Question

A unit vector perpendicular to the vectors \( \bar{a} = 2\bar{i} + 3\bar{j} + 4\bar{k} \) and \( \bar{b} = 3\bar{j} + 2\bar{k} \) is:

• A. \( \frac{3\bar{i} + 2\bar{j} - 2\bar{k}}{\sqrt{22}} \)
• B. \( \frac{3\bar{i} + 2\bar{j} - 3\bar{k}}{\sqrt{22}} \)
• C. \( \frac{3\bar{i} - 2\bar{j} + 3\bar{k}}{\sqrt{22}} \)
• D. \( \frac{3\bar{i} + 2\bar{j} + 3\bar{k}}{\sqrt{22}} \)

Hint:

To find a perpendicular unit vector, compute the cross product and divide by its magnitude.

Answer 1

The Correct Option is B

To find a unit vector perpendicular to the vectors \( \bar{a} = 2\bar{i} + 3\bar{j} + 4\bar{k} \) and \( \bar{b} = 3\bar{j} + 2
bar{k} \), we proceed as follows: Step 1: Compute the cross product \( \bar{a} \times \bar{b} \) The cross product of \( \bar{a} \) and \( \bar{b} \) is: 

\[\bar{a} \times \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 2 & 3 & 4 \\ 0 & 3 & 2 \end{vmatrix}.\]

 Expand the determinant: \[ \bar{a} \times \bar{b} = \bar{i} (3 \cdot 2 - 4 \cdot 3) - \bar{j} (2 \cdot 2 - 4 \cdot 0) + \bar{k} (2 \cdot 3 - 3 \cdot 0). \] Simplify: \[ \bar{a} \times \bar{b} = \bar{i} (6 - 12) - \bar{j} (4 - 0) + \bar{k} (6 - 0) = -6\bar{i} - 4\bar{j} + 6\bar{k}. \] Step 2: Normalize the cross product To find a unit vector perpendicular to \( \bar{a} \) and \( \bar{b} \), normalize \( \bar{a} \times \bar{b} \): \[ \mathbf{u} = \frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}. \] First, compute the magnitude of \( \bar{a} \times \bar{b} \): \[ |\bar{a} \times \bar{b}| = \sqrt{(-6)^2 + (-4)^2 + 6^2} = \sqrt{36 + 16 + 36} = \sqrt{88} = 2\sqrt{22}. \] Thus, the unit vector is: \[ \mathbf{u} = \frac{-6\bar{i} - 4\bar{j} + 6\bar{k}}{2\sqrt{22}} = \frac{-3\bar{i} - 2\bar{j} + 3\bar{k}}{\sqrt{22}}. \] Step 3: Compare with the options The unit vector \( \mathbf{u} = \frac{-3\bar{i}- 2\bar{j} + 3\bar{k}}{\sqrt{22}} \) matches option (3): \[ \frac{3\bar{i} - 2\bar{j} + 3\bar{k}}{\sqrt{22}}. \] Final Answer: \[ \boxed{\frac{3\bar{i} - 2\bar{j} + 3\bar{k}}{\sqrt{22}}} \]