Four dimensional cross product of THREE vectors












0












$begingroup$


There are many MSE posts about how to define a cross product in $mathbb{R^4}$. It is impossible to define a cross product of two vectors in $mathbb{R^4}$, since there are infinitely many directions perpendicular to those two vectors, and we don't know which direction to choose. However, If we are given THREE vectors $A,B,C$, it is possible to find a unique direction perpendicular to this three vectors, if $A,B,C$ are independent. However, finding this perpendicular vector involves solving a system of equations.



So my question is：can we define a Quasi Cross Product ${A,B,C}$ on $mathbb{R^4}$, so that we can find a direction perpendicular to $A,B,C$ without solving a system of equations?






linear-algebra systems-of-equations cross-product







share|cite|improve this question









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    0












    $begingroup$


    There are many MSE posts about how to define a cross product in $mathbb{R^4}$. It is impossible to define a cross product of two vectors in $mathbb{R^4}$, since there are infinitely many directions perpendicular to those two vectors, and we don't know which direction to choose. However, If we are given THREE vectors $A,B,C$, it is possible to find a unique direction perpendicular to this three vectors, if $A,B,C$ are independent. However, finding this perpendicular vector involves solving a system of equations.



    So my question is：can we define a Quasi Cross Product ${A,B,C}$ on $mathbb{R^4}$, so that we can find a direction perpendicular to $A,B,C$ without solving a system of equations?






    linear-algebra systems-of-equations cross-product







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      There are many MSE posts about how to define a cross product in $mathbb{R^4}$. It is impossible to define a cross product of two vectors in $mathbb{R^4}$, since there are infinitely many directions perpendicular to those two vectors, and we don't know which direction to choose. However, If we are given THREE vectors $A,B,C$, it is possible to find a unique direction perpendicular to this three vectors, if $A,B,C$ are independent. However, finding this perpendicular vector involves solving a system of equations.



      So my question is：can we define a Quasi Cross Product ${A,B,C}$ on $mathbb{R^4}$, so that we can find a direction perpendicular to $A,B,C$ without solving a system of equations?






      linear-algebra systems-of-equations cross-product







      share|cite|improve this question









      $endgroup$




      There are many MSE posts about how to define a cross product in $mathbb{R^4}$. It is impossible to define a cross product of two vectors in $mathbb{R^4}$, since there are infinitely many directions perpendicular to those two vectors, and we don't know which direction to choose. However, If we are given THREE vectors $A,B,C$, it is possible to find a unique direction perpendicular to this three vectors, if $A,B,C$ are independent. However, finding this perpendicular vector involves solving a system of equations.



      So my question is：can we define a Quasi Cross Product ${A,B,C}$ on $mathbb{R^4}$, so that we can find a direction perpendicular to $A,B,C$ without solving a system of equations?






      linear-algebra systems-of-equations cross-product




      linear-algebra systems-of-equations cross-product






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 2 '18 at 8:25









      Ma JoadMa Joad

      824316




      824316






















          3 Answers
          3






          active

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          2












          $begingroup$

          You have the "same" determinant formula. If $vec{a} = (a_1,a_2,a_3,a_4)$, similarly for $vec{b}$ and $vec{c}$, then $$vec{a}timesvec{b}times vec{c} = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ a_1 & a_2 & a_3 & a_4 \ b_1 & b_2 & b_3 & b_4 \ c_1 & c_2 & c_3 & c_4end{vmatrix},$$where $(vec{e}_1,ldots,vec{e}_4)$ is the standard basis for $Bbb R^4$. This does not require solving a system. Example: $$(1,1,0,0)times (0,1,1,0) times (0,0,1,1) = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1end{vmatrix} = (1,-1,1,-1).$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            An interested person could check this for determinant $vec x = vec a times vec b times vec c$ by computing the three dot products $vec x circ vec a$, $vec x circ vec b$, $vec x circ vec c$ and ensuring they are all zero.
            $endgroup$
            – DanielV
            Dec 2 '18 at 8:36



















