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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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FPQ=10kN   Δx=8;Δy=6;Δz=1 s=Δx2+Δy2+Δz2=10   l=0.8;m=0.6;n=0.1 Fx=8kN;Fy=6kN;Fz=1kN   ˉrSP=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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