- #1
- 147
- 0
Hi all,
This is a beginning step in proving aXb=|a||b|sin(theta)
thank you
This is a beginning step in proving aXb=|a||b|sin(theta)
thank you
Attachments
-
4 KB Views: 946
Assuming that expression with a's is the determinant, then yes: it must be the whole thing squared.Hi all,
![]()
This is a beginning step in proving aXb=|a||b|sin(theta)
thank you