Three forces $$\overrightarrow P ,\,\overrightarrow Q $$  and $$\overrightarrow R ,$$ each of 15 units, act along $$AB,\,BC$$   and $$CA$$  respectively. The position vectors of $$A,\,B$$  and $$C$$ are $$\overrightarrow {OA} = 2\overrightarrow i - \overrightarrow j + 3\overrightarrow k ,\,\overrightarrow {OB} = 5\overrightarrow i + 3\overrightarrow j - 2\overrightarrow k $$         and $$\overrightarrow {OC} = - 2\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k $$     respectively. The resultant force vector is :

Solution :
$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = 3\overrightarrow i + 4\overrightarrow j - 5\overrightarrow k \cr & \therefore \,\frac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} = \frac{1}{{5\sqrt 2 }}\left( {3\overrightarrow i + 4\overrightarrow j - 5\overrightarrow k } \right) \cr & \therefore \,\overrightarrow P = 15.\frac{1}{{5\sqrt 2 }}\left( {3\overrightarrow i + 4\overrightarrow j - 5\overrightarrow k } \right) = \frac{3}{{\sqrt 2 }}\left( {3\overrightarrow i + 4\overrightarrow j - 5\overrightarrow k } \right) \cr & {\text{Similarly, }}\overrightarrow Q = 15.\frac{{\overrightarrow {BC} }}{{\left| {\overrightarrow {BC} } \right|}} = 15.\frac{{ - 7\overrightarrow i - \overrightarrow j + 5\overrightarrow k }}{{5\sqrt 3 }} = \sqrt 3 \left( { - 7\overrightarrow i - \overrightarrow j + 5\overrightarrow k } \right){\text{, and}} \cr & \overrightarrow R = 15.\frac{{\overrightarrow {CA} }}{{\left| {\overrightarrow {CA} } \right|}} = 15.\frac{{4\overrightarrow i - 3\overrightarrow j }}{5} = 3\left( {4\overrightarrow i - 3\overrightarrow j } \right) \cr & \therefore \,{\text{the resultant}} = \overrightarrow P + \overrightarrow Q + \overrightarrow R \cr & = \left( {12 + \frac{9}{{\sqrt 2 }} - 7\sqrt 3 } \right)\overrightarrow i + \left( { - 9 + 6\sqrt 2 - \sqrt 3 } \right)\overrightarrow j + \left( {5\sqrt 3 - \frac{{15}}{{\sqrt 2 }}} \right)\overrightarrow k . \cr} $$