Question

Let \( \overrightarrow{a} = 2i + 3j - 4k, \quad \overrightarrow{b} = i + j - k, \quad \overrightarrow{c} = -i + 2j + 3k, \quad \overrightarrow{d} = i + j + k. \) Then \[ (\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{c} \times \overrightarrow{d}) = \]

• A. -5
• B. -4
• C. -3
• D. -6
• E. -8

Hint:

To find the volume or other scalar products involving vectors, always remember to first compute the cross product and then the dot product. Use the properties of determinants for cross product calculations.

Answer 1

The Correct Option is D

We are given the vectors: \[ \overrightarrow{a} = 2i + 3j - 4k, \quad \overrightarrow{b} = i + j - k, \quad \overrightarrow{c} = -i + 2j + 3k, \quad \overrightarrow{d} = i + j + k. \] We are asked to find the value of: \[ (\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{c} \times \overrightarrow{d}). \] Step 1: Compute the cross product \( \overrightarrow{a} \times \overrightarrow{b} \): \[ \overrightarrow{a} = 2i + 3j - 4k, \quad \overrightarrow{b} = i + j - k. \] The cross product is given by the determinant: \[ \overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -4 \\ 1 & 1 & -1 \end{vmatrix}. \] Expanding the determinant: \[ = \hat{i} \begin{vmatrix} 3 & -4 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -4 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix}. \] \[ = \hat{i} \left( 3(-1) - (-4)(1) \right) - \hat{j} \left( 2(-1) - (-4)(1) \right) + \hat{k} \left( 2(1) - 3(1) \right) \] \[ = \hat{i} \left( -3 + 4 \right) - \hat{j} \left( -2 + 4 \right) + \hat{k} \left( 2 - 3 \right) \] \[ = \hat{i} (1) - \hat{j} (2) + \hat{k} (-1) \] \[ = \hat{i} - 2\hat{j} - \hat{k}. \] Thus, \[ \overrightarrow{a} \times \overrightarrow{b} = \hat{i} - 2\hat{j} - \hat{k}. \] Step 2: Compute the cross product \( \overrightarrow{c} \times \overrightarrow{d} \): \[ \overrightarrow{c} = -i + 2j + 3k, \quad \overrightarrow{d} = i + j + k. \] The cross product is given by the determinant: \[ \overrightarrow{c} \times \overrightarrow{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix}. \] Expanding the determinant: \[ = \hat{i} \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix}. \] \[ = \hat{i} \left( 2(1) - (3)(1) \right) - \hat{j} \left( (-1)(1) - (3)(1) \right) + \hat{k} \left( (-1)(1) - (2)(1) \right) \] \[ = \hat{i} \left( 2 - 3 \right) - \hat{j} \left( -1 - 3 \right) + \hat{k} \left( -1 - 2 \right) \] \[ = \hat{i} (-1) - \hat{j} (-4) + \hat{k} (-3) \] \[ = -\hat{i} + 4\hat{j} - 3\hat{k}. \] Thus, \[ \overrightarrow{c} \times \overrightarrow{d} = -\hat{i} + 4\hat{j} - 3\hat{k}. \] Step 3: Now compute the dot product \( (\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{c} \times \overrightarrow{d}) \): \[ (\hat{i} - 2\hat{j} - \hat{k}) \cdot (-\hat{i} + 4\hat{j} - 3\hat{k}). \] This gives: \[ = (1)(-1) + (-2)(4) + (-1)(-3) \] \[ = -1 - 8 + 3 = -6. \] Thus, the value of \( (\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{c} \times \overrightarrow{d}) \) is \( -6 \). Therefore, the correct answer is option (D).