- #1

thechunk

- 11

- 0

[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'.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter thechunk
- Start date

- #1

thechunk

- 11

- 0

[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'.

- #2

quantumdude

Staff Emeritus

Science Advisor

Gold Member

- 5,575

- 23

- #3

amcavoy

- 665

- 0

- #4

thechunk

- 11

- 0

- #5

quantumdude

Staff Emeritus

Science Advisor

Gold Member

- 5,575

- 23

- #6

amcavoy

- 665

- 0

- #7

quantumdude

Staff Emeritus

Science Advisor

Gold Member

- 5,575

- 23

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!

- #8

es

- 70

- 0

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

Scroll down to history.

- #9

TD

Homework Helper

- 1,022

- 0

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.

- #10

john fairbanks

- 8

- 0

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

- #11

amcavoy

- 665

- 0

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

Share:

- Replies
- 7

- Views
- 511

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 459

- Replies
- 47

- Views
- 2K

- Replies
- 16

- Views
- 509

- Replies
- 9

- Views
- 559

- Replies
- 1

- Views
- 217

- Last Post

- Replies
- 3

- Views
- 576

MHB
Find the Product

- Last Post

- Replies
- 3

- Views
- 544

- Last Post

- Replies
- 4

- Views
- 587