4 2 1 | = | 6 + 0 + 0 + 0 0 + 4 + 0 + 0 0 + 0 − 2 + 0 0 + 0 + 0 + 1 | = | 6 4 −2 1 | Therefore the new coordinates for point B(6, 4, 2) will be B’(6, 4, -2) Refection relative to YZ Plane For point A(3, 4, 1) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | −1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | −1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 | | 6 4 2 1 | = | −3 + 0 + 0 + 0 0 + 4 + 0 + 0 0 + 0 + 1 + 0 0 + 0 + 0 + 1 | = | −3 4 1 1 | Therefore the new coordinates for point A(3, 4, 1) will be A’(-3, 4, 1)
- 10. For point B(6, 4, 2) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | −1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | −1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 | | 3 4 1 1 | = | −6 + 0 + 0 + 0 0 + 4 + 0 + 0 0 + 0 + 2 + 0 0 + 0 + 0 + 1 | = | −6 4 2 1 | Therefore the new coordinates for point B(6, 4, 2) will be B’(-6, 4, 2) Refection relative to XZ Plane For point A(3, 4, 1) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 0 −1 0 0 0 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 0 −1 0 0 0 0 1 0 0 0 0 1 | | 3 4 1 1 | = | 3 + 0 + 0 + 0 0 − 4 + 0 + 0 0 + 0 + 1 + 0 0 + 0 + 0 + 1 | = | 3 −4 1 1 | Therefore the new coordinates for point A(3, 4, 1) will be A’(3, -4, 1) For point B(6, 4, 2) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 0 −1 0 0 0 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 0 −1 0 0 0 0 1 0 0 0 0 1 | | 6 4 2 1 | = | 6 + 0 + 0 + 0 0 − 4 + 0 + 0 0 + 0 + 2 + 0 0 + 0 + 0 + 1 | = | 6 −4 2 1 | Therefore the new coordinates for point B(6, 4, 2) will be B’(6, -4, 2) 3D Shearing in Computer Graphics- In Computer graphics, 3D Shearing is an ideal technique to change the shape of an existing object in a three-dimensional plane. In a three-dimensional plane, the object size can be changed along X direction, Y direction as well as Z direction. So, there are three versions of shearing- • Shearing in X direction • Shearing in Y direction • Shearing in Z direction Consider a point object O has to be sheared in a 3D plane. Let- • Initial coordinates of the object O = (Xold, Yold, Zold) • Shearing parameter towards X direction = Shx
- 11. • Shearing parameter towards Y direction = Shy • Shearing parameter towards Z direction = Shz • New coordinates of the object O after shearing = (Xnew, Ynew, Znew) Shearing in X Axis- Shearing in X axis is achieved by using the following shearing equations- • Xnew = Xold • Ynew = Yold + Shy x Xold • Znew = Zold + Shz x Xold In Matrix form, the above shearing equations may be represented as- Shearing in Y Axis- Shearing in Y axis is achieved by using the following shearing equations- • Xnew = Xold + Shx x Yold • Ynew = Yold • Znew = Zold + Shz x Yold In Matrix form, the above shearing equations may be represented as- Shearing in Z Axis- Shearing in Z axis is achieved by using the following shearing equations- • Xnew = Xold + Shx x Zold
