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radou

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\textbf{i} & \textbf{j} & \textbf{k} \\

a1 & a2 & a3 \\

b1 & b2 & b3 \end{array} \right)[/tex], unless you were referring to something else?

(Of course,

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nicksauce

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"well that's kind of what I'm asking... is there a way to arrive at that determinant form without knowing a x b equals |a||b|sin(theta)?"

Well that's kind of the definition for the cross-product. Nevertheless, maybe this might help a bit:

Start with the following definitions for a right-handed co-ordinate system:

[tex]\hat{x}\times\hat{x} = \hat{y}\times\hat{y} = \hat{z}\times{z} = 0[/tex]

[tex]\hat{x}\times\hat{y} = -\hat{y}\times\hat{x} = \hat{z}[/tex]

[tex]\hat{y}\times\hat{z} = -\hat{z}\times\hat{y} = \hat{x}[/tex]

[tex]\hat{z}\times\hat{x} = -\hat{x}\times\hat{z} = \hat{y}[/tex]

So if you write

[tex]A\times B = (A_x\hat{x} + A_y\hat{y} + A_z\hat{z}) \times (B_x\hat{x} + B_y\hat{y} + B_z\hat{z})[/tex]

(Ie: [tex]A_y\hat{y}\times B_x\hat{x}=A_yB_x(\hat{y}\times\hat{x}) = -A_yB_x\hat{z}[/tex])

Expand, regroup, and this will lead you to the determinant form.

Well that's kind of the definition for the cross-product. Nevertheless, maybe this might help a bit:

Start with the following definitions for a right-handed co-ordinate system:

[tex]\hat{x}\times\hat{x} = \hat{y}\times\hat{y} = \hat{z}\times{z} = 0[/tex]

[tex]\hat{x}\times\hat{y} = -\hat{y}\times\hat{x} = \hat{z}[/tex]

[tex]\hat{y}\times\hat{z} = -\hat{z}\times\hat{y} = \hat{x}[/tex]

[tex]\hat{z}\times\hat{x} = -\hat{x}\times\hat{z} = \hat{y}[/tex]

So if you write

[tex]A\times B = (A_x\hat{x} + A_y\hat{y} + A_z\hat{z}) \times (B_x\hat{x} + B_y\hat{y} + B_z\hat{z})[/tex]

(Ie: [tex]A_y\hat{y}\times B_x\hat{x}=A_yB_x(\hat{y}\times\hat{x}) = -A_yB_x\hat{z}[/tex])

Expand, regroup, and this will lead you to the determinant form.

Last edited:

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HallsofIvy

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If you don't use "length of a x b= |a||b|sin(theta)" (and the fact that a x b is perpendicular to be a and b with the "right hand rule"- it is not correct that a x b= |a||b|sin(theta)!) the what definition of cross product ARE you using?

Obviously, you have to have some definition before you can derive a formula!

nicksause is using, as a definition, that [itex]\vec{i}\times \vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], and [itex]\vec{k}\times\vec{i}= \vec{j}[/itex] together with requiring that the cross product be associative, distribute over vector addition, and be anti-commutative.

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nicksauce

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Right, I should have mentioned that.nicksause is using, as a definition, that [itex]\vec{i}\times \vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], and [itex]\vec{k}\times\vec{i}= \vec{j}[/itex] together with requiring that the cross product be associative, distribute over vector addition, and be anti-commutative.

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That is the original quaternion-based definition of the cross product. Even the use of [itex]\vec{i}, \vec{j}, \vec{k}[/itex] as unit vectors comes straight from the quaternions. The determinant form is an easy mnemonic for some; I prefer the even/odd permutations of i,j,k (or whatever).nicksause is using, as a definition, that [itex]\vec{i}\times \vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], and [itex]\vec{k}\times\vec{i}= \vec{j}[/itex]

As an aside, the concept of vectors and vector spaces is a relatively recent invention (end of the 19th century). We are introduced to vectors in the first week of freshman physics and use vector-based calculations throughout. How did physicists do things, even very basic freshman-level physics things, before the invention of vectors and all that is associated with them?

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