          0












          $begingroup$

          Yes, it is the Hodge dual of the wegde product. Given three vectors $A=(a_1,a_2,a_3,a_4)$, $B=(b_1,b_2,b_3,b_4)$, and $C=(c_1,c_2,c_3,c_4)$, the triple-product is defined as
          $${A,B,C}= (det M_1, det M_2, det M_3, det M_4)$$
          with
          $$ M_1 = begin{pmatrix} a_2 & b_2 & c_2 \
          a_3 & b_3 & c_3 \
          a_4 & b_4 & c_4 end{pmatrix} $$
          where the other matrices $M_{2}$, $M_3$, $M_4$ are obtained by cyclically permuting the indices. Note that the product is antisymmetric with respect to exchanging any of its arguments.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            The short answer is yes. One way is to take the formal determinant
            $$left|begin{matrix}e_1&e_2&e_3&e_4\
            a_1&a_2&a_3&a_4\
            b_1&b_2&b_3&b_4\
            c_1&c_2&c_3&c_4\
            end{matrix}right|$$
            where $e_1,ldots,e_4$ are the standard unit vectors, and $a=sum a_ie_i$
            etc., are the three vectors.



            Or you can rephrase this in terms
            of exterior powers and the Hodge star operator.



            All this works in $n$ dimensions too.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              You have the "same" determinant formula. If $vec{a} = (a_1,a_2,a_3,a_4)$, similarly for $vec{b}$ and $vec{c}$, then $$vec{a}timesvec{b}times vec{c} = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ a_1 & a_2 & a_3 & a_4 \ b_1 & b_2 & b_3 & b_4 \ c_1 & c_2 & c_3 & c_4end{vmatrix},$$where $(vec{e}_1,ldots,vec{e}_4)$ is the standard basis for $Bbb R^4$. This does not require solving a system. Example: $$(1,1,0,0)times (0,1,1,0) times (0,0,1,1) = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1end{vmatrix} = (1,-1,1,-1).$$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                An interested person could check this for determinant $vec x = vec a times vec b times vec c$ by computing the three dot products $vec x circ vec a$, $vec x circ vec b$, $vec x circ vec c$ and ensuring they are all zero.
                $endgroup$
                – DanielV
                Dec 2 '18 at 8:36
















              2












              $begingroup$

              You have the "same" determinant formula. If $vec{a} = (a_1,a_2,a_3,a_4)$, similarly for $vec{b}$ and $vec{c}$, then $$vec{a}timesvec{b}times vec{c} = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ a_1 & a_2 & a_3 & a_4 \ b_1 & b_2 & b_3 & b_4 \ c_1 & c_2 & c_3 & c_4end{vmatrix},$$where $(vec{e}_1,ldots,vec{e}_4)$ is the standard basis for $Bbb R^4$. This does not require solving a system. Example: $$(1,1,0,0)times (0,1,1,0) times (0,0,1,1) = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1end{vmatrix} = (1,-1,1,-1).$$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                An interested person could check this for determinant $vec x = vec a times vec b times vec c$ by computing the three dot products $vec x circ vec a$, $vec x circ vec b$, $vec x circ vec c$ and ensuring they are all zero.
                $endgroup$
                – DanielV
                Dec 2 '18 at 8:36














              2












              2








              2





              $begingroup$

              You have the "same" determinant formula. If $vec{a} = (a_1,a_2,a_3,a_4)$, similarly for $vec{b}$ and $vec{c}$, then $$vec{a}timesvec{b}times vec{c} = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ a_1 & a_2 & a_3 & a_4 \ b_1 & b_2 & b_3 & b_4 \ c_1 & c_2 & c_3 & c_4end{vmatrix},$$where $(vec{e}_1,ldots,vec{e}_4)$ is the standard basis for $Bbb R^4$. This does not require solving a system. Example: $$(1,1,0,0)times (0,1,1,0) times (0,0,1,1) = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1end{vmatrix} = (1,-1,1,-1).$$






