If a and b are vectors perpendicular to each other Then is it okay to say that the general solution to r → × a → = b → is r → = r → ⋅ a → a → ⋅ a → a → + 1 a → ⋅ a → ( a → × b → ) ?I did cross product with a vector to the given vector equation to get the above result will we really say the solution is that only as such x has dependance on r again isn't ?

Question

If a and b are vectors perpendicular to each other Then is it okay to say that the general solution to             r      →        ×            a      →        =            b      →        is            r      →        =                              r          →                    ⋅                        a          →                                              a          →                    ⋅                        a          →                                a      →        +      1                            a          →                    ⋅                        a          →                      (            a      →        ×            b      →        )    ?I did cross product with a vector to the given vector equation to get the above result will we really say the solution is that only as such x has dependance on r again isn&#039;t ?

Dabbaghnn · Accepted Answer

Starting with             r      →        ×            a      →        =            b      →      where             a      →        ,            b      →        are given, we have from the properties of cross product that             r      →        must be perpendicular to             b      →       . Clearly we also must have             b      →        perpendicular to             a      →       . Therefore, our solution             r      →        lies in the plane spanned by             a      →       and             a      →        ×            b      →       , i.e.            r      →        =  t            a      →        +  s      (                  a        →              ×                  b        →              )  Plug this expression into the original equation      (    t                  a        →              +    s          (                        a          →                    ×                        b          →                    )        )    ×            a      →        =            b      →      Using the fact that             a      →        ×            a      →        =  0 and      (                  a        →              ×                  b        →              )    ×            a      →        =  −            a      →        (            a      →        ⋅            b      →        )  +            b      →        (            a      →        ⋅            a      →        )and using the fact that             a      →        ⋅            b      →        =  0 because they are pendicular, then  s  =            1                                    a            →                          ⋅                              a            →                              Therefore, the general solution is            r      →        =  t            a      →        +                                          a            →                          ×                              b            →                                                            a            →                          ⋅                              a            →                              where   t  ∈      R

Showing the general solution to the given vector in terms of other known vectors

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