Q 11-06-200JEE MainJEE Main 2020 (5 Sep, Shift 1)Easy
A force $\vec F = (\hat i + 2\hat j + 3\hat k)$ N acts at a point $(4\hat i + 3\hat j - \hat k)$ m. Then the magnitude of torque about the point $(\hat i + 2\hat j + \hat k)$ m will be $\sqrt x$ N m. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 195
$\vec r = (4 - 1)\hat i + (3 - 2)\hat j + (-1 - 1)\hat k = 3\hat i + \hat j - 2\hat k$.
$$\vec\tau = \vec r\times\vec F = \begin{vmatrix}\hat i & \hat j & \hat k\\ 3 & 1 & -2\\ 1 & 2 & 3\end{vmatrix} = 7\hat i - 11\hat j + 5\hat k$$
$|\vec\tau|^{2} = 49 + 121 + 25 = 195$, so $x = 195$.
Solution by Sreeraj P, M.Sc Physics