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 defined by the vectors \(\overline{OA}\), \(\overline{OB}\), and \(\overline{OC}\), we use the scalar triple product formula. The volume \(V\) can be calculated as: \[ V = |\overline{OA} \cdot (\overline{OB} \times \overline{OC})| \] First, let's find the vectors: \(\overline{OA} = 3\hat{i} - \hat{j} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}\) \(\overline{OB} = \hat{j} + 2\hat{k} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\) \(\overline{OC} = \hat{i} + 5\hat{j} + 4\hat{k} = \begin{pmatrix} 1 \\ 5 \\ 4 \end{pmatrix}\) Now, we compute the cross product \(\overline{OB} \times \overline{OC}\): \[ \overline{OB} \times \overline{OC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 5 & 4 \end{vmatrix} \] Calculating the determinant gives: \[ = \hat{i}(1 \cdot 4 - 2 \cdot 5) - \hat{j}(0 \cdot 4 - 2 \cdot 1) + \hat{k}(0 \cdot 5 - 1 \cdot 1) \] \[ = \hat{i}(4 - 10) - \hat{j}(0 - 2) + \hat{k}(0 - 1) \] \[ = -6\hat{i} + 2\hat{j} - \hat{k} \] Now, let's find the dot product of \(\overline{OA}\) with this result: \[ \overline{OA} \cdot (-6\hat{i} + 2\hat{j} - \hat{k}) = (3)(-6) + (-1)(2) + (0)(-1) \] \[ = -18 - 2 + 0 = -20 \] Finally, the volume \(V\) is the absolute value: \[ V = | -20 | = 20 \] So, the volume of the parallelepiped is \(20\) cubic units.