Three forces \(\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 3 }\) act on a rigid body. The forces \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) act through the points with position vectors \(\mathbf { r } _ { 1 }\) and \(\mathbf { r } _ { 2 }\) respectively.
\(\mathbf { r } _ { 1 } = ( - 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { m }\),
\(\mathbf { F } _ { 1 } = ( 3 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } ) \mathrm { N }\)
\(\mathbf { r } _ { 2 } = ( 3 \mathbf { i } + 2 \mathbf { k } ) \mathrm { m }\),
\(\mathbf { F } _ { 2 } = ( - 2 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \mathrm { N }\)
Given that the system \(\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 3 }\) is in equilibrium,
find \(\mathbf { F } _ { 3 }\),
find a vector equation of the line of action of \(\mathbf { F } _ { 3 }\), giving your answer in the form \(\mathbf { r } = \mathbf { a } + t \mathbf { b }\).
The force \(\mathbf { F } _ { 3 }\) is replaced by a force \(\mathbf { F } _ { 4 }\) acting through the point with position vector \(( \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } ) \mathrm { m }\). The system \(\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 4 }\) is equivalent to a single force ( \(3 \mathbf { i } + \mathbf { j } + \mathbf { k }\) ) N acting through the point with position vector \(( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }\) together with a couple.