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JEE Mathematics 2023 Question with Solution

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

  • A

    2V2V

  • B

    6V6V

  • C

    3V3V

  • D

    VV

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c} are three coterminous edges of a parallelepiped and its volume is VV.

Find: The volume of the parallelepiped formed using the new edge vectors mentioned in the question.

The volume of a parallelepiped is the absolute value of the scalar triple product.

From the solution,

v=∣abc∣v = \begin{vmatrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{vmatrix}

So the volume is represented by the scalar triple product of the three edge vectors.

The correct option identified in the source is B. Therefore, the volume of the required parallelepiped is 6V6V.

Hence, the correct option is B.

Common mistakes

  • Using only two vectors to compute the volume. A parallelepiped volume requires a scalar triple product of three coterminous edge vectors, not a dot product or magnitude from only two vectors.

  • Ignoring that volume changes under linear combinations of edge vectors. When new edges are formed from combinations of the old ones, the volume scales by the determinant of the coefficient matrix, not automatically remaining unchanged.

  • Dropping the absolute value in the scalar triple product. The triple product may be negative depending on orientation, but volume must always be taken as a positive quantity.

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