Question

The unit vector perpendicular to both vectors \( \hat{i} + \hat{k} \) and \( \hat{i} - \hat{k} \) is:

• A. \( 2 \hat{j} \)
• B. \( \hat{j} \)
• C. \( \frac{\hat{i} - \hat{k}}{\sqrt{2}} \)
• D. \( \frac{\hat{i} + \hat{k}}{\sqrt{2}} \)

Hint:

The cross product of two vectors results in a perpendicular vector. To find the unit vector, divide the cross product by its magnitude.

Answer 1

The Correct Option is B

Step 1: Cross product of two vectors.
The cross product of vectors \( \vec{A} = \hat{i} + \hat{k} \) and \( \vec{B} = \hat{i} - \hat{k} \) produces a vector perpendicular to both: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 0 & 1
1 & 0 & -1 \end{vmatrix}. \] To calculate this, expand the determinant: \[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 0 & 1
0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1
1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 0
1 & 0 \end{vmatrix}. \] Step 2: Simplify the minors.
- First minor: \[ \begin{vmatrix} 0 & 1
0 & -1 \end{vmatrix} = 0. \] - Second minor: \[ \begin{vmatrix} 1 & 1
1 & -1 \end{vmatrix} = (1)(-1) - (1)(1) = -1 - 1 = -2. \] - Third minor: \[ \begin{vmatrix} 1 & 0
1 & 0 \end{vmatrix} = (1)(0) - (1)(0) = 0. \] Thus, we have: \[ \vec{A} \times \vec{B} = -(-2)\hat{j} = 2\hat{j}. \] Step 3: Normalize the resulting vector.
The magnitude of \( \vec{A} \times \vec{B} \) is \( 2 \). Therefore, the unit vector in the direction of \( \vec{A} \times \vec{B} \) is: \[ \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{2\hat{j}}{2} = \hat{j}. \] Step 4: Final Answer.
The unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \) is: \[ \boxed{\hat{j}}. \]