- 12. • Ynew = Yold + Shy x Zold • Znew = Zold In Matrix form, the above shearing equations may be represented as- Problem Given a 3D triangle with points (0, 0, 0), (1, 1, 2) and (1, 1, 3). Apply shear parameter 2 on X axis, 2 on Y axis and 3 on Z axis and find out the new coordinates of the object. Solution Shearing along X-axis For point (0, 0, 0) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 𝑆ℎ𝑦 1 0 0 𝑆ℎ𝑧 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 2 1 0 0 3 0 1 0 0 0 0 1 | | 0 0 0 1 | = | 0 + 0 + 0 + 0 0 + 0 + 0 + 0 0 + 0 + 0 + 0 0 + 0 + 0 + 1 | = | 0 0 0 1 | Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0) For point (1, 1, 2) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 𝑆ℎ𝑦 1 0 0 𝑆ℎ𝑧 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 2 1 0 0 3 0 1 0 0 0 0 1 | | 1 1 2 1 | = | 1 + 0 + 0 + 0 2 + 1 + 0 + 0 3 + 0 + 2 + 0 0 + 0 + 0 + 1 | = | 1 3 5 1 | Therefore, the new coordinates for point (1, 1, 2) will be (1, 3, 5) For point (1, 1, 3) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 0 0 𝑆ℎ𝑦 1 0 0 𝑆ℎ𝑧 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 0 0 2 1 0 0 3 0 1 0 0 0 0 1 | | 1 1 3 1 | = | 1 + 0 + 0 + 0 2 + 1 + 0 + 0 3 + 0 + 3 + 0 0 + 0 + 0 + 1 | = | 1 3 6 1 |
- 13. Therefore, the new coordinates for point (1, 1, 3) will be (1, 3, 6) Shearing along Y-axis For point (0, 0, 0) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 𝑆ℎ𝑥 0 0 0 1 0 0 0 𝑆ℎ𝑧 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 2 0 0 0 1 0 0 0 3 1 0 0 0 0 1 | | 0 0 0 1 | = | 0 + 0 + 0 + 0 0 + 0 + 0 + 0 0 + 0 + 0 + 0 0 + 0 + 0 + 1 | = | 0 0 0 1 | Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0) For point (1, 1, 2) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 𝑆ℎ𝑥 0 0 0 1 0 0 0 𝑆ℎ𝑧 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 2 0 0 0 1 0 0 0 3 1 0 0 0 0 1 | | 1 1 2 1 | = | 1 + 2 + 0 + 0 0 + 1 + 0 + 0 0 + 3 + 2 + 0 0 + 0 + 0 + 1 | = | 3 1 5 1 | Therefore, the new coordinates for point (1, 1, 2) will be (3, 1, 5) For point (1, 1, 3) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 𝑆ℎ𝑥 0 0 0 1 0 0 0 𝑆ℎ𝑧 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 2 0 0 0 1 0 0 0 3 1 0 0 0 0 1 | | 1 1 3 1 | = | 1 + 2 + 0 + 0 0 + 1 + 0 + 0 0 + 3 + 3 + 0 0 + 0 + 0 + 1 | = | 3 1 6 1 | Therefore, the new coordinates for point (1, 1, 3) will be (3, 1, 6) Shearing along Z-axis For point (0, 0, 0) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 𝑆ℎ𝑥 0 0 1 𝑆ℎ𝑦 0 0 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 2 0 0 1 2 0 0 0 1 0 0 0 0 1 | | 0 0 0 1 | = | 0 + 0 + 0 + 0 0 + 0 + 0 + 0 0 + 0 + 0 + 0 0 + 0 + 0 + 1 | = | 0 0 0 1 | Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0) For point (1, 1, 2) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 𝑆ℎ𝑥 0 0 1 𝑆ℎ𝑦 0 0 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 2 0 0 1 2 0 0 0 1 0 0 0 0 1 | | 1 1 2 1 | = | 1 + 0 + 4 + 0 0 + 1 + 4 + 0 0 + 0 + 2 + 0 0 + 0 + 0 + 1 | = | 5 5 2 1 | Therefore, the new coordinates for point (1, 1, 2) will be (5, 5, 2)
- 14. For point (1, 1, 3) [ 𝑥𝑛𝑒𝑤 𝑦𝑛𝑒𝑤 𝑧𝑛𝑒𝑤 1 ] = | 1 0 𝑆ℎ𝑥 0 0 1 𝑆ℎ𝑦 0 0 0 1 0 0 0 0 1 | | 𝑥𝑜𝑙𝑑 𝑦𝑜𝑙𝑑 𝑧𝑜𝑙𝑑 1 | = | 1 0 2 0 0 1 2 0 0 0 1 0 0 0 0 1 | | 1 1 3 1 | = | 1 + 0 + 6 + 0 0 + 1 + 6 + 0 0 + 0 + 3 + 0 0 + 0 + 0 + 1 | = | 7 7 3 1 | Therefore, the new coordinates for point (0, 0, 0) will be (0, 0, 0)