Question:
Three concurrent edges of a parallelepiped are given by
\[
\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k},\quad
\vec{b} = \hat{i} - \hat{j} + 2\hat{k},\quad
\vec{c} = 2\hat{i} + \hat{j} - \hat{k}.
\]
The volume of the parallelepiped is:
Three concurrent edges of a parallelepiped are given by
\[
\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k},\quad
\vec{b} = \hat{i} - \hat{j} + 2\hat{k},\quad
\vec{c} = 2\hat{i} + \hat{j} - \hat{k}.
\]
The volume of the parallelepiped is:
Show Hint
Volume = absolute value of scalar triple product; sign indicates orientation.
Updated On: Apr 16, 2026
- 14
- 20
- 25
- 60
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The Correct Option is A
Solution and Explanation
Concept:
Volume = absolute value of scalar triple product \(|\vec{a} \cdot (\vec{b} \times \vec{c})|\).
Step 1: Compute \(\vec{b} \times \vec{c}\). \[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 1 & -1 & 2 2 & 1 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - 2\cdot1) - \hat{j}(1\cdot(-1) - 2\cdot2) + \hat{k}(1\cdot1 - (-1)\cdot2) \] \[ = \hat{i}(1 - 2) - \hat{j}(-1 - 4) + \hat{k}(1 + 2) = -\hat{i} + 5\hat{j} + 3\hat{k} \]
Step 2: Compute \(\vec{a} \cdot (\vec{b} \times \vec{c})\). \[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (2, -3, 1) \cdot (-1, 5, 3) \] \[ = 2(-1) + (-3)(5) + 1(3) = -2 -15 + 3 = -14 \]
Step 3: Volume. \[ V = |-14| = 14 \]
Step 1: Compute \(\vec{b} \times \vec{c}\). \[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 1 & -1 & 2 2 & 1 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - 2\cdot1) - \hat{j}(1\cdot(-1) - 2\cdot2) + \hat{k}(1\cdot1 - (-1)\cdot2) \] \[ = \hat{i}(1 - 2) - \hat{j}(-1 - 4) + \hat{k}(1 + 2) = -\hat{i} + 5\hat{j} + 3\hat{k} \]
Step 2: Compute \(\vec{a} \cdot (\vec{b} \times \vec{c})\). \[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (2, -3, 1) \cdot (-1, 5, 3) \] \[ = 2(-1) + (-3)(5) + 1(3) = -2 -15 + 3 = -14 \]
Step 3: Volume. \[ V = |-14| = 14 \]
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