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# Moment about the point $\overrightarrow i + 2\overrightarrow j - \overrightarrow k$of a force represented by $\overrightarrow i + 2\overrightarrow j + \overrightarrow k$ acting through the point $2\overrightarrow i + 3\overrightarrow j + \overrightarrow k$ is(A) $3\overrightarrow i + \overrightarrow j - \overrightarrow k$(B) $3\overrightarrow i - \overrightarrow j + \overrightarrow k$(C) $- 3\overrightarrow i + \overrightarrow j + \overrightarrow k$(D) $3\overrightarrow i + \overrightarrow j + \overrightarrow k$

Last updated date: 26th Feb 2024
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Hint: When a force is applied on a point lying on a line passing through another point then the force produces a torque on another point which lies on the line passing through the point of application of force. This torque is also known as the moment.
Formula used
$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$
$\overrightarrow \tau$ is the torque(moment of force)s, $\overrightarrow r$is distance vector, $\overrightarrow F$ is the force vector.

Complete Step-by-step solution
$\overrightarrow r$is the distance vector which is the difference of position vector of the points which represent the point of application of force and the point of reference.
$\Rightarrow \overrightarrow r = (2\overrightarrow i + 3\overrightarrow j + \overrightarrow k ) - (\overrightarrow i + 2\overrightarrow j - \overrightarrow k )$
$\Rightarrow \overrightarrow r = \overrightarrow i + \overrightarrow j + 2\overrightarrow k$
Given,
$\Rightarrow \overrightarrow F = \overrightarrow i + 2\overrightarrow j + \overrightarrow k$
$\overrightarrow F$ is the force vector.
We know that,
$\Rightarrow \overrightarrow \tau = \overrightarrow r \times \overrightarrow F$
$\overrightarrow \tau$ is the torque vector.
$\Rightarrow \overrightarrow \tau = (\overrightarrow i + \overrightarrow j + 2\overrightarrow k ) \times (\overrightarrow i + 2\overrightarrow j + \overrightarrow k )$
$\Rightarrow \overrightarrow \tau = - 3\overrightarrow i + \overrightarrow j + \overrightarrow k$

Hence the correct answer is (C) $- 3\overrightarrow i + \overrightarrow j + \overrightarrow k$.

In the above calculation, we wrote $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$ which means that the torque(moment of force) vector is the vector obtained through the cross product/vector product of $\overrightarrow r$ and $\overrightarrow F$.
$\overrightarrow \tau$ is the counterpart of $\overrightarrow F$ i.e. the significance of $\overrightarrow F$ in translation motion is similar to that of $\overrightarrow \tau$in rotational motion.