# Find the magnitude and direction cosines of the moment about the point $(1, -2 ,3)$ of a force $\overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{6k}$ Whose line of action passes through the origin.

Toolbox:
• The direction cosines of vector $\overrightarrow a=a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k$ are $l = \large\frac{a_1}{|\overrightarrow a|}, m = \large\frac{a_2}{|\overrightarrow a|}, n = \large\frac{a_3}{|\overrightarrow a|}$. The angles made by $\overrightarrow a$ with the coordinate axes are $\cos^{-1}l, \: \cos^{-1}m, \: \cos^{-1}n.$
• 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.
$\overrightarrow M = \overrightarrow r \times \overrightarrow F = \overrightarrow {PO} \times ( 2\overrightarrow i+3\overrightarrow j+6\overrightarrow k)$
$- {OP} \times (2\overrightarrow i+3\overrightarrow j+6\overrightarrow k)$
$= (-\overrightarrow i+\overrightarrow j-3\overrightarrow k) \times (2\overrightarrow i+3\overrightarrow j+6\overrightarrow k)$
$= \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ -1 & 1 & -3 \\ 2 & 3 & 6 \end{vmatrix} = (6+9)\overrightarrow i - (-6+6)\overrightarrow j + (-3-2) \overrightarrow k$
$= 15\overrightarrow i-5\overrightarrow k$
$= 5(3\overrightarrow i-\overrightarrow k)$
$|\overrightarrow M| = 5\sqrt{9+1} = 5\sqrt{10}$
The DC's are $\large\frac{15}{5\sqrt{10}}, \: 0, \: \large\frac{-5}{5\sqrt{10}}$
$= \large\frac{3}{\sqrt{10}}, \: 0, \: -\large\frac{1}{\sqrt{10}}$