Question

Find the torque of a force $ 7\vec{i}+3\vec{j}-5\vec{k} $ about the origin. The force acts on the particle whose position vector is $ \vec{i}+\vec{j}+\vec{k} $

Answer 1

Hint: Torque can be calculated by cross product of position vector and force. In terms of determinant, we can find the cross product using formula
 $ {a}\times {b=}\left| \begin{align}
  & {i j k} \\
 & {{{a}}_{1}}{ }{{{a}}_{2}}{ }{{{a}}_{3}} \\
 & {{{b}}_{1}}{ }{{{b}}_{2}}{ }{{{b}}_{3}} \\
\end{align} \right| $
This determinant is computed using cofactor expansion. It expands to
 ${a}\times {b} =\left( {{{a}}_{2}}{{{b}}_{3}}-{{{a}}_{3}}{{{b}}_{2}} \right){i}-\left( {{{a}}_{1}}{{{b}}_{3}}-{{{a}}_{3}}{{{b}}_{1}} \right){j}+\left( {{{a}}_{1}}{{{b}}_{2}}-{{{a}}_{2}}{{{b}}_{1}} \right){k} $
Formula used: $ \vec{\tau }=\vec{r}\times \vec{F} $ Torque is given by cross product of perpendicular distance and force.

Complete Step by step solution
Given:
 $ F=7\vec{i}+3\vec{j}-5\vec{k} $
 $ \vec{r}=\vec{i}+\vec{j}+\vec{k} $ $ $
We know that torque $ \vec{\tau } $ is given by:
 $ \vec{\tau }=\vec{r}\times \vec{F} $
 $ =\left( \begin{matrix}
   {\vec{i}} & {\vec{j}} & {\vec{k}} \\
   1 & 1 & 1 \\
   7 & 3 & -5 \\
\end{matrix} \right) $
     $ \begin{align}
  & =\vec{i}\left[ -5-3 \right]-\vec{j}\left[ -5-7 \right]+\vec{k}\left[ 3-7 \right] \\
 & =-8\vec{i}+12\vec{j}-4\vec{k} \\
\end{align} $

Additional Information
Torque is the rotational equivalent of linear force. It's S.I. unit is $ {N m} $ (Newton meter). In S.I. base unit, its unit is $ {kg }{{{m}}^{2}}{{{s}}^{-2}} $ . $ $
Its dimension is given by $ {M}{{{L}}^{2}}{{{T}}^{-2}} $ .
The magnitude of the torque of a rigid body depends up on three factors:
Force applied
Lever arm vector connecting the point about which the torque is being measured to the point of force application.
Angle between force and lever arm vectors.
The net torque on a body determines the rate of change of the body's angular momentum.
 $ \overrightarrow{{ }\!\!\tau\!\!{ }}=\overrightarrow{\frac{\alpha L}{\alpha t}} $
Where,
  $ \alpha = $ angular momentum vector
 t = Time
For motion of a point particle,
 $ {L = I}\omega $
Where, I = moment of inertia
 $ \omega { =} $ Orbital angular velocity pseudo vector.

Note
The student should know how to calculate cross product.
Torque is the measure of the force that can cause an object to rotate about an axis.
Torques cause an object to acquire angular acceleration.
It is a vector quantity.
Torque=force applied $ \times $ perpendicular distance from outermost layer to the canter.
 $ {\vec{ \tau}}={\vec{r}}\times {\vec{F}} $