Why Does the Determinant of a Matrix Represent the Cross Product?

[thechunk]

Does anyone know where I can find the derivation of the cross product. I know how to use it and the like but I do not understand why the norm of the matrix :
<br /> \left[ \begin{array}{ccc}i &amp; j &amp; k \\n1 &amp; n2 &amp; n3 \\m1 &amp; m2 &amp; m3 \\\end{array}\right]

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

 

[quantumdude]

I wouldn't prove that the cross product can be written as a determinant, I would simply define it that way. Then I would prove that the result is perpendicular to both of the vectors in the cross product. How would I do that? For vectors \vec{A} and \vec{B} I would form the dot products \vec{A}\cdot(\vec{A}\times\vec{B}) and \vec{B}\cdot(\vec{A}\times\vec{B}) and show that they both vanish identically.

 

[amcavoy]

Also, if you want to show that your definition above is equal to ABsinθ, you can find the determinant, square it, and then rearrange and use the definition of a dot product (you will see something similar after squaring).

 

[thechunk]

Yes I can see that, but what confuses me is why do those two expressions describe/yield the vector perpendicular to n and m.

 

[quantumdude]

Because the dot product of any vector and its cross product with any other vector vanishes. The "why" is in the proof.

 

[amcavoy]

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? Or was that the goal and then later shown that the determinant did the trick?

 

[quantumdude]

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!

 

[es]

I guess it was almost literally stumbled upon. The definition stuck because it is useful.
http://www.answers.com/topic/quaternion
Scroll down to history.

 

[TD]

Yes I can see that, but what confuses me is why do those two expressions describe/yield the vector perpendicular to n and m.

If
\vec c = \vec a \times \vec b = \left| {\begin{array}{*{20}c}<br /> {\vec 1_x } &amp; {\vec 1_y } &amp; {\vec 1_z } \\<br /> {a_1 } &amp; {a_2 } &amp; {a_3 } \\<br /> {b_1 } &amp; {b_2 } &amp; {b_3 } \\<br /> \end{array}} \right|
then
\left\langle {\vec a,\vec c} \right\rangle = \left| {\begin{array}{*{20}c}<br /> {a_1 } &amp; {a_2 } &amp; {a_3 } \\<br /> {a_1 } &amp; {a_2 } &amp; {a_3 } \\<br /> {b_1 } &amp; {b_2 } &amp; {b_3 } \\<br /> \end{array}} \right| = 0 \Rightarrow \vec a \bot \vec c

In the same way, \vec b \bot \vec c follows.

 

[john fairbanks]

derivation of Vector Cross Product

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

 

[amcavoy]

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

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).