# Show that the torque about the point $A(3, -1, 3 )$ of a force $\overrightarrow{4i}+\overrightarrow{2j}\overrightarrow{k}$ throught the point $\overrightarrow\;B(5, 2, 4)$ is $\overrightarrow{i}+\overrightarrow{2j}-\overrightarrow{8k}$.

Toolbox:
• If $\overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k, \: \overrightarrow b = b_1\overrightarrow i+b_2\overrightarrow j+b_3\overrightarrow k$ then $\overrightarrow a$ x $\overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
• Moment or torque of force $\overrightarrow F$ about the point 0 is $\overrightarrow M = \overrightarrow r \times \overrightarrow f$ where $\overrightarrow r = \overrightarrow {OP}, p$ being the point of application of the force.
$\overrightarrow F = 4\overrightarrow i+2\overrightarrow j+\overrightarrow k$
To find the moment of $\overrightarrow F$ acting at $B ( 5, 2, 4)$
$[ \overrightarrow {OB} = 5\overrightarrow i+2\overrightarrow j+4\overrightarrow k]$ about the point $A(3, -1, 3)$
$[ \overrightarrow {OA} = 3\overrightarrow i-\overrightarrow j+3\overrightarrow k]$
$\overrightarrow M = \overrightarrow r \times \overrightarrow F = \overrightarrow {AB} \times \overrightarrow F$
$= [ (5-3)\overrightarrow i + (2+1)\overrightarrow j+(4-3)\overrightarrow k ] \times [ 4\overrightarrow i+2\overrightarrow j+\overrightarrow k]$
$= (2\overrightarrow i+3\overrightarrow j+\overrightarrow k ) \times (4\overrightarrow i+2\overrightarrow j+\overrightarrow k)$
$= \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 2 & 3 & 1 & \\ 4 & 2 & 1 \end{vmatrix} = (3-2)\overrightarrow i-(2-4)\overrightarrow j+(4-12)\overrightarrow k$
$= \overrightarrow i+2\overrightarrow j-8\overrightarrow k$