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 \( O \)
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To find the volume of the parallelepiped formed by the vectors \(\overline{O A}\), \(\overline{O B}\), and \(\overline{O C}\), we can use the scalar triple product formula, which can be calculated as the absolute value of the determinant of the matrix formed by these vectors. The vectors are: \[ \overline{O A} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}, \quad \overline{O B} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}, \quad \overline{O C} = \begin{pmatrix} 1 \\ 5 \\ 4 \end{pmatrix} \] Now we arrange these vectors in a matrix: \[ \begin{vmatrix} 3 & 0 & 1 \\ -1 & 1 & 5 \\ 0 & 2 & 4 \end{vmatrix} \] Calculating this determinant: 1. Calculate the first row expansion: \[ 3 \begin{vmatrix} 1 & 5 \\ 2 & 4 \end{vmatrix} - 0 + 1 \begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} \] 2. Calculating the 2x2 determinants: \[ \begin{vmatrix} 1 & 5 \\ 2 & 4 \end{vmatrix} = (1)(4) - (5)(2) = 4 - 10 = -6 \] \[ \begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} = (-1)(2) - (1)(0) = -2 \] 3. Putting everything together: \[ 3(-6) + 1(-2) = -18 - 2 = -20 \] 4. Therefore, the volume \( V \) of the parallelepiped is: \[ V = | -20 | = 20 \] The volume of the parallelepiped is \( 20 \) cubic units.