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: Problem Definition

The problem provides the following parameters:

• A unit vector \( \hat{a} \) is orthogonal to vectors \( \mathbf{b} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \mathbf{c} = 2\hat{i} + 3\hat{j} - \hat{k} \).
  
• The angle between \( \hat{a} \) and \( \hat{i} + \alpha \hat{j} + \hat{k} \) is \( \cos^{-1}\left( -\frac{1}{3} \right) \).
  
• The angle between \( \hat{a} \) and \( \hat{i} + \alpha \hat{j} + \hat{k} \) is \( \frac{\pi}{3} \).
  

Step 2: Unit Vector Property

As \( \hat{a} \) is a unit vector, its magnitude is 1:

\[ |\hat{a}| = 1 \] This property ensures consistency in calculations involving \( \hat{a} \).

Step 3: Orthogonality Conditions

Since \( \hat{a} \) is orthogonal to \( \mathbf{b} \) and \( \mathbf{c} \), their dot products are zero:

\[ \hat{a} \cdot \mathbf{b} = 0 \quad \text{and} \quad \hat{a} \cdot \mathbf{c} = 0 \] These conditions will be used to establish a system of equations for the components of \( \hat{a} \).

Step 4: Angle Condition

The angle between \( \hat{a} \) and \( \hat{i} + \alpha \hat{j} + \hat{k} \) is \( \frac{\pi}{3} \). The cosine of this angle is:

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Using the dot product formula: \[ \hat{a} \cdot (\hat{i} + \alpha \hat{j} + \hat{k}) = \cos\left(\frac{\pi}{3}\right) \] Simplifying yields: \[ \hat{a} \cdot (\hat{i} + \alpha \hat{j} + \hat{k}) = \frac{1}{2} \]

Step 5: Solution for \( \alpha \)

By applying the orthogonality and angle conditions and solving the resulting system of equations, the value of \( \alpha \) is determined to be \( -\sqrt{6} \).

Conclusion

The determined value for \( \alpha \) is -√6.