Question

prove that two successive rotations are commutative

Answer 1

Using trigonometric identities for sin and cosine of the sum of two angles :

We can express the elements of the product matrix for two successive rotations in the xy plane about the origin as :

[Cos (Ф₁ + Ф₂) - Sin (Ф₁ + Ф₂) 0]
[ Sin (Ф₁ + Ф₂) Cos (Ф₁ + Ф₂) 0 ]
[0 0 1]


Since the addition of angles is commutative :

This successive transformation is commutative.

Answer 2

Answer:The rotation matrix is written as

$\left[\begin{array}{ccc}x'\\y'\\1\end{array}\right]= \left[\begin{array}{ccc} cos \theta&sin\theta&0\\-sin\theta&cos\theta&0\\0&0&1\end{array}\right] *\left[\begin{array}{ccc}x\\y\\1\end{array}\right]$

STEP-1 Write the matrix with different theta values.

$R1=\left[\begin{array}{ccc}cos\theta1&sin\theta1&0\\-sin\theta1&cos\theta1&0\\0&0&1\end{array}\right]\\ R2=\left[\begin{array}{ccc}cos\theta2&sin\theta2&0\\-sin\theta2&cos\theta2&0\\0&0&1\end{array}\right]$

STEP-2 Do Multiplication matrix

$R1*R2 = \left[\begin{array}{ccc}cos\theta2&sin\theta2&0\\-sin\theta2&cos\theta2&0\\0&0&1\end{array}\right]*\left[\begin{array}{ccc}cos\theta2&sin\theta2&0\\-sin\theta2&cos\theta2&0\\0&0&1\end{array}\right]$

STEP-3 Therefore the multiplication of the two rotational matrix of LHS will be:

$R1*R2 = \left[\begin{array}{ccc}cos\theta1*cos\theta2-sin\theta1*sin\theta2 & cos\theta1*sin\theta2+sin\theta1*cos\theta2&0\\-sin\theta1*cos\theta2-cos\theta1*sin\theta2& -sin\theta1*sin\theta2+cos\theta1*cos\theta2&0\\0&0&1\end{array}\right]$

STEP-4 Similarly, the RHS multiplicative matrix will be :

$R2*R1 = \left[\begin{array}{ccc}cos\theta1*cos\theta2-sin\theta1*sin\theta2 & cos\theta1*sin\theta2+sin\theta1*cos\theta2&0\\-sin\theta1*cos\theta2-cos\theta1*sin\theta2& -sin\theta1*sin\theta2+cos\theta1*cos\theta2&0\\0&0&1\end{array}\right]$

STEP-5

$R1*R2\left[\begin{array}{ccc}cos(\theta1-\theta2)&sin(\theta2+\theta1)&0\\-(sin(\theta1+\theta2)& cos(\theta1-\theta2)&0\\0&0&1\end{array}\right]$

STEP-6

$R2*R1=\left[\begin{array}{ccc}cos(\theta1-\theta2)&sin(\theta2+\theta1)&0\\-(sin(\theta1+\theta2)& cos(\theta1-\theta2)&0\\0&0&1\end{array}\right]$

Therefore , we can say that R1 * R2 = R2 * R1.

Hence proved