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. \( \sqrt{3} \)
• D. \( \sqrt{6} \)

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: Calculate the cross product of \( \mathbf{b} \text{and} \mathbf{c} \). We find the cross product as follows:

\[\mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = -7\hat{i} + 7\hat{j} + 7\hat{k}\]

Step 2: Normalizing the vector \( \mathbf{v} \). Now, we define the unit vector \( \hat{a} \), which is perpendicular to both vectors, and calculate: \[ \hat{a} = \frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|} = \frac{-7\hat{i} + 7\hat{j} + 7\hat{k}}{\sqrt{(-7)^2 + 7^2 + 7^2}} = \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k}) \] 

Step 3: Finding the angle between the vectors Now, we use the formula for the cosine of the angle between two vectors: \[ \cos \theta = \frac{\hat{a} \cdot (\hat{i} + \alpha \hat{j} + \hat{k})}{|\hat{a}| |\hat{i} + \alpha \hat{j} + \hat{k}|} \] \[ \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} = \frac{(-1 + \alpha + 1)}{\sqrt{3} \cdot \sqrt{1 + \alpha^2 + 1}} \] Simplifying: \[ \frac{1}{2} = \frac{\alpha}{\sqrt{3}(\sqrt{\alpha^2 + 2})} = \frac{\sqrt{3}}{2} \] \[ \sqrt{3} + 2 = 2\alpha^2 \] \[ \alpha^2 = 6 \quad \Rightarrow \quad \alpha = \sqrt{6} \] 

Step 4: Final Answer \[ \boxed{\alpha = \sqrt{6}} \]