2D and 3D rotation matrices

This note builds rotation matrices from the ground up and shows the arithmetic in detail.

Convention used in this note

• right-handed coordinate system
• column vectors
• active rotations: the vector rotates while the coordinate frame stays fixed
• positive rotations follow the right-hand rule

Other conventions exist, so always check the one used by a robotics library or lecture.

What a rotation matrix is trying to do

Suppose a robot has a vector pointing in some direction:

$$v= \begin{bmatrix} x\\ y \end{bmatrix} \qquad\text{or in 3D}\qquad v= \begin{bmatrix} x\\ y\\ z \end{bmatrix}$$

A rotation matrix changes the direction of that vector without changing its length. The basic operation is:

$$\boxed{v'=Rv}$$

where $v$ is the original vector, $R$ is the rotation matrix and $v'$ is the rotated vector.

The matrix does not mean “multiply matching-looking numbers together”. Matrix multiplication follows a specific row-by-column rule.

Matrix multiplication refresher

Suppose:

$$A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}, \qquad v= \begin{bmatrix} x\\ y \end{bmatrix}$$

The first answer value comes from the first row of the matrix dotted with the vector, $ax+by$. The second comes from the second row, $cx+dy$. So:

$$\boxed{ Av= \begin{bmatrix} ax+by\\ cx+dy \end{bmatrix} }$$

Worked mini-example

$$A= \begin{bmatrix} 2&3\\ 4&5 \end{bmatrix}, \qquad v= \begin{bmatrix} 10\\ 20 \end{bmatrix}$$

First row times the column:

$$(2)(10)+(3)(20)=20+60=80$$

Second row times the column:

$$(4)(10)+(5)(20)=40+100=140$$

Therefore:

$$Av= \begin{bmatrix} 80\\ 140 \end{bmatrix}$$

You always use addition between the products. For a rotation matrix, the negative signs are already built into the matrix entries.

The standard 2D rotation matrix

The standard counterclockwise 2D rotation matrix is:

$$\boxed{ R(\theta)= \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} }$$

where $\theta$ is the rotation angle. Multiplying it out row-by-column:

$$\begin{bmatrix} x'\\ y' \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$$

The first row gives:

$$\boxed{x'=x\cos\theta-y\sin\theta}$$

and the second:

$$\boxed{y'=x\sin\theta+y\cos\theta}$$

Worked 2D examples

Rotate (1, 0) by 90°

We expect a vector pointing along positive $x$ to rotate counterclockwise until it points along positive $y$. At $90^\circ$, $\cos90^\circ=0$ and $\sin90^\circ=1$, so:

$$R(90^\circ)= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix}$$

Multiply:

$$v'= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} \begin{bmatrix} 1\\ 0 \end{bmatrix}$$ $$x'=(0)(1)+(-1)(0)=0 \qquad y'=(1)(1)+(0)(0)=1$$ $$\boxed{ v'= \begin{bmatrix} 0\\ 1 \end{bmatrix} }$$

That is exactly the $(1,0)\rightarrow(0,1)$ we expected before calculating. This kind of geometric sanity check is extremely useful in robotics.

Rotate (2, 1) by 90°

Using the same $R(90^\circ)$:

$$x'=(0)(2)+(-1)(1)=-1 \qquad y'=(1)(2)+(0)(1)=2$$ $$\boxed{ v'= \begin{bmatrix} -1\\ 2 \end{bmatrix} }$$

Check the length before and after:

$$\|v\|=\sqrt{2^2+1^2}=\sqrt5 \qquad \|v'\|=\sqrt{(-1)^2+2^2}=\sqrt5$$

The direction changed and the length did not, which is exactly what a pure rotation should do.

