Vector Calculus

Let represent the magnitude of vector , where: We are given the following conditions: Now, let's prove each statement.
Statement A: We know that: We also have: Thus, Hence, statement A is false because the magnitude of the vector is less than or equal to the maximum of the absolute values of its components, not necessarily equal.
Statement B: This statement is true because the magnitude of the vector is always less than or equal to the maximum of the absolute values of the components.
Thus, statement (B) is the correct statement.

Step 1: Understand the Problem

The problem provides the following information:

• We have a unit vector that is perpendicular to the vectors and .
  
• The vector makes an angle of with the vector .
  
• Additionally, the vector makes an angle of with the vector .
  

Step 2: Unit Vector Condition for

Since is a unit vector, we know that its magnitude is 1. Hence, we can write: This will be important in maintaining consistency in the equation involving .

Step 3: Use the Perpendicularity Condition

Since is perpendicular to and , we can apply the condition for perpendicularity, which means the dot products of with and must be zero: Using these conditions, we will set up a system of equations for the components of .

Step 4: Angle Condition with

From the given angle condition, we know that the angle between and the vector is . The cosine of this angle is given by: Using the dot product formula, we can write: Simplifying the equation:

Step 5: Solve for

Using the dot product conditions and solving the system of equations derived from the perpendicularity and angle conditions, we find that the value of is .

Conclusion

The value of is -√6.