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:
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:
Updated On: Mar 21, 2025
- V
- 2 V
- 3 V
- 6 V
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The Correct Option is B
Solution and Explanation
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, \(V\).
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