              share|cite|improve this answer









              $endgroup$



              You have the "same" determinant formula. If $vec{a} = (a_1,a_2,a_3,a_4)$, similarly for $vec{b}$ and $vec{c}$, then $$vec{a}timesvec{b}times vec{c} = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ a_1 & a_2 & a_3 & a_4 \ b_1 & b_2 & b_3 & b_4 \ c_1 & c_2 & c_3 & c_4end{vmatrix},$$where $(vec{e}_1,ldots,vec{e}_4)$ is the standard basis for $Bbb R^4$. This does not require solving a system. Example: $$(1,1,0,0)times (0,1,1,0) times (0,0,1,1) = begin{vmatrix} vec{e}_1 & vec{e}_2 & vec{e}_3 & vec{e}_4 \ 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1end{vmatrix} = (1,-1,1,-1).$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 2 '18 at 8:31









              Ivo TerekIvo Terek

              45.7k952141




              45.7k952141








              • 1




                $begingroup$
                An interested person could check this for determinant $vec x = vec a times vec b times vec c$ by computing the three dot products $vec x circ vec a$, $vec x circ vec b$, $vec x circ vec c$ and ensuring they are all zero.
                $endgroup$
                – DanielV
                Dec 2 '18 at 8:36














              • 1




                $begingroup$
                An interested person could check this for determinant $vec x = vec a times vec b times vec c$ by computing the three dot products $vec x circ vec a$, $vec x circ vec b$, $vec x circ vec c$ and ensuring they are all zero.
                $endgroup$
                – DanielV
                Dec 2 '18 at 8:36








              1




              1




              $begingroup$
              An interested person could check this for determinant $vec x = vec a times vec b times vec c$ by computing the three dot products $vec x circ vec a$, $vec x circ vec b$, $vec x circ vec c$ and ensuring they are all zero.
              $endgroup$
              – DanielV
              Dec 2 '18 at 8:36




              $begingroup$
              An interested person could check this for determinant $vec x = vec a times vec b times vec c$ by computing the three dot products $vec x circ vec a$, $vec x circ vec b$, $vec x circ vec c$ and ensuring they are all zero.
              $endgroup$
              – DanielV
              Dec 2 '18 at 8:36











              0












              $begingroup$

              Yes, it is the Hodge dual of the wegde product. Given three vectors $A=(a_1,a_2,a_3,a_4)$, $B=(b_1,b_2,b_3,b_4)$, and $C=(c_1,c_2,c_3,c_4)$, the triple-product is defined as
              $${A,B,C}= (det M_1, det M_2, det M_3, det M_4)$$
              with
              $$ M_1 = begin{pmatrix} a_2 & b_2 & c_2 \
              a_3 & b_3 & c_3 \
              a_4 & b_4 & c_4 end{pmatrix} $$
              where the other matrices $M_{2}$, $M_3$, $M_4$ are obtained by cyclically permuting the indices. Note that the product is antisymmetric with respect to exchanging any of its arguments.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Yes, it is the Hodge dual of the wegde product. Given three vectors $A=(a_1,a_2,a_3,a_4)$, $B=(b_1,b_2,b_3,b_4)$, and $C=(c_1,c_2,c_3,c_4)$, the triple-product is defined as
                $${A,B,C}= (det M_1, det M_2, det M_3, det M_4)$$
                with
                $$ M_1 = begin{pmatrix} a_2 & b_2 & c_2 \
                a_3 & b_3 & c_3 \
                a_4 & b_4 & c_4 end{pmatrix} $$
                where the other matrices $M_{2}$, $M_3$, $M_4$ are obtained by cyclically permuting the indices. Note that the product is antisymmetric with respect to exchanging any of its arguments.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Yes, it is the Hodge dual of the wegde product. Given three vectors $A=(a_1,a_2,a_3,a_4)$, $B=(b_1,b_2,b_3,b_4)$, and $C=(c_1,c_2,c_3,c_4)$, the triple-product is defined as
                  $${A,B,C}= (det M_1, det M_2, det M_3, det M_4)$$
                  with
                  $$ M_1 = begin{pmatrix} a_2 & b_2 & c_2 \
                  a_3 & b_3 & c_3 \
                  a_4 & b_4 & c_4 end{pmatrix} $$
                  where the other matrices $M_{2}$, $M_3$, $M_4$ are obtained by cyclically permuting the indices. Note that the product is antisymmetric with respect to exchanging any of its arguments.






