Question

Let \( \mathbf{A} = 2\hat{i} + \hat{j} - 2\hat{k} \) and \( \mathbf{B} = \hat{i} + \hat{j} \). If \( \mathbf{C} \) is a vector such that \( |\mathbf{C} - \mathbf{A}| = 3 \) and the angle between \( \mathbf{A} \times \mathbf{B} \) and \( \mathbf{C} \) is \( 30^\circ \), then \( [(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}] = 3 \), the value of \( \mathbf{A} \cdot \mathbf{C} \) is equal to:
 

• A. \( \frac{25}{8} \)
• B. 2
• C. 5
• D. \( \frac{1}{8} \)

Hint:

When dealing with vector cross products and dot products, always use the magnitude formulas and trigonometric identities for angles between vectors to simplify the problem.

Answer 1

The Correct Option is B

We are given the vectors: \[ \mathbf{A} = 2\hat{i} + \hat{j} - 2\hat{k}, \mathbf{B} = \hat{i} + \hat{j} \] Step 1: Calculate \( \mathbf{A} \times \mathbf{B} \) The cross product of \( \mathbf{A} \) and \( \mathbf{B} \) is: \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 1 & -2 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -2 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \] \[ = \hat{i} (1(0) - (-2)(1)) - \hat{j} (2(0) - (-2)(1)) + \hat{k} (2(1) - 1(1)) \] \[ = \hat{i}(2) - \hat{j}(2) + \hat{k}(1) \] \[ = 2\hat{i} - 2\hat{j} + \hat{k} \] Thus, \[ \mathbf{A} \times \mathbf{B} = 2\hat{i} - 2\hat{j} + \hat{k} \] Step 2: Use the given information about the angle between \( \mathbf{A} \times \mathbf{B} \) and \( \mathbf{C} \) We are given that the angle between \( \mathbf{A} \times \mathbf{B} \) and \( \mathbf{C} \) is \( 30^\circ \), and: \[ [(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}] = 3 \] This implies: \[ |\mathbf{A} \times \mathbf{B}| |\mathbf{C}| \sin(30^\circ) = 3 \] The magnitude of \( \mathbf{A} \times \mathbf{B} \) is: \[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \] Thus, we have: \[ 3 \cdot |\mathbf{C}| \cdot \frac{1}{2} = 3 \] \[ |\mathbf{C}| = 2 \] Step 3: Find \( \mathbf{A} \cdot \mathbf{C} \) We are also given that \( |\mathbf{C} - \mathbf{A}| = 3 \). Using the distance formula, we have: \[ |\mathbf{C} - \mathbf{A}| = \sqrt{(x_C - 2)^2 + (y_C - 1)^2 + (z_C + 2)^2} = 3 \] Solving this equation will give us the value of \( \mathbf{A} \cdot \mathbf{C} \). After solving, we find that the correct value of \( \mathbf{A} \cdot \mathbf{C} \) is: \[ \boxed{2} \]