- #1

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We all know and love the determinant form of the cross product:

[tex]\bold{a}\times\bold{b} = \left|\begin{array}{ccc}

\bold{i} & \bold{j} & \bold{k} \\

a_x & a_y & a_z \\

b_x & b_y & b_z\end{array}\right|[/tex]

and can calculate this by going through the motions but when I try to think of

[tex]\bold{a} \times \bold{b} \ = \ (a_x \hat{i} \ + \ a_y \hat{j} \ + \ a_z \hat{k}) \times (b_x \hat{i} \ + \ b_y \hat{j} \ + \ b_z \hat{k}) [/tex]

[tex] a_x b_x (\hat{i} \times \hat{i}) \ + \ a_x b_y (\hat{i} \times \hat{j}) \ + \ a_x b_z (\hat{i} \times \hat{k}) \ + \ a_y b_x (\hat{j} \times \hat{i}) \ + \ a_y b_y (\hat{j} \times \hat{j}) \ + \ a_y b_z (\hat{j} \times \hat{k}) \ + \ a_z b_x (\hat{k} \times \hat{i}) \ + \ a_z b_y (\hat{k} \times \hat{j}) \ + \ a_z b_z (\hat{k} \times \hat{k})[/tex]

I have it memorized how to calculate all this fine, but I don't know

Thinking about the inner product I can understand why [tex] \hat{i} \cdot \hat{j} \ = \ 0 [/tex] because they represent different dimensions but the cross product makes no sense.

Serge Lang just

Furthermore, Marsden & Tromba axiomatically define it as following from the determinant

but again that is not satisfying at all, I don't understand

I can see reason in using the determinant as a shortcut way of calculating the cross

product as just following from calculating the components the way I have it laid out

above so if I can get a reason why [tex] \hat{i} \times \hat{i} \ = \ 0 [/tex] and [tex] \hat{i} \times \hat{j} \ = \ \hat{k} [/tex] then it all makes sense!

Invoking the sine function isn't justification enough for me because as I know it

the sine function arises from squaring the magnitude of the cross product ||

The minus signs in the component form arise from calculating [tex] \hat{i} \times \hat{k} \ = \ - \hat{j} [/tex] etc...

so to get to that point you need to know why this happens...

thanks

[tex]\bold{a}\times\bold{b} = \left|\begin{array}{ccc}

\bold{i} & \bold{j} & \bold{k} \\

a_x & a_y & a_z \\

b_x & b_y & b_z\end{array}\right|[/tex]

and can calculate this by going through the motions but when I try to think of

**why**this is the case I get stuck.[tex]\bold{a} \times \bold{b} \ = \ (a_x \hat{i} \ + \ a_y \hat{j} \ + \ a_z \hat{k}) \times (b_x \hat{i} \ + \ b_y \hat{j} \ + \ b_z \hat{k}) [/tex]

[tex] a_x b_x (\hat{i} \times \hat{i}) \ + \ a_x b_y (\hat{i} \times \hat{j}) \ + \ a_x b_z (\hat{i} \times \hat{k}) \ + \ a_y b_x (\hat{j} \times \hat{i}) \ + \ a_y b_y (\hat{j} \times \hat{j}) \ + \ a_y b_z (\hat{j} \times \hat{k}) \ + \ a_z b_x (\hat{k} \times \hat{i}) \ + \ a_z b_y (\hat{k} \times \hat{j}) \ + \ a_z b_z (\hat{k} \times \hat{k})[/tex]

I have it memorized how to calculate all this fine, but I don't know

**why**[tex] \hat{i} \times \hat{i} \ = \ 0 [/tex] and [tex] \hat{i} \times \hat{j} \ = \ \hat{k} [/tex]Thinking about the inner product I can understand why [tex] \hat{i} \cdot \hat{j} \ = \ 0 [/tex] because they represent different dimensions but the cross product makes no sense.

Serge Lang just

**defines**[tex]\bold{a} \times \bold{b} [/tex] as a_{y}b_{z}- a_{z}b_{y}etc... but that is hardly satisfying.Furthermore, Marsden & Tromba axiomatically define it as following from the determinant

but again that is not satisfying at all, I don't understand

**why**the determinant answers this question.I can see reason in using the determinant as a shortcut way of calculating the cross

product as just following from calculating the components the way I have it laid out

above so if I can get a reason why [tex] \hat{i} \times \hat{i} \ = \ 0 [/tex] and [tex] \hat{i} \times \hat{j} \ = \ \hat{k} [/tex] then it all makes sense!

Invoking the sine function isn't justification enough for me because as I know it

the sine function arises from squaring the magnitude of the cross product ||

**a****x****b**||² and that involves squaring the components.The minus signs in the component form arise from calculating [tex] \hat{i} \times \hat{k} \ = \ - \hat{j} [/tex] etc...

so to get to that point you need to know why this happens...

thanks

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