Question

Find the torque about the point (1, -2, 1) due to the forceF = 2 i - j + 3 kacting at the point ( 1, 1, -3)

Short answer

The torque is found to be $\mathrm{\tau }=5\mathrm{i}-8\mathrm{j}-6\mathrm{k}$.

Step-by-step solution

Given Information.

It has been given that the point is (1, -2, 1) and ( 1, 1, -3) and the force is F = 2 i - j + 3 k

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

Use the formula for torque.

The torque is given by the formula.

$\tau =r\times F$

Here is the vector pointing from the point of question to the point at which the force is acting.

$\mathrm{r}=(1,1,-3)-(1,-2,1)=3\mathrm{j}-4\mathrm{k}$

Given $\mathrm{F}=2\mathrm{i}-\mathrm{j}+3\mathrm{k}$, calculate as simple cross product.

$$\begin{gathered} \mathrm{\tau }=\mathrm{r}\times \mathrm{F}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 0& 3& -4\\ 2& -1& 3\end{array}\right| \\ \mathrm{\tau }=\left(\right(3\left)\right(3)-(-4\left)\right(-1\left)\right)\mathrm{i}-\left(\right(0\left)\right(3)-(-4\left)\right(2\left)\right)\mathrm{j}+\left(\right(0\left)\right(-1)-(3\left)\right(2\left)\right)\mathrm{k} \\ \mathrm{\tau }=5\mathrm{i}-8\mathrm{j}-6\mathrm{k} \end{gathered}$$

Hence, the torque is found to be $\mathrm{\tau }=5\mathrm{i}-8\mathrm{j}-6\mathrm{k}$.