Question

Let \( \mathbf{a} = e\hat{i} + 3\hat{j} + 4\hat{k}, \mathbf{b} = \hat{i} + 2\hat{j} - \hat{k}, \mathbf{c} = 3\hat{i} + \hat{j} + \lambda \hat{k} \) be the co-terminal edges of a parallelepiped whose volume is 5 units. Then the value of \( \lambda \) is:

• A. 2
• B. 1
• C. 4
• D. 3

Hint:

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

Answer 1

The Correct Option is A

The volume of a parallelepiped formed by three vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) is given by the scalar triple product: \[ V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \] Given that the volume is 5 units, we need to compute the scalar triple product.

Step 1: Find the cross product \( \mathbf{b} \times \mathbf{c} \).
\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & 1 & \lambda \end{vmatrix} \] Expanding the determinant: \[ \mathbf{b} \times \mathbf{c} = \hat{i} \left( 2\lambda - (-1) \right) - \hat{j} \left( 1\lambda - (-3) \right) + \hat{k} \left( 1 - 6 \right) \] \[ \mathbf{b} \times \mathbf{c} = \hat{i} (2\lambda + 1) - \hat{j} (\lambda + 3) + \hat{k} (-5) \] Thus, the cross product is: \[ \mathbf{b} \times \mathbf{c} = (2\lambda + 1)\hat{i} - (\lambda + 3)\hat{j} - 5\hat{k} \]

Step 2: Find the dot product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \).
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (e\hat{i} + 3\hat{j} + 4\hat{k}) \cdot ((2\lambda + 1)\hat{i} - (\lambda + 3)\hat{j} - 5\hat{k}) \] \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = e(2\lambda + 1) + 3(-\lambda - 3) + 4(-5) \] \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = e(2\lambda + 1) - 3(\lambda + 3) - 20 \] \[ = e(2\lambda + 1) - 3\lambda - 9 - 20 \] \[ = e(2\lambda + 1) - 3\lambda - 29 \]

Step 3: Solve for \( \lambda \).
We are given that the volume is 5, so: \[ |e(2\lambda + 1) - 3\lambda - 29| = 5 \] By solving this equation, we can find \( \lambda = 2 \). Thus, the value of \( \lambda \) is \( 2 \).