Rotate (1, 1) by 45°

Recall $\cos45^\circ=\sin45^\circ=\frac{\sqrt2}{2}\approx0.7071$, so:

$$R(45^\circ)= \begin{bmatrix} 0.7071&-0.7071\\ 0.7071&0.7071 \end{bmatrix}$$ $$x'=(0.7071)(1)+(-0.7071)(1)=0 \qquad y'=(0.7071)(1)+(0.7071)(1)=1.4142$$ $$\boxed{ v'\approx \begin{bmatrix} 0\\ 1.4142 \end{bmatrix} }$$

$(1,1)$ already points at $45^\circ$. Rotating it another $45^\circ$ should make it point straight up, and it does.

Direction convention

With this convention a positive angle is counterclockwise and a negative angle is clockwise. A clockwise rotation by $\theta$ is $R(-\theta)$, and using $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$:

$$\boxed{ R(-\theta)= \begin{bmatrix} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{bmatrix} }$$

Example: clockwise 90°

$$R(-90^\circ)= \begin{bmatrix} 0&1\\ -1&0 \end{bmatrix}$$

Applied to $(1, 0)$:

$$x'=(0)(1)+(1)(0)=0 \qquad y'=(-1)(1)+(0)(0)=-1$$

which is exactly a $90^\circ$ clockwise turn from positive $x$.

Moving from 2D to 3D

In 3D the rotation matrix is $3\times3$. The key idea:

When rotating around one axis, the coordinate along that axis stays unchanged.

Rotating around the $z$-axis changes $x$ and $y$ and leaves $z$ alone. That is why the 2D rotation pattern appears inside each 3D matrix.

Rotation about the x-axis

$$\boxed{ R_x(\phi)= \begin{bmatrix} 1&0&0\\ 0&\cos\phi&-\sin\phi\\ 0&\sin\phi&\cos\phi \end{bmatrix} }$$

$\phi$ is commonly used for roll. The first row, $\begin{bmatrix}1&0&0\end{bmatrix}$, guarantees $x'=x$ while the $y$-$z$ plane rotates.

Example: 90° about x

$$R_x(90^\circ)= \begin{bmatrix} 1&0&0\\ 0&0&-1\\ 0&1&0 \end{bmatrix}$$

Rotate the vector along positive $y$, $v=(0,1,0)$:

$$x'=(1)(0)+(0)(1)+(0)(0)=0$$ $$y'=(0)(0)+(0)(1)+(-1)(0)=0$$ $$z'=(0)(0)+(1)(1)+(0)(0)=1$$

So a positive $90^\circ$ rotation about $x$ sends positive $y$ to positive $z$, following the right-hand rule.

Rotation about the y-axis

$$\boxed{ R_y(\theta)= \begin{bmatrix} \cos\theta&0&\sin\theta\\ 0&1&0\\ -\sin\theta&0&\cos\theta \end{bmatrix} }$$

The middle row guarantees $y'=y$.

Example: 90° about y

$$R_y(90^\circ)= \begin{bmatrix} 0&0&1\\ 0&1&0\\ -1&0&0 \end{bmatrix}$$

Rotate positive $x$, $v=(1,0,0)$:

$$x'=0 \qquad y'=0 \qquad z'=(-1)(1)+(0)(0)+(0)(0)=-1$$

So positive $x$ rotates toward negative $z$. The sign placement is the reverse of $R_x$ and $R_z$, which is why $R_y$ often feels less intuitive at first.

Rotation about the z-axis

$$\boxed{ R_z(\psi)= \begin{bmatrix} \cos\psi&-\sin\psi&0\\ \sin\psi&\cos\psi&0\\ 0&0&1 \end{bmatrix} }$$

$\psi$ is commonly used for yaw. This is the 2D rotation matrix placed in the $x$-$y$ corner of a $3\times3$ matrix, and $z$ stays unchanged.

A 3D example

Rotate $v=(2,1,3)$ by $90^\circ$ about $z$:

$$\begin{bmatrix} 0&-1&0\\ 1&0&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 2\\ 1\\ 3 \end{bmatrix} = \begin{bmatrix} (0)(2)+(-1)(1)+(0)(3)\\ (1)(2)+(0)(1)+(0)(3)\\ (0)(2)+(0)(1)+(1)(3) \end{bmatrix} = \begin{bmatrix} -1\\ 2\\ 3 \end{bmatrix}$$

$z=3$ did not change, because $z$ was the rotation axis. This is a useful way to reconstruct the three standard matrices rather than memorising them: the axis row and column are zeros with a $1$ where they cross, and the other four entries are the 2D pattern.

