I have two 2D unit vectors a and b. I'd like to find the rotation matrix that rotates a to b. The formulas I see online are for a rotation matrix are ( cos ⁡ θ − sin ⁡ θ sin ⁡ θ cos ⁡ θ ) And I can get the angle between a and b with θ = cos − 1 ⁡ ( a ⋅ b )My problem is that that doesn't give me the direction. For example, if θ is π / 2 when maybe the matrix should use − π / 2

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I have two 2D unit vectors a and b. I&#039;d like to find the rotation matrix that rotates a to b. The formulas I see online are for a rotation matrix are      (                            cos          ⁡          θ                          −          sin          ⁡          θ                                      sin          ⁡          θ                          cos          ⁡          θ                      )  And I can get the angle between a and b with  θ  =      cos          −      1        ⁡  (      a    ⋅      b    )My problem is that that doesn&#039;t give me the direction. For example, if   θ is   π      /    2 when maybe the matrix should use   −  π      /    2

ab8s1k28q · Accepted Answer

You don’t need to compute the angle explicitly, or indeed refer to an angle at all.Observe that the result of rotating any vector   (  x  ,  y      )    T   90 degrees counterclockwise is   (  −  y  ,  x      )    T  . Then, using the fact that the columns of a transformation matrix are the images of the basis vectors, the matrix      [                            x                          −          y                                      y                          x                      ]  represents a rotation that maps   (  1  ,  0      )    T   onto the unit vector   (  x  ,  y      )    T  . Therefore, a rotation that takes the unit vector       a    =  (      x    a    ,      y    a        )    T   to the unit vector       b    =  (      x    b    ,      y    b        )    T   has the matrix      [                                        x            b                                    −                      y            b                                                            y            b                                                x            b                                ]              [                                                  x              a                                            −                          y              a                                                                          y              a                                                          x              a                                          ]              −      1        =      [                                        x            b                                    −                      y            b                                                            y            b                                                x            b                                ]        [                                        x            a                                                y            a                                                −                      y            a                                                x            a                                ]    =      [                                        x            a                                x            b                    +                      y            a                                y            b                                                x            b                                y            a                    −                      x            a                                y            b                                                            x            a                                y            b                    −                      x            b                                y            a                                                x            a                                x            b                    +                      y            a                                y            b                                ]    .

Darian Hubbard · Accepted Answer

My solution is to regard both vectors as complex numbers. Thus         a    =      a    x    +      a    y    i   and         b    =      b    x    +      b    y    i  .  . If both are unit vectors, then to rotate a to b you need to multiply by            a              −        1              b    =                  a        ¯              b    =  (      a    x      −        a    y    i  )  (      b    x      +        b    y    i  )  =  (      a    x        b    x    +      a    y        b    y    )  +  (      a    x        b    y    −      a    y        b    x    )  i  .Since this is a unit vector it is of the form     cos  ⁡  (  θ  )  +  sin  ⁡  (  θ  )    i   and you can get the rotation matrix      (                            cos          ⁡          θ                          −          sin          ⁡          θ                                      sin          ⁡          θ                          cos          ⁡          θ                      )  directly from the complex number form.

I have two 2D unit vectors a and b. I'd like to find the rotation matrix that rotates a to b. ((cos theta,- sin theta),(sin theta, cos theta))

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