18. A triangular prism has side lengths \( O A=8, O B=10 \), and \( O E=12 \). \( O \) is the origin. (a) Find the position vectors of \( C \) and \( D \). (b) \( F \) and \( G \) are the midpoints of the edges \( A B \) and \( C D \), respectively. Find their position vectors. (c) Find the vectors \( \overrightarrow{A G} \) and \( \overrightarrow{F D} \) and explain your results.

To solve the problem, we will follow the steps outlined in the prompt. ### Known Conditions - The triangular prism has vertices \( O, A, B, C, D, E, F, G \). - The lengths of the edges are given as: - \( OA = 8 \) - \( OB = 10 \) - \( OE = 12 \) - \( O \) is the origin. ### Part (a): Find the position vectors of \( C \) and \( D \) Assuming the triangular prism is positioned in a 3D coordinate system, we can assign the following coordinates based on the given lengths: - Let \( O = (0, 0, 0) \) - Let \( A = (8, 0, 0) \) (along the x-axis) - Let \( B = (0, 10, 0) \) (along the y-axis) - Let \( E = (0, 0, 12) \) (along the z-axis) To find the position vectors of \( C \) and \( D \), we need to determine their coordinates. Since \( C \) and \( D \) are directly above \( A \) and \( B \) respectively, we can assign: - \( C = (8, 0, 12) \) (directly above \( A \)) - \( D = (0, 10, 12) \) (directly above \( B \)) Thus, the position vectors are: - \( \vec{C} = \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} \) - \( \vec{D} = \begin{pmatrix} 0 \\ 10 \\ 12 \end{pmatrix} \) ### Part (b): Find the position vectors of midpoints \( F \) and \( G \) **Midpoint \( F \) of edge \( AB \)**: The position vector of \( F \) can be calculated as: \[ \vec{F} = \frac{\vec{A} + \vec{B}}{2} = \frac{\begin{pmatrix} 8 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 10 \\ 0 \end{pmatrix}}{2} = \frac{\begin{pmatrix} 8 \\ 10 \\ 0 \end{pmatrix}}{2} = \begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix} \] **Midpoint \( G \) of edge \( CD \)**: The position vector of \( G \) can be calculated as: \[ \vec{G} = \frac{\vec{C} + \vec{D}}{2} = \frac{\begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} + \begin{pmatrix} 0 \\ 10 \\ 12 \end{pmatrix}}{2} = \frac{\begin{pmatrix} 8 \\ 10 \\ 12 \end{pmatrix}}{2} = \begin{pmatrix} 4 \\ 5 \\ 12 \end{pmatrix} \] ### Part (c): Find the vectors \( \overrightarrow{AG} \) and \( \overrightarrow{FD} \) **Vector \( \overrightarrow{AG} \)**: \[ \overrightarrow{AG} = \vec{G} - \vec{A} = \begin{pmatrix} 4 \\ 5 \\ 12 \end{pmatrix} - \begin{pmatrix} 8 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 - 8 \\ 5 - 0 \\ 12 - 0 \end{pmatrix} = \begin{pmatrix} -4 \\ 5 \\ 12 \end{pmatrix} \] **Vector \( \overrightarrow{FD} \)**: \[ \overrightarrow{FD} = \vec{D} - \vec{F} = \begin{pmatrix} 0 \\ 10 \\ 12 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 - 4 \\ 10 - 5 \\ 12 - 0 \end{pmatrix} = \begin{pmatrix} -4 \\ 5 \\ 12 \end{pmatrix} \] ### Explanation of Results Both vectors \( \overrightarrow{AG} \) and \( \overrightarrow{FD} \) are equal: \[ \overrightarrow{AG} = \overrightarrow{FD} = \begin{pmatrix} -4 \\ 5 \\ 12 \end{pmatrix} \] This indicates that the displacement from point \( A \) to midpoint \( G \) is the same as the displacement from midpoint \( F \) to point \( D \). This is consistent with the symmetry of the triangular prism, where midpoints of corresponding edges are equidistant in the vertical direction.