2005年10月17日 · 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 in which C is chosen C=1 giving ixi=0 ixj=k ixk=-j jxi=-k jxj=0 jxk=i kxi=j kxj=-i kxk=0 Similar reasoning applies to higher products a.bxc=axb.c is the only ...

2013年9月27日 · 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 5k x(3i+ 4j)= 15 kxi+ 20 kxj.

2019年9月2日 · 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 vector A+B on question 1b and then followed by finding theta, but I'm not sure if that's what I'm supposed to do...

2013年8月13日 · 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 perpendicular vectors, defined with cross product, would not change when rigidly rotated.

2020年6月13日 · 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