Answer:
The torque about the origin is [tex]2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k} [/tex]
Explanation:
Torque [tex] \overrightarrow{\tau}[/tex] is the cross product between force [tex]\overrightarrow{F} [/tex] and vector position [tex] \overrightarrow{r} [/tex] respect a fixed point (in our case the origin):
[tex]\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F} [/tex]
There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:
[tex] \overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right] [/tex]
[tex]\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k}) [/tex]
[tex]\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k}) [/tex]
[tex] \overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau} [/tex]
