Question

Let the vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) represent three coterminous edges of a parallelepiped of volume \(V\). Then the volume of the parallelepiped, whose coterminous edges are represented by \(\mathbf{a} + \mathbf{b} + \mathbf{c}\) and \(\mathbf{a} + 2\mathbf{b} + 3\mathbf{c}\), is equal to:

• A. V
• B. 2 V
• C. 3 V
• D. 6 V

Answer 1

The Correct Option is B

We are given the vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) as three coterminous edges of a parallelepiped. The volume of the parallelepiped is given by the scalar triple product of these vectors. The volume of the new parallelepiped formed by the vectors \(\mathbf{a} + \mathbf{b} + \mathbf{c}\) and \(\mathbf{a} + 2\mathbf{b} + 3\mathbf{c}\) is: \[ v = \begin{vmatrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{vmatrix} \] This gives us that the volume remains the same, \(V\).