This article is a derivative of Homogeneous Coordinates and Matrix Derivation of Basic Geometric Transformations in 2D . Understanding 2D transformations makes it easy to derive 3D transformations.
Basic Geometric Transformations in 3D#
Translation#
Describes the transformation from point (x, y, z) to (x + dx, y + dy, z + dz).
Introducing homogeneous coordinates, it can be expressed as (x, y, z, 1) transformed to (x + dx, y + dy, z + dz, 1).
Given:
Copy ( a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 ) ( x y z 1 ) = ( x + d x y + d y z + d z 1 ) \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z \\
1
\end{pmatrix}
=
\begin{pmatrix}
x + d_x \\
y + d_y \\
z + d_z \\
1
\end{pmatrix} a 11 a 21 a 31 a 41 a 12 a 22 a 32 a 42 a 13 a 23 a 33 a 43 a 14 a 24 a 34 a 44 x y z 1 = x + d x y + d y z + d z 1 We have:
Copy { a 11 x + a 12 y + a 13 z + a 14 = x + d x a 21 x + a 22 y + a 23 z + a 24 = y + d y a 31 x + a 32 y + a 33 z + a 34 = z + d z a 41 x + a 42 y + a 43 z + a 44 = 1 \begin{cases}
a_{11}x + a_{12}y + a_{13}z + a_{14} = x + d_x \\
a_{21}x + a_{22}y + a_{23}z + a_{24} = y + d_y \\
a_{31}x + a_{32}y + a_{33}z + a_{34} = z + d_z \\
a_{41}x + a_{42}y + a_{43}z + a_{44} = 1 \\
\end{cases} ⎩ ⎨ ⎧ a 11 x + a 12 y + a 13 z + a 14 = x + d x a 21 x + a 22 y + a 23 z + a 24 = y + d y a 31 x + a 32 y + a 33 z + a 34 = z + d z a 41 x + a 42 y + a 43 z + a 44 = 1 Solving for:
Copy ( 1 0 0 d x 0 1 0 d y 0 0 1 d z 0 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 & dx \\
0 & 1 & 0 & dy \\
0 & 0 & 1 & dz \\
0 & 0 & 0 & 1
\end{pmatrix} 1 0 0 0 0 1 0 0 0 0 1 0 d x d y d z 1 Scaling#
Describes the transformation from point (x, y, z) to (sx * x, sy * y, sz * z), where sx, sy, sz are constants.
Introducing homogeneous coordinates, it can be expressed as (x, y, z, 1) transformed to (sx * x, sy * y, sz * z, 1).
Given:
Copy ( a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 ) ( x y z 1 ) = ( s x ∗ x s y ∗ y s z ∗ z 1 ) \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z \\
1
\end{pmatrix}
=
\begin{pmatrix}
s_x * x \\
s_y * y \\
s_z * z \\
1
\end{pmatrix} a 11 a 21 a 31 a 41 a 12 a 22 a 32 a 42 a 13 a 23 a 33 a 43 a 14 a 24 a 34 a 44 x y z 1 = s x ∗ x s y ∗ y s z ∗ z 1 We have:
Copy { a 11 x + a 12 y + a 13 z + a 14 = s x ∗ x a 21 x + a 22 y + a 23 z + a 24 = s y ∗ y a 31 x + a 32 y + a 33 z + a 34 = s z ∗ z a 41 x + a 42 y + a 43 z + a 44 = 1 \begin{cases}
a_{11}x + a_{12}y + a_{13}z + a_{14} = s_x * x \\
a_{21}x + a_{22}y + a_{23}z + a_{24} = s_y * y \\
a_{31}x + a_{32}y + a_{33}z + a_{34} = s_z * z \\
a_{41}x + a_{42}y + a_{43}z + a_{44} = 1 \\
\end{cases} ⎩ ⎨ ⎧ a 11 x + a 12 y + a 13 z + a 14 = s x ∗ x a 21 x + a 22 y + a 23 z + a 24 = s y ∗ y a 31 x + a 32 y + a 33 z + a 34 = s z ∗ z a 41 x + a 42 y + a 43 z + a 44 = 1 Solving for:
Copy ( s x 0 0 0 0 s y 0 0 0 0 s z 0 0 0 0 1 ) \begin{pmatrix}
s_x & 0 & 0 & 0 \\
0 & s_y & 0 & 0 \\
0 & 0 & s_z & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} s x 0 0 0 0 s y 0 0 0 0 s z 0 0 0 0 1 Rotation#
For 3D rotation, it can be decomposed into the product of three matrices (rotation around the x-axis, rotation around the y-axis, rotation around the z-axis).
Rotation around the z-axis —— Rz (β)#
Do you remember the derivation of 2D rotation? When we rotate on the plane formed by the x and y axes, it can also be viewed as rotating around the z-axis. So the rotation in 3D is easy to understand, it's just adding an additional dimension that remains fixed.
