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

The problem requires us to calculate the volume of a new parallelepiped formed by the given vectors. Let's approach this systematically:

1. First, understand the basics of vector operations in the context of volumes: The volume \( V \) of a parallelepiped with coterminous edges given by vectors \mathbf{a}, \mathbf{b}, \mathbf{c} is computed using the scalar triple product as V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| .
2. The given question involves a transformation of these vectors. We have new vectors as: \mathbf{A} = \mathbf{a} + \mathbf{b} + \mathbf{c} and \mathbf{B} = \mathbf{a} + 2\mathbf{b} + 3\mathbf{c} along with the original vector \mathbf{c}.
3. We need to determine the volume of the parallelepiped with edges \mathbf{A}, \mathbf{B}, and \mathbf{c}. The volume formula for this new parallelepiped is:

V_{\text{new}} = |\mathbf{A} \cdot (\mathbf{B} \times \mathbf{c})|

4. To compute this, let's expand these vectors in terms of the original vectors:

\mathbf{A} = \mathbf{a} + \mathbf{b} + \mathbf{c} and \mathbf{B} = \mathbf{a} + 2\mathbf{b} + 3\mathbf{c}

5. Calculate the vector product \mathbf{B} \times \mathbf{c} using the distributive property of the cross product:

\mathbf{B} \times \mathbf{c} = (\mathbf{a} + 2\mathbf{b} + 3\mathbf{c}) \times \mathbf{c} = \mathbf{a} \times \mathbf{c} + 2(\mathbf{b} \times \mathbf{c}) + 3(\mathbf{c} \times \mathbf{c})

• Note: \mathbf{c} \times \mathbf{c} = \mathbf{0}, so it does not contribute to the product.

6. Substitute this back into the volume formula:

V_{\text{new}} = |\mathbf{A} \cdot (\mathbf{a} \times \mathbf{c} + 2(\mathbf{b} \times \mathbf{c}))|

7. Now distribute \mathbf{A} into the scalar (dot) product:

=\left| (\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{a} \times \mathbf{c}) + 2(\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} \times \mathbf{c})\right|

8. Apply the distributive property:

• (\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{a} \times \mathbf{c}) = \mathbf{a} \cdot (\mathbf{a} \times \mathbf{c}) + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{c}) + \mathbf{c} \cdot (\mathbf{a} \times \mathbf{c})
• Here, \mathbf{a} \cdot (\mathbf{a} \times \mathbf{c}) and \mathbf{c} \cdot (\mathbf{a} \times \mathbf{c}) are zero due to the property of triple products. Only \mathbf{b} \cdot (\mathbf{a} \times \mathbf{c}) remains, which contributes V.
• 2(\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} \times \mathbf{c}) = 2(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \cdot (\mathbf{b} \times \mathbf{c}) + \mathbf{c} \cdot (\mathbf{b} \times \mathbf{c}))
• Here, the term \mathbf{b} \cdot (\mathbf{b} \times \mathbf{c}) is zero, contributing only 2V from the a \cdot (b \times c) term.

9. Thus, the simplification results in:

V_{\text{new}} = |V + 2V| = 2V

10. Thus, the volume of the parallelepiped formed by the vectors \mathbf{A}, \mathbf{B}, and \mathbf{C} is 2V.

Therefore, the correct answer is 2 V.