Properties of rotation matrices

A valid pure rotation matrix satisfies:

$$R^TR=I$$

which means its rows and columns are mutually perpendicular unit vectors. It follows that:

$$\boxed{R^{-1}=R^T} \qquad \boxed{\det(R)=1}$$

A determinant of $-1$ would indicate a reflection rather than a proper rotation.

Checking it for 90° about z

$$R^TR= \begin{bmatrix} 0&1&0\\ -1&0&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 0&-1&0\\ 1&0&0\\ 0&0&1 \end{bmatrix}$$

The top-left entry is row 1 of $R^T$ dotted with column 1 of $R$: $(0)(0)+(1)(1)+(0)(0)=1$. The entry beside it is $(0)(-1)+(1)(0)+(0)(0)=0$. Continuing gives the identity, confirming the matrix preserves lengths and angles.

Undoing a rotation

If $v'=Rv$, then $v=R^{-1}v'$, and because $R^{-1}=R^T$:

$$\boxed{v=R^Tv'}$$

For example, $R_z(90^\circ)$ took $(1,0,0)$ to $(0,1,0)$, and its transpose takes $(0,1,0)$ straight back to $(1,0,0)$.

Robotics interpretation

Rotation matrices mostly appear when moving a vector between coordinate frames. Suppose a camera sees a point as $p_C=(1,0,2)$, where the subscript $C$ means “expressed in the camera frame”. A matrix ${}^BR_C$ describes how the camera axes are oriented relative to the robot body frame $B$, and:

$$p_B={}^BR_C\,p_C$$

expresses the same physical vector in the body frame. This is where careful frame notation becomes important.

Project connection: the balancing robot’s IMU

The balancing-car firmware gives the MPU6050 DMP this orientation matrix:

$$S= \begin{bmatrix} -1&0&0\\ 0&-1&0\\ 0&0&1 \end{bmatrix}$$

For a sensor-frame vector $v_s=(x_s,y_s,z_s)$, the body-frame components are $v_b=Sv_s$:

$$x_b=(-1)x_s+(0)y_s+(0)z_s=-x_s$$ $$y_b=(0)x_s+(-1)y_s+(0)z_s=-y_s$$ $$z_b=(0)x_s+(0)y_s+(1)z_s=z_s$$ $$\boxed{ v_b= \begin{bmatrix} -x_s\\ -y_s\\ z_s \end{bmatrix} }$$

The sensor’s $x$ and $y$ are reversed while $z$ stays unchanged, and that is exactly $R_z(180^\circ)$, because $\cos180^\circ=-1$ and $\sin180^\circ=0$. The sensor is mounted rotated half a turn about the vertical axis.

Common mistakes

Multiplying element-by-element. $Rv$ is row-by-column multiplication, not matching entries.

Changing the plus signs. A row $\begin{bmatrix}0&-1&0\end{bmatrix}$ times $(0,1,0)$ is $(0)(0)+(-1)(1)+(0)(0)$. The negative comes from the matrix entry, not from switching to subtraction.

Mixing degrees and radians. Hand calculations use degrees; robotics software almost always uses radians, $90^\circ=\frac{\pi}{2}$. In NumPy:

theta = np.deg2rad(90)

Mixing active and passive rotations. An active rotation rotates the vector. A passive rotation changes the frame used to describe the same vector. They are related by the transpose, so a sign or transpose difference can appear between textbooks.

Summary

• The main operation is $v'=Rv$, and one output number is one matrix row dotted with the vector column.
• Rotation preserves length.
• The rotation-axis coordinate stays unchanged; the other two behave like a 2D rotation.
• The transpose reverses a proper rotation.
• Multiplication order matters when combining rotations, which is where Euler angles come in next.