Question

The unit vector perpendicular to each of the vectors $ (2\hat{i} - \hat{j} + \hat{k}) \text{ and } (3\hat{i} + 4\hat{j}) \text{ is} $

• A. \( \frac{1}{\sqrt{146}} (4\hat{i} - 3\hat{j} + 11\hat{k}) \)
• B. \( \frac{1}{\sqrt{146}} (-4\hat{i} + 3\hat{j} + 11\hat{k}) \)
• C. \( \frac{1}{\sqrt{146}} (4\hat{i} + 3\hat{j} + 11\hat{k}) \)
• D. \( \frac{1}{146} (-4\hat{i} + 3\hat{j} + 11\hat{k}) \)

Hint:

To find a unit vector perpendicular to two given vectors, compute their cross product and then divide by the magnitude of the resulting vector.

Answer 1

The Correct Option is B

We are given two vectors: \[ \vec{A} = 2\hat{i} - \hat{j} + \hat{k}, \quad \vec{B} = 3\hat{i} + 4\hat{j} \]
Step 1: Find the cross product of \( \vec{A} \) and \( \vec{B} \)
The cross product of two vectors \( \vec{A} \) and \( \vec{B} \) gives a vector perpendicular to both: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 4 & 0 \end{vmatrix} \] Using the determinant method, we compute the cross product: \[ \vec{A} \times \vec{B} = \hat{i} \left( \begin{vmatrix} -1 & 1 \\ 4 & 0 \end{vmatrix} \right) - \hat{j} \left( \begin{vmatrix} 2 & 1 \\ 3 & 0 \end{vmatrix} \right) + \hat{k} \left( \begin{vmatrix} 2 & -1 \\ 3 & 4 \end{vmatrix} \right) \] \[ \vec{A} \times \vec{B} = \hat{i} (-4 - 4) - \hat{j} (0 - 3) + \hat{k} (8 + 3) \] \[ \vec{A} \times \vec{B} = -8\hat{i} + 3\hat{j} + 11\hat{k} \] 
Step 2: Find the unit vector
To get the unit vector, we divide the cross product by its magnitude: \[ |\vec{A} \times \vec{B}| = \sqrt{(-8)^2 + 3^2 + 11^2} = \sqrt{64 + 9 + 121} = \sqrt{146} \] Thus, the unit vector is: \[ \hat{n} = \frac{1}{\sqrt{146}} (-8\hat{i} + 3\hat{j} + 11\hat{k}) \] This corresponds to option (b). 
Step 3: Conclusion
The correct unit vector perpendicular to both vectors is \( \frac{1}{\sqrt{146}} (-4\hat{i} + 3\hat{j} + 11\hat{k}) \).