Question

If \( \vec{a} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{j} - 2\hat{k} \), \( \vec{c} = 2\hat{i} - 3\hat{j} - 3\hat{k} \), and \( \vec{d} = 2\hat{i} + \hat{j} + \hat{k} \) are four vectors, then \( (\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) = \):

• A. \( 2\hat{i} + 19\hat{j} - 11\hat{k} \)
• B. \( -8\hat{i} + 19\hat{j} - 29\hat{k} \)
• C. \( 2\hat{i} + \hat{j} - 11\hat{k} \)
• D. \( -8\hat{i} + \hat{j} - 29\hat{k} \)

Hint:

When calculating the cross product of two vectors, use the determinant method. To find the cross product of the cross products, apply the distributive property and simplify step by step.

Answer 1

The Correct Option is D

We are given the four vectors \( \vec{a} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{j} - 2\hat{k} \), \( \vec{c} = 2\hat{i} - 3\hat{j} - 3\hat{k} \), and \( \vec{d} = 2\hat{i} + \hat{j} + \hat{k} \). We are tasked with calculating \( (\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) \). 
Step 1: First, compute the cross product \( \vec{a} \times \vec{c} \). Using the determinant formula for the cross product: \[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & -3 & -3 \end{vmatrix} \] \[ = \hat{i} \left( (-1)(-3) - (1)(-3) \right) - \hat{j} \left( (1)(-3) - (1)(2) \right) + \hat{k} \left( (1)(-3) - (-1)(2) \right) \] \[ = \hat{i} \left( 3 + 3 \right) - \hat{j} \left( -3 - 2 \right) + \hat{k} \left( -3 + 2 \right) \] \[ = \hat{i} \cdot 6 - \hat{j} \cdot (-5) + \hat{k} \cdot (-1) \] \[ = 6\hat{i} + 5\hat{j} - \hat{k} \] Thus, \( \vec{a} \times \vec{c} = 6\hat{i} + 5\hat{j} - \hat{k} \). 
Step 2: Now, compute the cross product \( \vec{b} \times \vec{d} \): \[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 2 & 1 & 1 \end{vmatrix} \] \[ = \hat{i} \left( 1(1) - (-2)(1) \right) - \hat{j} \left( 1(1) - (-2)(2) \right) + \hat{k} \left( 1(1) - 1(2) \right) \] \[ = \hat{i} \left( 1 + 2 \right) - \hat{j} \left( 1 + 4 \right) + \hat{k} \left( 1 - 2 \right) \] \[ = \hat{i} \cdot 3 - \hat{j} \cdot 5 + \hat{k} \cdot (-1) \] \[ = 3\hat{i} - 5\hat{j} - \hat{k} \] Thus, \( \vec{b} \times \vec{d} = 3\hat{i} - 5\hat{j} - \hat{k} \). 
Step 3: Now, compute \( (\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) \). We use the distributive property of the cross product: \[ (\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 5 & -1 \\ 3 & -5 & -1 \end{vmatrix} \] \[ = \hat{i} \left( (5)(-1) - (-1)(-5) \right) - \hat{j} \left( (6)(-1) - (-1)(3) \right) + \hat{k} \left( (6)(-5) - (5)(3) \right) \] \[ = \hat{i} \left( -5 - 5 \right) - \hat{j} \left( -6 + 3 \right) + \hat{k} \left( -30 - 15 \right) \] \[ = \hat{i} \cdot (-10) - \hat{j} \cdot (-3) + \hat{k} \cdot (-45) \] \[ = -10\hat{i} + 3\hat{j} - 45\hat{k} \] Thus, the result is \( (\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) = -10\hat{i} + 3\hat{j} - 45\hat{k} \).