1. **Problem statement:** Calculate the coordinates of point I given points E and F of a cuboid and vectors \( \mathbf{u} = \begin{pmatrix}-1 \\ 2 \\ 1.5\end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix}-2 \\ -2 \\ 3\end{pmatrix} \) such that I lies along these vectors from E and F respectively. 2. **Given data:** - Point \( E = (6, 0, 4) \) - Point \( F \) is not given explicitly but can be found from the cuboid coordinates (not required here since we only need to use vectors from E and F). - Vector \( \mathbf{u} = (-1, 2, 1.5) \) from E to I - Vector \( \mathbf{v} = (-2, -2, 3) \) from F to I 3. **Formula and approach:** Point I lies on both lines: \[ I = E + \lambda \mathbf{u} = F + \mu \mathbf{v} \] where \( \lambda \) and \( \mu \) are scalars. 4. **Find coordinates of F:** Since the cuboid has points A(6,0,0), B(6,6,0), C(0,6,0), E(6,0,4), and F is the point above B at height 4, so \[ F = (6, 6, 4) \] 5. **Set up equations:** \[ E + \lambda \mathbf{u} = F + \mu \mathbf{v} \] \[ (6, 0, 4) + \lambda (-1, 2, 1.5) = (6, 6, 4) + \mu (-2, -2, 3) \] This gives the system: \[ 6 - \lambda = 6 - 2\mu \\ 0 + 2\lambda = 6 - 2\mu \\ 4 + 1.5\lambda = 4 + 3\mu \] 6. **Simplify each equation:** - From the x-coordinate: \[ 6 - \lambda = 6 - 2\mu \implies -\lambda = -2\mu \implies \lambda = 2\mu \] - From the y-coordinate: \[ 2\lambda = 6 - 2\mu \] - From the z-coordinate: \[ 4 + 1.5\lambda = 4 + 3\mu \implies 1.5\lambda = 3\mu \] 7. **Substitute \( \lambda = 2\mu \) into y and z equations:** - y: \[ 2(2\mu) = 6 - 2\mu \implies 4\mu = 6 - 2\mu \implies 4\mu + 2\mu = 6 \implies 6\mu = 6 \implies \mu = 1 \] - z: \[ 1.5(2\mu) = 3\mu \implies 3\mu = 3\mu \] This confirms consistency. 8. **Calculate \( \lambda \):** \[ \lambda = 2\mu = 2 \times 1 = 2 \] 9. **Find coordinates of I:** \[ I = E + \lambda \mathbf{u} = (6, 0, 4) + 2(-1, 2, 1.5) = (6 - 2, 0 + 4, 4 + 3) = (4, 4, 7) \] **Final answer:** \[ \boxed{I = (4, 4, 7)} \]