Vector `1/3 (2hati - 2hatj +hatk) ` is

Options

A. a unit vector
B. makes an angle `pi//3` with vector `(2hati- 4hatj + 3hatk) `
C. parallel to vector `( -hati + hatj - 1/2 hatk)`
D. perpendicular to vector `3hati + 2hatj - 2hatk`

Explanation

To solve the problem, we need to analyze the vector given by:

\[ \mathbf{A} = \frac{1}{3} (2\hat{i} - 2\hat{j} + \hat{k}) \]

### Step 1: Determine if it is a unit vector

A vector is a unit vector if its magnitude is equal to 1. We first calculate the magnitude of the vector \(\mathbf{A}\).

1. **Calculate the magnitude**:
\[ |\mathbf{A}| = \left| \frac{1}{3} (2\hat{i} - 2\hat{j} + \hat{k}) \right| = \frac{1}{3} \sqrt{(2)^2 + (-2)^2 + (1)^2} \]
\[ = \frac{1}{3} \sqrt{4 + 4 + 1} = \frac{1}{3} \sqrt{9} = \frac{1}{3} \times 3 = 1 \]

Since the magnitude is 1, \(\mathbf{A}\) is indeed a unit vector.

### Step 2: Check if it makes an angle of \(\frac{\pi}{3}\) with another vector

Let’s assume the other vector is \(\mathbf{B} = 2\hat{i} - 4\hat{j} + 3\hat{k}\). We will find the angle between \(\mathbf{A}\) and \(\mathbf{B}\) using the dot product.

2. **Calculate the dot product**:
\[ \mathbf{A} \cdot \mathbf{B} = \left(\frac{1}{3}(2)\right)(2) + \left(\frac{1}{3}(-2)\right)(-4) + \left(\frac{1}{3}(1)\right)(3) \]
\[ = \frac{2}{3} \cdot 2 + \frac{2}{3} \cdot 4 + \frac{1}{3} \cdot 3 = \frac{4}{3} + \frac{8}{3} + 1 = \frac{4 + 8 + 3}{3} = \frac{15}{3} = 5 \]

3. **Find the magnitudes**:
\[ |\mathbf{A}| = 1 \quad \text{(as calculated earlier)} \]
\[ |\mathbf{B}| = \sqrt{(2)^2 + (-4)^2 + (3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \]

4. **Use the dot product to find the angle**:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \]
\[ 5 = 1 \cdot \sqrt{29} \cos(\theta) \implies \cos(\theta) = \frac{5}{\sqrt{29}} \]

Since \(\cos(\frac{\pi}{3}) = \frac{1}{2}\), we can see that \(\frac{5}{\sqrt{29}} \neq \frac{1}{2}\). Therefore, \(\mathbf{A}\) does not make an angle of \(\frac{\pi}{3}\) with \(\mathbf{B}\).

### Step 3: Check if it is parallel to another vector

Let’s check if \(\mathbf{A}\) is parallel to \(\mathbf{C} = -\hat{i} + \hat{j} - \frac{1}{2}\hat{k}\).

5. **Check for parallelism**:
For \(\mathbf{A}\) to be parallel to \(\mathbf{C}\), there must exist a scalar \(\lambda\) such that:
\[ \mathbf{A} = \lambda \mathbf{C} \]

If we multiply \(\mathbf{C}\) by \(-2\):
\[ -2\mathbf{C} = 2\hat{i} - 2\hat{j} + \hat{k} \]

This shows that:
\[ \mathbf{A} = \frac{1}{3}(-2\mathbf{C}) \implies \mathbf{A} \text{ is parallel to } \mathbf{C}. \]

### Step 4: Check if it is perpendicular to another vector

Let’s check if \(\mathbf{A}\) is perpendicular to \(\mathbf{D} = 3\hat{i} + 2\hat{j} - 2\hat{k}\).

6. **Calculate the dot product**:
\[ \mathbf{A} \cdot \mathbf{D} = \left(\frac{1}{3}(2)\right)(3) + \left(\frac{1}{3}(-2)\right)(2) + \left(\frac{1}{3}(1)\right)(-2) \]
\[ = \frac{2}{3} \cdot 3 + \frac{-2}{3} \cdot 2 + \frac{-2}{3} \cdot 1 = 2 - \frac{4}{3} - \frac{2}{3} = 2 - 2 = 0 \]

Since the dot product is 0, \(\mathbf{A}\) is perpendicular to \(\mathbf{D}\).

### Conclusion

The correct options are:
- \(\mathbf{A}\) is a unit vector.
- \(\mathbf{A}\) is parallel to \(\mathbf{C}\).
- \(\mathbf{A}\) is perpendicular to \(\mathbf{D}\).