The force F = i - 2jacts at the point (0, 1, 2) Find the torque of F about the line $\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{i}\mathbf{-}\mathbf{j}\mathbf{)}\mathbf{t}$.

It has been given that the force is given by the vector $\mathrm{F}=\mathrm{i}-2\mathrm{j}-2\mathrm{k}$.

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.

The torque $\tau$is given by the formula.

$\tau =r\times F$

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

$$\begin{gathered} \mathrm{r}=(0,1,2)-(0,0,0) \\ =\mathrm{j}+2\mathrm{k} \end{gathered}$$

Given F = i - 2j - 2k calculate as simple cross product.

$$\begin{gathered} \mathrm{\tau }=\mathrm{r}\times \mathrm{F} \\ =\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& -2& 4\\ 1& -5& 2\end{array}\right| \\ =\left(\left(-2\right)\left(2\right)-\left(-5\right)\left(4\right)\right)\mathrm{i}-\left(\left(1\right)\left(2\right)-\left(1\right)\left(4\right)\right)\mathrm{j}+\left(\left(1\right)\left(-5\right)-\left(1\right)\left(-2\right)\right)\mathrm{k} \\ \mathrm{\tau }=-24\mathrm{i}+2\mathrm{j}-3\mathrm{k} \end{gathered}$$

Get a unit vector in the direction, let it be $r_1$of the line and calculate the component of the total torque in that direction using the dot product between $\tau$and$r_1$

Use

r(t)

localid="1659244540124" $$\begin{gathered} {\mathrm{r}}_1=\frac{\mathrm{r}\left(1\right)-\mathrm{r}\left(\mathrm{o}\right)}{\left|\mathrm{r}\left(1\right)-\mathrm{r}\left(\mathrm{o}\right)\right|} \\ =\frac{2\mathrm{i}-\mathrm{j}}{\sqrt{{\left(2-\mathrm{o}\right)}^2+{\left(-1-0\right)}^2+{\left(0-0\right)}^2}} \\ =\frac{1}{\sqrt{5}}\left(2\mathrm{i}-\mathrm{j}\right) \end{gathered}$$

So the torque about the axis ${\tau }_{r1}$is given by formula mentioned below

localid="1659244630963" role="math" $$\begin{gathered} {\mathrm{\tau }}_{\mathrm{r}1}=\mathrm{\tau }.\mathrm{r}1 \\ =\left(2\mathrm{i}+2\mathrm{j}-3\mathrm{k}\right).\left(\frac{1}{\sqrt{5}}\left(2\mathrm{i}-\mathrm{j}\right)\right) \\ =\frac{1}{\sqrt{5}}\left[\left(2\right)\left(2\right)+\left(2\right)\left(-1\right)+\left(-3\right)\left(0\right)\right] \\ {\mathrm{\tau }}_{\mathrm{r}1}=\frac{2}{\sqrt{5}} \end{gathered}$$

Hence, the torque is found by the equation mentioned below.

localid="1659244615896" $$\begin{gathered} \tau =2i+2j-3k \\ {\tau }_{r1}=\frac{2}{\sqrt{5}} \end{gathered}$$