A force of magnitude 60 kN acts at point A(1, 2, 3) m and is directed towards point B(4, −2, 15) m. Find the components of the force along the X, Y and Z axes and write it in vector form.

Answer: \(\vec F=13.85\,\hat i-18.46\,\hat j+55.38\,\hat k\) kN, i.e. \(F_x=13.85\) kN, \(F_y=-18.46\) kN, \(F_z=55.38\) kN.

Position vector from A to B: \(\overrightarrow{AB}=(4-1)\hat i+(-2-2)\hat j+(15-3)\hat k=3\hat i-4\hat j+12\hat k\) m.

\(|\overrightarrow{AB}|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\) m.

Unit vector: \(\hat e_{AB}=\dfrac{3\hat i-4\hat j+12\hat k}{13}\).

\[\vec F=60\,\hat e_{AB}=\frac{60}{13}(3\hat i-4\hat j+12\hat k)=13.85\,\hat i-18.46\,\hat j+55.38\,\hat k\ \text{kN}\]

Check: \(\sqrt{13.85^2+18.46^2+55.38^2}=60\) kN. ✓