Sign in to follow this  

Deriving the matrix form of the Rodrigues formula

Recommended Posts


I'm trying to derive the matrix form of Rodrigues rotation formula but I do something really bad.

R{N}(V) = cos(a)*V+(1-cos(a))*(N*V)*N+sin(a)*(NxV)


  • N: Unit vector of the rotation angle.
  • V: The vector to be rotated around N
  • a: The angle of rotation (deg).
  • (N*V): The dot product of N and V vectors.
  • (NxV): The cross product of N and V vectors.
What I have found is that I have to apply the i(1,0,0) j(0,1,0) and k(0,0,1) to this formula to get the desired form. So I did chose to express the first row vector of the matrix form which should be:

R{1,*} = (cos(a)+(1-cos(a))*x2 , (1-cos(a))*x*y-sin(a)*z , (1-cos(a))*x*z-sin(a)*y)

Here's my best try so far:


n = i => n(1,0,0)


N*V = 1*x+0*y+0*z = x

NxV = (0,-z,y)

|cos(a)*x+(1-cos(a))*x+0 | => cos(a)*x + (1-cos(a))*x ?= cos(a)+(1-cos(a))*x2

|cos(a)*y+(1-cos(a))*x-sin(a)*z | => how will cos(a) disappear?

| cos(a)*z+(1-cos(a))*x+sin(a)*y | => again no cos(a) in the final matrix :-/

I can't really understand the concept of the first element of this vector. How can I actually calculate cos(a)+(1-cos(a))*x2 from the actual cos(a)*x+(1-cos(a))*x? Everytime I stuck at:

cos(a)*x+(1-cos(a))*x+0 = cos(a)*x+(1-cos(a))*x = (cos(a) + (1 - cos(a)) * x

which does not look equals to me. Edited by Wrath87

Share this post

Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this