Can't we factor out a constant in the cross product?












0














I have the vectors $A=ahat e_x$ and $B=ahat e_y$, so
$$
Atimes B =
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix}=hat za^2
$$

Q1: But why is the following wrong
begin{align}
Atimes B &=ahat e_xtimes ahat e_y\
&=(ahat e_x)times (ahat e_y) tag 1\
&=(ahat e_x)times (hat e_ya) tag 2\
&=a(hat e_x)times (hat e_y)a tag 3\
&=a((hat e_x)times (hat e_y)) tag 4\
&=a(hat e_xtimes hat e_y) tag 5\
&=ahat e_z quad ?
end{align}



Q2: Also, why is this wrong
begin{align}
Atimes B &=
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} \
&=
a
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}\
&=ahat z quad ?
end{align}










share|cite|improve this question






















  • They are wrong because the cross-product isn't defined that way.
    – Bernard Massé
    Dec 4 '18 at 11:00






  • 1




    You can take out a common factor of a single row in a determinant. Here you take out the common factor from two rows. That's two factors, each equal to $a$. Therefore you get $a^2$ altogether.
    – Jyrki Lahtonen
    Dec 4 '18 at 11:06










  • @Bernard Massé Why do you say that ? it is a form that is used rather usually in physics. Have a look at physics.stackexchange.com/q/133311
    – Jean Marie
    Dec 4 '18 at 18:11










  • @ Jean Marie. What I meant was the cross-product is defined in such a way (be it done by determinants or another way) that product by a constant is not distributive over it.
    – Bernard Massé
    Dec 5 '18 at 16:20
















0














I have the vectors $A=ahat e_x$ and $B=ahat e_y$, so
$$
Atimes B =
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix}=hat za^2
$$

Q1: But why is the following wrong
begin{align}
Atimes B &=ahat e_xtimes ahat e_y\
&=(ahat e_x)times (ahat e_y) tag 1\
&=(ahat e_x)times (hat e_ya) tag 2\
&=a(hat e_x)times (hat e_y)a tag 3\
&=a((hat e_x)times (hat e_y)) tag 4\
&=a(hat e_xtimes hat e_y) tag 5\
&=ahat e_z quad ?
end{align}



Q2: Also, why is this wrong
begin{align}
Atimes B &=
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} \
&=
a
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}\
&=ahat z quad ?
end{align}










share|cite|improve this question






















  • They are wrong because the cross-product isn't defined that way.
    – Bernard Massé
    Dec 4 '18 at 11:00






  • 1




    You can take out a common factor of a single row in a determinant. Here you take out the common factor from two rows. That's two factors, each equal to $a$. Therefore you get $a^2$ altogether.
    – Jyrki Lahtonen
    Dec 4 '18 at 11:06










  • @Bernard Massé Why do you say that ? it is a form that is used rather usually in physics. Have a look at physics.stackexchange.com/q/133311
    – Jean Marie
    Dec 4 '18 at 18:11










  • @ Jean Marie. What I meant was the cross-product is defined in such a way (be it done by determinants or another way) that product by a constant is not distributive over it.
    – Bernard Massé
    Dec 5 '18 at 16:20














0












0








0







I have the vectors $A=ahat e_x$ and $B=ahat e_y$, so
$$
Atimes B =
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix}=hat za^2
$$

Q1: But why is the following wrong
begin{align}
Atimes B &=ahat e_xtimes ahat e_y\
&=(ahat e_x)times (ahat e_y) tag 1\
&=(ahat e_x)times (hat e_ya) tag 2\
&=a(hat e_x)times (hat e_y)a tag 3\
&=a((hat e_x)times (hat e_y)) tag 4\
&=a(hat e_xtimes hat e_y) tag 5\
&=ahat e_z quad ?
end{align}



Q2: Also, why is this wrong
begin{align}
Atimes B &=
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} \
&=
a
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}\
&=ahat z quad ?
end{align}










share|cite|improve this question













I have the vectors $A=ahat e_x$ and $B=ahat e_y$, so
$$
Atimes B =
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix}=hat za^2
$$

Q1: But why is the following wrong
begin{align}
Atimes B &=ahat e_xtimes ahat e_y\
&=(ahat e_x)times (ahat e_y) tag 1\
&=(ahat e_x)times (hat e_ya) tag 2\
&=a(hat e_x)times (hat e_y)a tag 3\
&=a((hat e_x)times (hat e_y)) tag 4\
&=a(hat e_xtimes hat e_y) tag 5\
&=ahat e_z quad ?
end{align}



Q2: Also, why is this wrong
begin{align}
Atimes B &=
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} \
&=
a
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}\
&=ahat z quad ?
end{align}







linear-algebra vectors cross-product






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asked Dec 4 '18 at 10:57









Donsert

1339




1339












  • They are wrong because the cross-product isn't defined that way.
    – Bernard Massé
    Dec 4 '18 at 11:00






