If the angle between the vectors `x vec i + vec j - vec k` and `vec i + x vec j + vec k` is equal to `pi/3` then x=_______.

Answer

To solve the problem, we need to find the value of \( x \) such that the angle between the vectors \( \vec{A} = x \hat{i} + \hat{j} - \hat{k} \) and \( \vec{B} = \hat{i} + x \hat{j} + \hat{k} \) is equal to \( \frac{\pi}{3} \).

### Step-by-Step Solution:

1. **Identify the Vectors**:
\[ \vec{A} = x \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{B} = \hat{i} + x \hat{j} + \hat{k} \]

2. **Use the Cosine Formula for the Angle Between Two Vectors**:
The cosine of the angle \( \theta \) between two vectors \( \vec{A} \) and \( \vec{B} \) is given by:
\[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
Given \( \theta = \frac{\pi}{3} \), we have:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

3. **Calculate the Dot Product \( \vec{A} \cdot \vec{B} \)**:
\[ \vec{A} \cdot \vec{B} = (x)(1) + (1)(x) + (-1)(1) = x + x - 1 = 2x - 1 \]

4. **Calculate the Magnitudes \( |\vec{A}| \) and \( |\vec{B}| \)**:
\[ |\vec{A}| = \sqrt{x^2 + 1^2 + (-1)^2} = \sqrt{x^2 + 1 + 1} = \sqrt{x^2 + 2} \]
\[ |\vec{B}| = \sqrt{1^2 + x^2 + 1^2} = \sqrt{1 + x^2 + 1} = \sqrt{x^2 + 2} \]

5. **Substitute into the Cosine Formula**:
\[ \frac{1}{2} = \frac{2x - 1}{\sqrt{x^2 + 2} \cdot \sqrt{x^2 + 2}} = \frac{2x - 1}{x^2 + 2} \]

6. **Cross Multiply to Eliminate the Fraction**:
\[ 1 \cdot (x^2 + 2) = 2 \cdot (2x - 1) \]
\[ x^2 + 2 = 4x - 2 \]

7. **Rearrange the Equation**:
\[ x^2 - 4x + 2 + 2 = 0 \]
\[ x^2 - 4x + 4 = 0 \]

8. **Factor the Quadratic Equation**:
\[ (x - 2)(x - 2) = 0 \]

9. **Solve for \( x \)**:
\[ x - 2 = 0 \implies x = 2 \]

### Final Answer:
\[ x = 2 \]