Question

a body is in equilibrium under the action of three force vectors a,b & c simultaneously. show that a cross b = b cross c=c cross a.

Answer 1

One of the multiple ways to do it analytically:

Let be,
\[a=(a_x,a_y,a_z)\\
b=(b_x,b_y,b_z)\\
c=(c_x,c_y,c_z)\]
If the body is in equilibrim then,
\[a+b+c=0\Rightarrow\\
a_x+b_x+c_x=0\\
a_y+b_y+c_y=0\\
a_z+b_z+c_z=0\] In particular,
\[a_x+c_x=-b_x\\
a_y+c_y=-b_y\\
a_z+c_z=-b_z \Rightarrow a+c=-b\] Also
\[a\times b=b\times c\Rightarrow (a+c)\times b=0\] But we have shown that
\[(a+c)\times b=-b\times b\] But b is parallel to itself, so the product is 0. And it is demostrated. It cab done the same way for the others.

Answer 2

Well, I do a useless things, but I think the reasoning is correct.

Answer 3

thanks a lot John_ES.

Answer 4

thanks for viewing cherry17ann.