Answer with Explanation:
We are given that
Force acts on a pebble=[tex]2\hat{i}-3\hat{k}[/tex] N
Position vector=[tex]r=0.5\hat{j}-2\hat{k}[/tex] m
a.We have to find the resulting torque on the pebble about origin.
Torque=[tex]r\times F[/tex]
Substitute the values then we get
[tex]Torque= (0.5j-2k)\times (2i-3k)[/tex]
Torque=[tex]-k-1.5i-4j[/tex]N-m
By using [tex]i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0[/tex]
b.[tex]r=2i-3k[/tex]
[tex]r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k[/tex]
Torque about point (2,0,-3)
[tex]\tau=(-2i+0.5j+k)\times (2i-3k)[/tex]
[tex]\tau=-6j-k-1.5i+2j=-1.5i-4j-k[/tex]N-m
