Question

Find the cross product of the vectors: \[ \left( \hat{i} + 3\hat{j} - 2\hat{k} \right) \times \left( -\hat{i} + 3\hat{k} \right) \]

• A. \( 9\hat{i} - \hat{j} + 3\hat{k} \)
• B. \( 9\hat{i} + \hat{j} - 3\hat{k} \)
• C. \( \hat{i} - \hat{j} + 3\hat{k} \)
• D. \( \hat{i} + \hat{j} - 3\hat{k} \)

Hint:

To calculate the cross product of two vectors, use the determinant formula involving unit vectors \( \hat{i}, \hat{j}, \hat{k} \). This will give the components of the resulting vector.

Answer 1

The Correct Option is A

The formula for the cross product is: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}. \]

Given: \[ \vec{A} = (1, 3, -2), \quad \vec{B} = (-1, 0, 3). \]

Substitute these values into the determinant: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix}. \]

Expanding the determinant: \[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & -2 \\ 0 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 3 \\ -1 & 0 \end{vmatrix}. \]

Compute each minor determinant: \[ = \hat{i}[(3)(3) - (-2)(0)] - \hat{j}[(1)(3) - (-2)(-1)] + \hat{k}[(1)(0) - (3)(-1)]. \]

Simplify: \[ = \hat{i}(9) - \hat{j}(3 - 2) + \hat{k}(0 + 3). \]

Hence, \[ \boxed{ \vec{A} \times \vec{B} = 9\hat{i} - \hat{j} + 3\hat{k}. } \]

Final Answer:

\( \vec{A} \times \vec{B} = 9\hat{i} - \hat{j} + 3\hat{k} \)