Question

If \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \), find the magnitude of \( \vec{a} \times \vec{b} \).

• A. \( \sqrt{75} \)
• B. \( \sqrt{86} \)
• C. \( \sqrt{110} \)
• D. \( \sqrt{102} \)

Hint:

To compute \( \vec{a} \times \vec{b} \) quickly, remember the determinant expansion along the first row and carefully track the negative sign in the \( \hat{j} \) component.

Answer 1

The Correct Option is A

Concept: The cross product of two vectors \( \vec{a} \) and \( \vec{b} \) is given by the determinant \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} a_1 & a_2 & a_3 b_1 & b_2 & b_3 \end{vmatrix} \] The magnitude of the cross product is \[ |\vec{a} \times \vec{b}| = \sqrt{(C_1)^2 + (C_2)^2 + (C_3)^2} \] where \(C_1, C_2, C_3\) are the components of the cross product.

Step 1: Write the vectors in component form. \[ \vec{a} = (2,-1,1), \qquad \vec{b} = (1,2,-3) \]

Step 2: Compute the cross product using determinant form. \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 2 & -1 & 1 1 & 2 & -3 \end{vmatrix} \] \[ = \hat{i} \begin{vmatrix} -1 & 1 2 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 1 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 1 & 2 \end{vmatrix} \] \[ = \hat{i}((-1)(-3) - (1)(2)) - \hat{j}((2)(-3) - (1)(1)) + \hat{k}((2)(2) - (-1)(1)) \] \[ = \hat{i}(3-2) - \hat{j}(-6-1) + \hat{k}(4+1) \] \[ = \hat{i} + 7\hat{j} + 5\hat{k} \]

Step 3: Find the magnitude of the resulting vector. \[ |\vec{a} \times \vec{b}| = \sqrt{1^2 + 7^2 + 5^2} \] \[ = \sqrt{1 + 49 + 25} \] \[ = \sqrt{75} \]