  • 1




    You can take out a common factor of a single row in a determinant. Here you take out the common factor from two rows. That's two factors, each equal to $a$. Therefore you get $a^2$ altogether.
    – Jyrki Lahtonen
    Dec 4 '18 at 11:06










  • @Bernard Massé Why do you say that ? it is a form that is used rather usually in physics. Have a look at physics.stackexchange.com/q/133311
    – Jean Marie
    Dec 4 '18 at 18:11










  • @ Jean Marie. What I meant was the cross-product is defined in such a way (be it done by determinants or another way) that product by a constant is not distributive over it.
    – Bernard Massé
    Dec 5 '18 at 16:20


















  • They are wrong because the cross-product isn't defined that way.
    – Bernard Massé
    Dec 4 '18 at 11:00






  • 1




    You can take out a common factor of a single row in a determinant. Here you take out the common factor from two rows. That's two factors, each equal to $a$. Therefore you get $a^2$ altogether.
    – Jyrki Lahtonen
    Dec 4 '18 at 11:06










  • @Bernard Massé Why do you say that ? it is a form that is used rather usually in physics. Have a look at physics.stackexchange.com/q/133311
    – Jean Marie
    Dec 4 '18 at 18:11










  • @ Jean Marie. What I meant was the cross-product is defined in such a way (be it done by determinants or another way) that product by a constant is not distributive over it.
    – Bernard Massé
    Dec 5 '18 at 16:20
















They are wrong because the cross-product isn't defined that way.
– Bernard Massé
Dec 4 '18 at 11:00




They are wrong because the cross-product isn't defined that way.
– Bernard Massé
Dec 4 '18 at 11:00




1




1




You can take out a common factor of a single row in a determinant. Here you take out the common factor from two rows. That's two factors, each equal to $a$. Therefore you get $a^2$ altogether.
– Jyrki Lahtonen
Dec 4 '18 at 11:06




You can take out a common factor of a single row in a determinant. Here you take out the common factor from two rows. That's two factors, each equal to $a$. Therefore you get $a^2$ altogether.
– Jyrki Lahtonen
Dec 4 '18 at 11:06












@Bernard Massé Why do you say that ? it is a form that is used rather usually in physics. Have a look at physics.stackexchange.com/q/133311
– Jean Marie
Dec 4 '18 at 18:11




@Bernard Massé Why do you say that ? it is a form that is used rather usually in physics. Have a look at physics.stackexchange.com/q/133311
– Jean Marie
Dec 4 '18 at 18:11












@ Jean Marie. What I meant was the cross-product is defined in such a way (be it done by determinants or another way) that product by a constant is not distributive over it.
– Bernard Massé
Dec 5 '18 at 16:20




@ Jean Marie. What I meant was the cross-product is defined in such a way (be it done by determinants or another way) that product by a constant is not distributive over it.
– Bernard Massé
Dec 5 '18 at 16:20










3 Answers
3






active

oldest

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0















$$=a(hat e_x)times (hat e_y)a tag 3$$
$$=a((hat e_x)times (hat e_y)) tag 4$$




What happens here...?



You may be mixing it up with the distributive property of the cross product over addition.




Q2: Also, why is this wrong
begin{align}
Atimes B &=
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
a & 0 & 0\
0 & a &0
end{vmatrix} \
&=
a
begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\ tag 6
0 & 1 &0
end{vmatrix}\
&=ahat z quad ?
end{align}




For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example:
$$begin{vmatrix}
a & b & c \
color{purple}{k}d& color{purple}{k}e & color{purple}{k}f \
g & h & i
end{vmatrix}=color{purple}{k}begin{vmatrix}
a & b & c \
d& e & f \
g & h & i
end{vmatrix}$$

This means that if you want to factor out the $a$, you do that for the second and the third row:
$$begin{vmatrix}
hat e_x &hat e_y & hat e_z \
color{blue}{a}& 0 & 0\
0 & color{red}{a} &0
end{vmatrix}=color{blue}{a}color{red}{a}begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\
0 & 1 &0
end{vmatrix}=a^2begin{vmatrix}
hat e_x &hat e_y & hat e_z \
1 & 0 & 0\
0 & 1 &0
end{vmatrix}$$






share|cite|improve this answer





























    0














    We have that by cross product definition



    $$Atimes B =ahat e_xtimes ahat e_y=a^2( e_xtimes hat e_y)$$



    and by the properties of the determinant



    $$Atimes B =
    begin{vmatrix}
    hat e_x &hat e_y & hat e_z \
    a & 0 & 0\
    0 & a &0
    end{vmatrix} =
    a^2
    begin{vmatrix}
    hat e_x &hat e_y & hat e_z \
    1 & 0 & 0\ tag 6
    0 & 1 &0
    end{vmatrix}$$






    share|cite|improve this answer





























      0














      Your first argument is wrong because you lost an $a$ between lines (3) and (4).



      Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
        3






        active

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        active

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        0















        $$=a(hat e_x)times (hat e_y)a tag 3$$
        $$=a((hat e_x)times (hat e_y)) tag 4$$




        What happens here...?



        You may be mixing it up with the distributive property of the cross product over addition.




        Q2: Also, why is this wrong
        begin{align}
        Atimes B &=
        begin{vmatrix}
        hat e_x &hat e_y & hat e_z \
        a & 0 & 0\
        0 & a &0
        end{vmatrix} \
        &=
        a
        begin{vmatrix}
        hat e_x &hat e_y & hat e_z \
        1 & 0 & 0\ tag 6
        0 & 1 &0
        end{vmatrix}\
        &=ahat z quad ?
        end{align}




        For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example:
        $$begin{vmatrix}
        a & b & c \
        color{purple}{k}d& color{purple}{k}e & color{purple}{k}f \
        g & h & i
        end{vmatrix}=color{purple}{k}begin{vmatrix}
        a & b & c \
        d& e & f \
        g & h & i
        end{vmatrix}$$

        This means that if you want to factor out the $a$, you do that for the second and the third row:
        $$begin{vmatrix}
        hat e_x &hat e_y & hat e_z \
        color{blue}{a}& 0 & 0\
        0 & color{red}{a} &0
        end{vmatrix}=color{blue}{a}color{red}{a}begin{vmatrix}
        hat e_x &hat e_y & hat e_z \
        1 & 0 & 0\
        0 & 1 &0
        end{vmatrix}=a^2begin{vmatrix}
        hat e_x &hat e_y & hat e_z \
        1 & 0 & 0\
        0 & 1 &0
        end{vmatrix}$$






        share|cite|improve this answer


























          0















          $$=a(hat e_x)times (hat e_y)a tag 3$$
          $$=a((hat e_x)times (hat e_y)) tag 4$$




          What happens here...?



          You may be mixing it up with the distributive property of the cross product over addition.




          Q2: Also, why is this wrong
          begin{align}
          Atimes B &=
          begin{vmatrix}
          hat e_x &hat e_y & hat e_z \
          a & 0 & 0\
          0 & a &0
          end{vmatrix} \
          &=
          a
          begin{vmatrix}
          hat e_x &hat e_y & hat e_z \
          1 & 0 & 0\ tag 6
          0 & 1 &0
          end{vmatrix}\
          &=ahat z quad ?
          end{align}




          For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example:
          $$begin{vmatrix}
          a & b & c \
          color{purple}{k}d& color{purple}{k}e & color{purple}{k}f \
          g & h & i
          end{vmatrix}=color{purple}{k}begin{vmatrix}
          a & b & c \
          d& e & f \
          g & h & i
          end{vmatrix}$$

          This means that if you want to factor out the $a$, you do that for the second and the third row:
          $$begin{vmatrix}
          hat e_x &hat e_y & hat e_z \
          color{blue}{a}& 0 & 0\
          0 & color{red}{a} &0
          end{vmatrix}=color{blue}{a}color{red}{a}begin{vmatrix}
          hat e_x &hat e_y & hat e_z \
          1 & 0 & 0\
          0 & 1 &0
          end{vmatrix}=a^2begin{vmatrix}
          hat e_x &hat e_y & hat e_z \
          1 & 0 & 0\
          0 & 1 &0
          end{vmatrix}$$






          share|cite|improve this answer
























            0












            0








            0







            $$=a(hat e_x)times (hat e_y)a tag 3$$
            $$=a((hat e_x)times (hat e_y)) tag 4$$




            What happens here...?



            You may be mixing it up with the distributive property of the cross product over addition.




            Q2: Also, why is this wrong
            begin{align}
            Atimes B &=
            begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            a & 0 & 0\
            0 & a &0
            end{vmatrix} \
            &=
            a
            begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            1 & 0 & 0\ tag 6
            0 & 1 &0
            end{vmatrix}\
            &=ahat z quad ?
            end{align}




            For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example:
            $$begin{vmatrix}
            a & b & c \
            color{purple}{k}d& color{purple}{k}e & color{purple}{k}f \
            g & h & i
            end{vmatrix}=color{purple}{k}begin{vmatrix}
            a & b & c \
            d& e & f \
            g & h & i
            end{vmatrix}$$

            This means that if you want to factor out the $a$, you do that for the second and the third row:
            $$begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            color{blue}{a}& 0 & 0\
            0 & color{red}{a} &0
            end{vmatrix}=color{blue}{a}color{red}{a}begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            1 & 0 & 0\
            0 & 1 &0
            end{vmatrix}=a^2begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            1 & 0 & 0\
            0 & 1 &0
            end{vmatrix}$$






            share|cite|improve this answer













            $$=a(hat e_x)times (hat e_y)a tag 3$$
            $$=a((hat e_x)times (hat e_y)) tag 4$$




            What happens here...?



