EXERCISE 5.1 1. Find the volume of the parallelopiped with adjacent sides. \[ \overline{O A}=3 \hat{i}-\hat{j}, \quad \overline{O B}=\hat{j}+2 \hat{k} \text {, and } \overline{O C}=\hat{i}+5 \hat{j}+4 \hat{k} \] extending from the origin of co-ordinates \( O \)

To find the volume of the parallelepiped formed by the vectors \(\overline{O A}\), \(\overline{O B}\), and \(\overline{O C}\), you can use the scalar triple product, which is given by the absolute value of the determinant of the matrix formed by the vectors. The vectors are: \[ \overline{O A} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}, \quad \overline{O B} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}, \quad \overline{O C} = \begin{pmatrix} 1 \\ 5 \\ 4 \end{pmatrix} \] The volume \( V \) is calculated as follows: \[ V = \left| \overline{O A} \cdot (\overline{O B} \times \overline{O C}) \right| \] First, calculate the cross product \( \overline{O B} \times \overline{O C} \): \[ \overline{O B} \times \overline{O C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 5 & 4 \end{vmatrix} \] Calculating this determinant, we get: \[ = \hat{i} \begin{vmatrix} 1 & 2 \\ 5 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 2 \\ 1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ 1 & 5 \end{vmatrix} \] Calculating the minors: \[ = \hat{i} (1 \cdot 4 - 2 \cdot 5) - \hat{j} (0 \cdot 4 - 2 \cdot 1) + \hat{k} (0 \cdot 5 - 1 \cdot 1) \] \[ = \hat{i} (4 - 10) - \hat{j} (0 - 2) + \hat{k} (0 - 1) \] \[ = -6\hat{i} + 2\hat{j} - \hat{k} \] Now, compute the dot product \( \overline{O A} \cdot (\overline{O B} \times \overline{O C}) \): \[ \overline{O A} \cdot (-6\hat{i} + 2\hat{j} - \hat{k}) = (3)(-6) + (-1)(2) + (0)(-1) = -18 - 2 + 0 = -20 \] Thus, the volume \( V \) is: \[ V = | -20 | = 20 \] The volume of the parallelepiped is \( 20 \) cubic units.