Question

Let \( \hat{a} \) be a unit vector perpendicular to the vectors \[ \mathbf{b} = \hat{i} - 2\hat{j} + 3\hat{k} \quad \text{and} \quad \mathbf{c} = 2\hat{i} + 3\hat{j} - \hat{k}, \] \(\text{and makes an angle of}\) \( \cos\left( -\frac{1}{3} \right) \) \(\text{with the vector}\) \( \hat{i} + \alpha \hat{j} + \hat{k} \). \(\text{If}\) \( \hat{a} \) \(\text{makes an angle of}\) \( \frac{\pi}{3} \) \(\text{with the vector}\) \( \hat{i} + \alpha \hat{j} + \hat{k} \), then the value of \( \alpha \) \(\text{is:}\)

• A. −\( \sqrt{3} \)
• B. \( \sqrt{6} \)
• C. −\( \sqrt{6} \)
• D. \( \sqrt{3} \)

Hint:

When working with unit vectors and angles, using dot product and cross product properties allows for simplification of complex vector relationships.

Answer 1

The Correct Option is C

Step 1: Understand the Problem

The problem provides the following information:

• We have a unit vector \( \hat{a} \) that is perpendicular to the vectors \( \mathbf{b} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \mathbf{c} = 2\hat{i} + 3\hat{j} - \hat{k} \).
  
• The vector \( \hat{a} \) makes an angle of \( \cos^{-1}\left( -\frac{1}{3} \right) \) with the vector \( \hat{i} + \alpha \hat{j} + \hat{k} \).
  
• Additionally, the vector \( \hat{a} \) makes an angle of \( \frac{\pi}{3} \) with the vector \( \hat{i} + \alpha \hat{j} + \hat{k} \).
  

Step 2: Unit Vector Condition for \( \hat{a} \)

Since \( \hat{a} \) is a unit vector, we know that its magnitude is 1. Hence, we can write: \[ |\hat{a}| = 1 \] This will be important in maintaining consistency in the equation involving \( \hat{a} \).

Step 3: Use the Perpendicularity Condition

Since \( \hat{a} \) is perpendicular to \( \mathbf{b} \) and \( \mathbf{c} \), we can apply the condition for perpendicularity, which means the dot products of \( \hat{a} \) with \( \mathbf{b} \) and \( \mathbf{c} \) must be zero: \[ \hat{a} \cdot \mathbf{b} = 0 \quad \text{and} \quad \hat{a} \cdot \mathbf{c} = 0 \] Using these conditions, we will set up a system of equations for the components of \( \hat{a} \).

Step 4: Angle Condition with \( \hat{i} + \alpha \hat{j} + \hat{k} \)

From the given angle condition, we know that the angle between \( \hat{a} \) and the vector \( \hat{i} + \alpha \hat{j} + \hat{k} \) is \( \frac{\pi}{3} \). The cosine of this angle is given by: \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Using the dot product formula, we can write: \[ \hat{a} \cdot (\hat{i} + \alpha \hat{j} + \hat{k}) = \cos\left(\frac{\pi}{3}\right) \] Simplifying the equation: \[ \hat{a} \cdot (\hat{i} + \alpha \hat{j} + \hat{k}) = \frac{1}{2} \]

Step 5: Solve for \( \alpha \)

Using the dot product conditions and solving the system of equations derived from the perpendicularity and angle conditions, we find that the value of \( \alpha \) is \( -\sqrt{6} \).

Conclusion

The value of \( \alpha \) is -√6.