            You may be mixing it up with the distributive property of the cross product over addition.




            Q2: Also, why is this wrong
            begin{align}
            Atimes B &=
            begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            a & 0 & 0\
            0 & a &0
            end{vmatrix} \
            &=
            a
            begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            1 & 0 & 0\ tag 6
            0 & 1 &0
            end{vmatrix}\
            &=ahat z quad ?
            end{align}




            For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example:
            $$begin{vmatrix}
            a & b & c \
            color{purple}{k}d& color{purple}{k}e & color{purple}{k}f \
            g & h & i
            end{vmatrix}=color{purple}{k}begin{vmatrix}
            a & b & c \
            d& e & f \
            g & h & i
            end{vmatrix}$$

            This means that if you want to factor out the $a$, you do that for the second and the third row:
            $$begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            color{blue}{a}& 0 & 0\
            0 & color{red}{a} &0
            end{vmatrix}=color{blue}{a}color{red}{a}begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            1 & 0 & 0\
            0 & 1 &0
            end{vmatrix}=a^2begin{vmatrix}
            hat e_x &hat e_y & hat e_z \
            1 & 0 & 0\
            0 & 1 &0
            end{vmatrix}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 11:05









            StackTD

            22.3k2049




            22.3k2049























                0














                We have that by cross product definition



                $$Atimes B =ahat e_xtimes ahat e_y=a^2( e_xtimes hat e_y)$$



                and by the properties of the determinant



                $$Atimes B =
                begin{vmatrix}
                hat e_x &hat e_y & hat e_z \
                a & 0 & 0\
                0 & a &0
                end{vmatrix} =
                a^2
                begin{vmatrix}
                hat e_x &hat e_y & hat e_z \
                1 & 0 & 0\ tag 6
                0 & 1 &0
                end{vmatrix}$$






                share|cite|improve this answer


























                  0














                  We have that by cross product definition



                  $$Atimes B =ahat e_xtimes ahat e_y=a^2( e_xtimes hat e_y)$$



                  and by the properties of the determinant



                  $$Atimes B =
                  begin{vmatrix}
                  hat e_x &hat e_y & hat e_z \
                  a & 0 & 0\
                  0 & a &0
                  end{vmatrix} =
                  a^2
                  begin{vmatrix}
                  hat e_x &hat e_y & hat e_z \
                  1 & 0 & 0\ tag 6
                  0 & 1 &0
                  end{vmatrix}$$






                  share|cite|improve this answer
























                    0












                    0








                    0






                    We have that by cross product definition



                    $$Atimes B =ahat e_xtimes ahat e_y=a^2( e_xtimes hat e_y)$$



                    and by the properties of the determinant



                    $$Atimes B =
                    begin{vmatrix}
                    hat e_x &hat e_y & hat e_z \
                    a & 0 & 0\
                    0 & a &0
                    end{vmatrix} =
                    a^2
                    begin{vmatrix}
                    hat e_x &hat e_y & hat e_z \
                    1 & 0 & 0\ tag 6
                    0 & 1 &0
                    end{vmatrix}$$






                    share|cite|improve this answer












                    We have that by cross product definition



                    $$Atimes B =ahat e_xtimes ahat e_y=a^2( e_xtimes hat e_y)$$



                    and by the properties of the determinant



                    $$Atimes B =
                    begin{vmatrix}
                    hat e_x &hat e_y & hat e_z \
                    a & 0 & 0\
                    0 & a &0
                    end{vmatrix} =
                    a^2
                    begin{vmatrix}
                    hat e_x &hat e_y & hat e_z \
                    1 & 0 & 0\ tag 6
                    0 & 1 &0
                    end{vmatrix}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 4 '18 at 11:00









                    gimusi

                    1




                    1























                        0














                        Your first argument is wrong because you lost an $a$ between lines (3) and (4).



                        Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.






                        share|cite|improve this answer


























                          0














                          Your first argument is wrong because you lost an $a$ between lines (3) and (4).



                          Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            Your first argument is wrong because you lost an $a$ between lines (3) and (4).



                            Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.






                            share|cite|improve this answer












                            Your first argument is wrong because you lost an $a$ between lines (3) and (4).



                            Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 4 '18 at 11:02









                            bubba

                            30.1k33086




                            30.1k33086






























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