Find angle `theta`between the vectors ` veca= hat i+ hat j- hat k`and ` vecb= hat i- hat j+ hat k`.

Options

A. `cos^(-1)(1/3)`
B. `cos^(-1)(-1/2)`
C. `cos^(-1)(-1/3)`
D. `cos^(-1)(1/2)`

Explanation

To find the angle \( \theta \) between the vectors \( \vec{a} = \hat{i} + \hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \), we will use the dot product formula and the magnitudes of the vectors.

### Step-by-Step Solution:

1. **Write down the vectors:**
\[ \vec{a} = \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]

2. **Calculate the magnitudes of both vectors:**
- For \( \vec{a} \):
\[ |\vec{a}| = \sqrt{(1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]
- For \( \vec{b} \):
\[ |\vec{b}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

3. **Calculate the dot product \( \vec{a} \cdot \vec{b} \):**
\[ \vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1 \]

4. **Use the dot product formula:**
The dot product is also given by:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the values we found:
\[ -1 = (\sqrt{3})(\sqrt{3}) \cos \theta \]
\[ -1 = 3 \cos \theta \]
\[ \cos \theta = -\frac{1}{3} \]

5. **Find the angle \( \theta \):**
\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]

### Final Answer:
The angle \( \theta \) between the vectors \( \vec{a} \) and \( \vec{b} \) is:
\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]