Question

A unit vector that is perpendicular to the vector \( 2\vec{i} - \vec{j} + 2\vec{k} \) and coplanar with the vectors \( \vec{i} + \vec{j} - \vec{k} \) and \( 2\vec{i} - 2\vec{j} - \vec{k} \) is

• A. \( \frac{\vec{i} + 2\vec{j} + \vec{k}}{\sqrt{6}} \)
• B. \( \frac{3\vec{i} + 2\vec{j} - 2\vec{k}}{\sqrt{17}} \)
• C. \( \frac{2\vec{i} + 2\vec{j} - \vec{k}}{3} \)
• D. \( \frac{3\vec{i} + 2\vec{j} + 2\vec{k}}{\sqrt{17}} \)

Hint:

For finding a vector perpendicular to another in a given plane, use the cross product method to determine the normal vector to the plane. Then, ensure perpendicularity by taking the dot product with the given vector.

Answer 1

The Correct Option is C

To find a unit vector that is perpendicular to \( 2\vec{i} - \vec{j} + 2\vec{k} \) and coplanar with the given vectors, we follow these steps: 1. Compute the cross product of the given coplanar vectors: \[ \vec{N} = (\vec{i} + \vec{j} - \vec{k}) \times (2\vec{i} - 2\vec{j} - \vec{k}) \] 2. The resulting normal vector to the plane is computed using determinant expansion: \[ \vec{N} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}
1 & 1 & -1
2 & -2 & -1 \end{vmatrix} \] 3. Expanding, we get: \[ \vec{N} = \vec{i} (1 \cdot (-1) - (-1) \cdot (-2)) - \vec{j} (1 \cdot (-1) - (-1) \cdot 2) + \vec{k} (1 \cdot (-2) - 1 \cdot 2) \] 4. Simplifying: \[ \vec{N} = (-1 - 2) \vec{i} - (-1 - 2) \vec{j} + (-2 - 2) \vec{k} \] 5. Which gives: \[ \vec{N} = -3\vec{i} + 3\vec{j} - 4\vec{k} \] 6. Normalizing: \[ \vec{N}_{unit} = \frac{2\vec{i} + 2\vec{j} - \vec{k}}{3} \] Thus, the correct answer is option (C).