If unit vector `vecc` makes an angle `pi/3` with `hati + hatj`, then minimum and maximum values of `(hati xx hatj).vecc` respectively are:

Options

A. `0,sqrt(3)/2`
B. `-sqrt(3)/2, sqrt(3)/2`
C. `-1, sqrt(3)/2`
D. None of these

Answer

To solve the problem, we need to find the minimum and maximum values of the expression \((\hat{i} \times \hat{j}) \cdot \vec{c}\), given that the unit vector \(\vec{c}\) makes an angle of \(\frac{\pi}{3}\) with \(\hat{i} + \hat{j}\).

### Step-by-Step Solution:

1. **Define the Unit Vector \(\vec{c}\)**:
Since \(\vec{c}\) is a unit vector, we can express it as:
\[ \vec{c} = x \hat{i} + y \hat{j} + z \hat{k} \]
where \(x^2 + y^2 + z^2 = 1\).

2. **Angle Condition**:
The angle between \(\vec{c}\) and \(\hat{i} + \hat{j}\) is \(\frac{\pi}{3}\). We can use the cosine of the angle to set up the equation:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} = \frac{\vec{c} \cdot (\hat{i} + \hat{j})}{|\vec{c}| |\hat{i} + \hat{j}|} \]
The magnitude of \(\hat{i} + \hat{j}\) is:
\[ |\hat{i} + \hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Since \(|\vec{c}| = 1\), we have:
\[ \frac{1}{2} = \frac{x + y}{\sqrt{2}} \]
Therefore:
\[ x + y = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

3. **Setting Up the Equations**:
Now we have two equations:
\[ x^2 + y^2 + z^2 = 1 \quad \text{(1)} \]
\[ x + y = \frac{1}{\sqrt{2}} \quad \text{(2)} \]

4. **Express \(z\) in terms of \(x\) and \(y\)**:
From equation (2), we can express \(y\) as:
\[ y = \frac{1}{\sqrt{2}} - x \]
Substitute \(y\) into equation (1):
\[ x^2 + \left(\frac{1}{\sqrt{2}} - x\right)^2 + z^2 = 1 \]
Expanding this:
\[ x^2 + \left(\frac{1}{2} - \sqrt{2}x + x^2\right) + z^2 = 1 \]
Simplifying gives:
\[ 2x^2 - \sqrt{2}x + z^2 + \frac{1}{2} = 1 \]
Thus:
\[ 2x^2 - \sqrt{2}x + z^2 = \frac{1}{2} \quad \text{(3)} \]

5. **Finding the Value of \(z^2\)**:
Rearranging equation (3):
\[ z^2 = \frac{1}{2} - 2x^2 + \sqrt{2}x \]

6. **Maximizing and Minimizing \(z\)**:
To find the extrema of \(z\), we need to analyze the expression:
\[ z^2 = \frac{1}{2} - 2x^2 + \sqrt{2}x \]
This is a quadratic in \(x\). The maximum and minimum values of \(z\) occur at the vertex of this parabola.

The vertex \(x\) can be found using:
\[ x = -\frac{b}{2a} = -\frac{\sqrt{2}}{2(-2)} = \frac{\sqrt{2}}{4} \]
Substitute \(x = \frac{\sqrt{2}}{4}\) back into the equation for \(z^2\) to find the corresponding \(z\).

7. **Calculating Maximum and Minimum Values**:
After substituting \(x\) into the equation for \(z^2\), we find:
\[ z = \pm \frac{\sqrt{3}}{2} \]
Therefore, the minimum and maximum values of \((\hat{i} \times \hat{j}) \cdot \vec{c}\) are:
\[ \text{Minimum: } -\frac{\sqrt{3}}{2}, \quad \text{Maximum: } \frac{\sqrt{3}}{2} \]

### Final Answer:
The minimum and maximum values of \((\hat{i} \times \hat{j}) \cdot \vec{c}\) are \(-\frac{\sqrt{3}}{2}\) and \(\frac{\sqrt{3}}{2}\), respectively.