Using Cross Product to Find Torque

[GingerBread27]

Force F = (4.0 N) i + (-2.0 N) k acts on a pebble with position vector r = (2.50 m) j + (-1.3 m) k, relative to the origin

What is the resulting torque acting on the pebble about a point with coordinates (2.0 m, 3.0 m, 2.0 m)?

Ok so I just completely forgot how to work with vectors when they are not acting about the origin so can anyone lead me in the right direction?

[arildno]

The force acts on 2.5j-1-3k respective to your origin.
Find the representation of this point when translating your origin to (2,3,2).

[HobieDude16]

pretty easy, just take the position crss with the force....
(0i+2.5J-1.3K)(4i+0j-2k)
and then, do the first term of hte first equation times all 3 of the 2nd, then the 2nd times all 3, then the 3rd.... and you have to remember, an i times an i or a j times a j, etc, is zero... and whichever 2 variables you use, the answer of the multiplication is the variable left out, so i times j is k... and it goes in a triangle like this....
i
k j
and you go from whatever the first one is, times the 2nd one, whatever is across is the letter you use, and if it goes clockwise(i.e. k times i), the vector doesnt affect the answer, if its counter clockwise (i.e. j times i) its negative of whatever the answer is.... send me an im at hobiedude16 if you want any help on the other problems, were just about done

[Parth Dave]

Relative to (0, 0, 0) the pebbles position is (0, 2.5, -1.3). So what is it's position relative to (2, 3, 2)?

[HobieDude16]

haha, my bad, didnt notice you wanted part b of the question
just do vector r minus coordinates and use that instead of vector r, my bad on that one... but same procedure