Description of matrices and geometric rotation with examples
A polar coordinate can be described by a pair of numbers\( (x, y)\). The numbers are the distance from the y axis (x), and from the x axis (y) in the coordinate system. Any points to the left of the y-axis will have a negative x coordinate. Any points below the x axis, will have a negative y coordinate.
Instead with the term \((x, y)\) , we can describe the point with radius \(r\) and the angle \(θ\) as \((r, θ)\).
\(\displaystyle r=\sqrt{x^2+y^2}\) \(\displaystyle tan(θ)=\frac{y}{x}\)
In the diagram above, \(r\) is the hypotenuse of a right-angled triangle.
The x-position can be calculated from the radius \(r\) and the angle \(θ\) according to the following formula.
\(x=r·cos(θ)\)
The y-position is calculated accordingly from the formula
\(y=r·sin(θ)\)
In the following figure, we have rotated the point (x, y) by the angle \(φ\). So, we have now
\(x'=r·cos(θ+φ)\) \(y'=r·sin(θ+φ)\)
When we write this in a matrix form it looks like this
\(\displaystyle \left[\matrix{x'\\y'}\right] = \left[\matrix{cos(φ) & - sin(φ) \\ sin(φ) & cos(φ)}\right] \left[\matrix{x \\ y }\right] \)
The example below shows a matrix that rotates the vector by an angle of \(φ = 30°\).
\(\displaystyle \left[\matrix{cos(30) & - sin(30) \\ sin(30) & cos(30)}\right] = \left[\matrix{0.866 & - 0.5 \\ 0.5 & 0.866}\right] \)
With this Matrix the position vector for the point \((1,0)\) becomes
\(\displaystyle \left[\matrix{x' \\ y'}\right] = \left[\matrix{0.866 & - 0.5 \\ 0.5 & 0.866}\right] · \left[\matrix{1 \\ 0}\right] =\left[\matrix{0.866 \\ 0.5}\right] \)
A rotation in 3-space, counter clockwise shows the matrices below
About the x-axis
\(\displaystyle \left[\matrix{1 & 0 & 0 \\ 0 & cos(φ) & - sin(φ) \\ 0 & sin(φ) & cos(φ) }\right] \)
About the y-axis
\(\displaystyle \left[\matrix{cos(φ) & 0 & sin(φ) \\ 0 & 1 & 0 \\ - sin(φ) & 0 & cos(φ) }\right] \)
About the z-axis
\(\displaystyle \left[\matrix{cos(φ) & - sin(φ) & 0 \\ sin(φ) & cos(φ) & 0 \\ 0 & 0 & 1}\right] \)