                  share|cite|improve this answer









                  $endgroup$



                  Yes, it is the Hodge dual of the wegde product. Given three vectors $A=(a_1,a_2,a_3,a_4)$, $B=(b_1,b_2,b_3,b_4)$, and $C=(c_1,c_2,c_3,c_4)$, the triple-product is defined as
                  $${A,B,C}= (det M_1, det M_2, det M_3, det M_4)$$
                  with
                  $$ M_1 = begin{pmatrix} a_2 & b_2 & c_2 \
                  a_3 & b_3 & c_3 \
                  a_4 & b_4 & c_4 end{pmatrix} $$
                  where the other matrices $M_{2}$, $M_3$, $M_4$ are obtained by cyclically permuting the indices. Note that the product is antisymmetric with respect to exchanging any of its arguments.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 8:31









                  FabianFabian

                  19.7k3674




                  19.7k3674























                      0












                      $begingroup$

                      The short answer is yes. One way is to take the formal determinant
                      $$left|begin{matrix}e_1&e_2&e_3&e_4\
                      a_1&a_2&a_3&a_4\
                      b_1&b_2&b_3&b_4\
                      c_1&c_2&c_3&c_4\
                      end{matrix}right|$$
                      where $e_1,ldots,e_4$ are the standard unit vectors, and $a=sum a_ie_i$
                      etc., are the three vectors.



                      Or you can rephrase this in terms
                      of exterior powers and the Hodge star operator.



                      All this works in $n$ dimensions too.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        The short answer is yes. One way is to take the formal determinant
                        $$left|begin{matrix}e_1&e_2&e_3&e_4\
                        a_1&a_2&a_3&a_4\
                        b_1&b_2&b_3&b_4\
                        c_1&c_2&c_3&c_4\
                        end{matrix}right|$$
                        where $e_1,ldots,e_4$ are the standard unit vectors, and $a=sum a_ie_i$
                        etc., are the three vectors.



                        Or you can rephrase this in terms
                        of exterior powers and the Hodge star operator.



                        All this works in $n$ dimensions too.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The short answer is yes. One way is to take the formal determinant
                          $$left|begin{matrix}e_1&e_2&e_3&e_4\
                          a_1&a_2&a_3&a_4\
                          b_1&b_2&b_3&b_4\
                          c_1&c_2&c_3&c_4\
                          end{matrix}right|$$
                          where $e_1,ldots,e_4$ are the standard unit vectors, and $a=sum a_ie_i$
                          etc., are the three vectors.



                          Or you can rephrase this in terms
                          of exterior powers and the Hodge star operator.



                          All this works in $n$ dimensions too.






                          share|cite|improve this answer









                          $endgroup$



                          The short answer is yes. One way is to take the formal determinant
                          $$left|begin{matrix}e_1&e_2&e_3&e_4\
                          a_1&a_2&a_3&a_4\
                          b_1&b_2&b_3&b_4\
                          c_1&c_2&c_3&c_4\
                          end{matrix}right|$$
                          where $e_1,ldots,e_4$ are the standard unit vectors, and $a=sum a_ie_i$
                          etc., are the three vectors.



                          Or you can rephrase this in terms
                          of exterior powers and the Hodge star operator.



                          All this works in $n$ dimensions too.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 2 '18 at 8:32









                          Lord Shark the UnknownLord Shark the Unknown

                          103k1160132




                          103k1160132






























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