Answer:
[tex]\vec{\tau} = 6\hat{i} + 80\hat{j}+63\hat{k}[/tex]
Explanation:
given,
F=(13)i + (-12)j + 14(k)
position vector of vector A = r = 2 i + 3 j - 4 k
moment of force about origin
[tex]\vec{\tau} = \vec{F} \times \vec{r}[/tex]
[tex]\vec{\tau} = \begin{bmatrix}\hat{i} &\hat{j}&\hat{j} \\ 13 & -12 &14 \\2& 3 & -4\end{bmatrix}[/tex]
[tex]\vec{\tau} = 6\hat{i} + 80\hat{j}+63\hat{k}[/tex]