That is:
Copy ( a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 ) ( x y z 1 ) = ( x c o s β − y s i n β y c o s β + x s i n β z 1 ) \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z \\
1
\end{pmatrix}
=
\begin{pmatrix}
xcosβ - ysinβ \\
ycosβ + xsinβ \\
z \\
1
\end{pmatrix} a 11 a 21 a 31 a 41 a 12 a 22 a 32 a 42 a 13 a 23 a 33 a 43 a 14 a 24 a 34 a 44 x y z 1 = x cos β − ys in β ycos β + x s in β z 1 We have:
Copy { a 11 x + a 12 y + a 13 z + a 14 = x c o s β − y s i n β a 21 x + a 22 y + a 23 z + a 24 = y c o s β + x s i n β a 31 x + a 32 y + a 33 z + a 34 = z a 41 x + a 42 y + a 43 z + a 44 = 1 \begin{cases}
a_{11}x + a_{12}y + a_{13}z + a_{14} = xcosβ - ysinβ \\
a_{21}x + a_{22}y + a_{23}z + a_{24} = ycosβ + xsinβ \\
a_{31}x + a_{32}y + a_{33}z + a_{34} = z \\
a_{41}x + a_{42}y + a_{43}z + a_{44} = 1 \\
\end{cases} ⎩ ⎨ ⎧ a 11 x + a 12 y + a 13 z + a 14 = x cos β − ys in β a 21 x + a 22 y + a 23 z + a 24 = ycos β + x s in β a 31 x + a 32 y + a 33 z + a 34 = z a 41 x + a 42 y + a 43 z + a 44 = 1 Solving for:
Copy R z β = ( c o s β − s i n β 0 0 s i n β c o s β 0 0 0 0 1 0 0 0 0 1 ) R_{z}β =
\begin{pmatrix}
cosβ & -sinβ & 0 & 0 \\
sinβ & cosβ & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} R z β = cos β s in β 0 0 − s in β cos β 0 0 0 0 1 0 0 0 0 1 Rotation around the x-axis —— Rx (α)#
Similarly (fixing x):
Copy R x α = ( 1 0 0 0 0 c o s α − s i n α 0 0 s i n α c o s α 0 0 0 0 1 ) R_{x}α =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & cosα & -sinα & 0 \\
0 & sinα & cosα & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} R x α = 1 0 0 0 0 cos α s in α 0 0 − s in α cos α 0 0 0 0 1 Rotation around the y-axis —— Ry (γ)#
Similarly (fixing y):
Copy R y γ = ( c o s γ 0 s i n γ 0 0 1 0 0 − s i n γ 0 c o s γ 0 0 0 0 1 ) R_{y}γ =
\begin{pmatrix}
cosγ & 0 & sinγ & 0 \\
0 & 1 & 0 & 0 \\
-sinγ & 0 & cosγ & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} R y γ = cos γ 0 − s inγ 0 0 1 0 0 s inγ 0 cos γ 0 0 0 0 1 So we have the rotation matrix in 3D:
Copy R = R x α R y γ R z β R = R_{x}αR_{y}γR_{z}β R = R x α R y γ R z β But you may wonder how to decompose θ angle around a certain axis into α, β, γ? Let's leave this question for another article.
However, currently the rotation axes are limited to passing through the origin. If the rotation axis does not pass through the origin, it needs to be translated to the origin first, then rotated, and finally translated back to its original position. Any complex transformation can be decomposed into the composition of several simple transformations.
Shearing#
Shearing along the X-axis#
Describes the transformation from (x, y, z) to (x + m * y + n * z, y, z), where m and n are tangent functions.
Copy ( a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 ) ( x y z 1 ) = ( x + m ∗ y + n ∗ z y z 1 ) \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z \\
1
\end{pmatrix}
=
\begin{pmatrix}
x + m * y + n * z \\
y \\
z \\
1
\end{pmatrix} a 11 a 21 a 31 a 41 a 12 a 22 a 32 a 42 a 13 a 23 a 33 a 43 a 14 a 24 a 34 a 44 x y z 1 = x + m ∗ y + n ∗ z y z 1 Solving for:
Copy ( 1 m n 0 0 1 0 0 0 0 1 0 0 0 0 1 ) \begin{pmatrix}
1 & m & n & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix} 1 0 0 0 m 1 0 0 n 0 1 0 0 0 0 1 Shearing along the Y-axis#
Describes the transformation from (x, y, z) to (x, y + m * x + n * z, z).
Skipping the solution, I believe smart readers can understand the meaning of the row and column terms.
Copy ( 1 0 0 0 m 1 n 0 0 0 1 0 0 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 & 0 \\
m & 1 & n & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix} 1 m 0 0 0 1 0 0 0 n 1 0 0 0 0 1 Shearing along the Z-axis#
Describes the transformation from (x, y, z) to (x, y, z + m * x + n * y).
Copy ( 1 0 0 0 0 1 0 0 m n 1 0 0 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
m & n & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix} 1 0 m 0 0 1 n 0 0 0 1 0 0 0 0 1