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A force of 10 kN acts at a point P(2,3,5) m and has its line of action passing through Q(10,-3,4)m. Calculate moment of this force about a point S(1,-10,3)m.
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$F_{PQ}=10kN$   $\Delta x=8;\Delta y=-6;\Delta z=-1$ $\therefore s=\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}=10$   $\therefore l=0.8;m=-0.6;n=-0.1$ $\therefore F_x=8kN;F_y=-6kN;F_z=-1kN$   $\bar r_{SP}=1i+13j+2k$ $\begin{align} \therefore M_S^F&=\bar F\times \bar r\[2ex] &=\begin{vmatrix}i&j&k\8&-6&-1\1&13&2 \end{vmatrix}\[2ex] &=i(-6\times 2+13\times 1)-j(8\times 2+1\times 1)+k(8\times 13+6\times 1)\[2ex] &=i(1)-j(17)+k(110)\[2ex] &=i-17j+110k \ kNm\end{align}$

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