Let F = i - 5j + 2kact at the point (2, 1, 0) Find the torque of F about the line -(3j + 4k)- 2it .

It has been given that the force is given by the 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.

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

$\mathrm{\tau }=\mathrm{r}\times \mathrm{F}$

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

r(0)=3j - 4k

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

$$\begin{gathered} \mathrm{r}=(2,1,0)-(0,3,-4) \\ =2\mathrm{i}-2\mathrm{j}+4\mathrm{k} \end{gathered}$$

Given F = i -5j + 2k calculate $\tau$as simple cross product.

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

Get a unit vector in the direction, let it be ${\mathrm{r}}_1$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)

$$\begin{gathered} {\mathrm{r}}_1=\frac{\mathrm{r}\left(1\right)-\mathrm{r}\left(\mathrm{o}\right)}{\left|\mathrm{r}\left(1\right)-\mathrm{r}\left(\mathrm{o}\right)\right|} \\ =\frac{-2\mathrm{i}+3\mathrm{j}-4\mathrm{k}-3\mathrm{j}+4\mathrm{k}}{\sqrt{{\left(-2-0\right)}^2+{\left(3-3\right)}^2+{\left(-4+4\right)}^2}} \\ =-\mathrm{i} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{\tau }}_{\mathrm{r}1}=\mathrm{\tau }.\mathrm{r}1 \\ =\left(-24\mathrm{i}+2\mathrm{j}-3\mathrm{k}\right).\left(-i\right) \\ =\left[\left(-24\right)\left(-1\right)+\left(2\right)\left(0\right)+\left(-3\right)\left(0\right)\right] \\ {\mathrm{\tau }}_{\mathrm{r}1}=24 \end{gathered}$$

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

$$\begin{gathered} \mathrm{\tau }=-24\mathrm{i}+2\mathrm{j}-3\mathrm{k} \\ {\mathrm{\tau }}_{\mathrm{r}1}=24 \end{gathered}$$.