How was the vector cross product derived? - Physics Forums
Oct 17, 2005 · jxk=Ci j'xi'=-Ck' kxj=-Ci k'xk'=0 ixi=0 collecting things up again ixi=0 ixj=Ck ixk=-Cj jxi=-Ck jxj=0 jxk=Ci kxi=Cj kxj=-Ci kxk=0 So any such product is a multiple of the standard one …
What is the cross product of 5k and 3i+4j? - Physics Forums
Sep 27, 2013 · The basic rules for the cross product is that ixj= k, jxk= i, kxi= j, the cross product is "anti- symmetric" (uxv= -vxu), and linear. The fact that the cross product is linear means that …
Stuck on a few Vector homework problems - Physics Forums
Sep 2, 2019 · I'm stuck on a few Vector homework problems. I don't quite understand how to write vectors A+B and A-B for questions 1b and 2b. I tried starting with calculating the magnitude for …
Why Is the Cross Product Perpendicular to Its Constituent Vectors?
Aug 13, 2013 · This is basically by convention. ixj = k, jxk = i, kxi=j. If you reversed one sign, you would need to reverse the others. Essentially you need this so that a set of mutually …
What are these formulas? (WX=...cm^3 and Jx=....cm^4) - Physics …
Jun 13, 2020 · Jx is probably 2nd moment of area (aka moment of inertia) about x-axis , seeing as its units are cm4. and Wx is probably section modulus about x axis