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[tex]

\left[ \begin{array}{ccc}i & j & k \\n1 & n2 & n3 \\m1 & m2 & m3 \\\end{array}\right] [/tex]

yields the vector perpendicular to 'n' and 'm'.

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- Thread starter thechunk
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[tex]

\left[ \begin{array}{ccc}i & j & k \\n1 & n2 & n3 \\m1 & m2 & m3 \\\end{array}\right] [/tex]

yields the vector perpendicular to 'n' and 'm'.

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Tom Mattson

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Tom Mattson

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Tom Mattson

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apmcavoy said:Just out of curiosity: was this stumbled upon just like you said? (defining the cross product as a determinant) and then later shown that the resulting vector orthogonal to both?

I don't know, but it sure seems easier than doing it the other way around!

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http://www.answers.com/topic/quaternion

Scroll down to history.

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TD

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thechunk said:

If

[tex]\vec c = \vec a \times \vec b = \left| {\begin{array}{*{20}c}

{\vec 1_x } & {\vec 1_y } & {\vec 1_z } \\

{a_1 } & {a_2 } & {a_3 } \\

{b_1 } & {b_2 } & {b_3 } \\

\end{array}} \right|[/tex]

then

[tex]\left\langle {\vec a,\vec c} \right\rangle = \left| {\begin{array}{*{20}c}

{a_1 } & {a_2 } & {a_3 } \\

{a_1 } & {a_2 } & {a_3 } \\

{b_1 } & {b_2 } & {b_3 } \\

\end{array}} \right| = 0 \Rightarrow \vec a \bot \vec c[/tex]

In the same way, [itex]\vec b \bot \vec c[/itex] follows.

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Using determinants to describe cross products I think was discovered by a mathamatician named William Rowan Hamiliton. He came up with the algebraic forms in dot and cross products, but I have no idea how he did it. Does anybody know. I just memorize the cross product formula but don't know where it comes from. Thanks

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Have you taken linear algebra? You learn a lot about how determinants equal the volume of the parallelopiped made by the three vectors (or other dimensions).john fairbanks said:Using determinants to describe cross products I think was discovered by a mathamatician named William Rowan Hamiliton. He came up with the algebraic forms in dot and cross products, but I have no idea how he did it. Does anybody know. I just memorize the cross product formula but don't know where it comes from. Thanks

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