Given u → = ( u 1 u 2 ) , if it is rotated sixty degrees anti-clockwise prove that the resulting vector will be v → = 1 2 ( u 1 − 3 u 2 3 u 1 + u 2 ) I think there are better methods out there but one idea I had was to use cos ⁡ 60 = 1 2 = u → ⋅ v → | u → | | v → | and solve for v → but obviously this seems impossible to me with the | v → | and I see no where to force a 3 to appear.Is this true? Or if this method works please share. I have other ideas which probably work but this one also came to mind but with no fruitition

Question

Given             u      →        =                    (                            u        1                    u        2                            )            , if it is rotated sixty degrees anti-clockwise prove that the resulting vector will be             v      →        =      1    2                      (                                      u          1                −                  3                          u          2                                      3                          u          1                +                  u          2                                    )            I think there are better methods out there but one idea I had was to use  cos  ⁡  60  =      1    2    =                              u          →                    ⋅                        v          →                                    |                              u            →                          |                    |                              v            →                          |              and solve for             v      →       but obviously this seems impossible to me with the       |              v      →            |   and I see no where to force a       3   to appear.Is this true? Or if this method works please share. I have other ideas which probably work but this one also came to mind but with no fruitition

Royce Morrison · Accepted Answer

Your method actually could work but there is an easier approach involving the Rotation matrix which is defined as      R    θ    =      (                            cos          ⁡          θ                          −          sin          ⁡          θ                                      sin          ⁡          θ                          cos          ⁡          θ                      )  it can be easily proven that      R    θ        (                            cos          ⁡          x                                      sin          ⁡          x                      )    =      (                            cos          ⁡          (          x          +          θ          )                                      sin          ⁡          (          x          +          θ          )                      )  Let       (                                        u            1                                                            u            2                                )    =  r      (                            cos          ⁡          x                                      sin          ⁡          x                      )  since rotating sixty degrees anti-clockwise will result same as rotating three hundred clockwise let   θ  =  300      R    θ        (                                        u            1                                                            u            2                                )    =  r      R    θ        (                            cos          ⁡          x                                      sin          ⁡          x                      )    =  r      (                            cos          ⁡          (          x          +          300          )                                      sin          ⁡          (          x          +          300          )                      )    r      (                            cos          ⁡          (          x          +          300          )                                      sin          ⁡          (          x          +          300          )                      )    =  r      (                            cos          ⁡          (          x          )          cos          ⁡          300          −          sin          ⁡          x          sin          ⁡          300                                      sin          ⁡          x          cos          ⁡          300          +          cos          ⁡          x          sin          ⁡          300                      )    =      (                                        r            2                    cos          ⁡          x          −          r                                                    3                            sin              ⁡              x                        2                                                            r            2                    sin          ⁡          x          +                                    r                              3                                      2                    cos          ⁡          x                      )    =      1    2                      (                                      u          1                −                  3                          u          2                                      3                          u          1                +                  u          2                                    )

Given vec(u) =((u_1),(u_2)), if it is rotated sixty degrees anti-clockwise prove that the resulting vector will be vec(v) =1/2((u_1−sqrt(3)u_2),(sqrt(3)u_1+u_2))

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