Find the sine of the angle between the vectors a and b, if `a=i-j+k" and "b=-i+j+2k`.

Answer

To find the sine of the angle between the vectors **a** and **b**, we will follow these steps:

### Step 1: Identify the vectors
Given:
\[ \mathbf{a} = \mathbf{i} - \mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} = -\mathbf{i} + \mathbf{j} + 2\mathbf{k} \]

### Step 2: Calculate the dot product \(\mathbf{a} \cdot \mathbf{b}\)
The dot product of two vectors \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) and \(\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}\) is given by:
\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]
For our vectors:
- \(a_1 = 1\), \(a_2 = -1\), \(a_3 = 1\)
- \(b_1 = -1\), \(b_2 = 1\), \(b_3 = 2\)

Calculating the dot product:
\[ \mathbf{a} \cdot \mathbf{b} = (1)(-1) + (-1)(1) + (1)(2) = -1 - 1 + 2 = 0 \]

### Step 3: Calculate the magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\)
The magnitude of a vector \(\mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}\) is given by:
\[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
Calculating the magnitude of \(\mathbf{a}\):
\[ |\mathbf{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

Calculating the magnitude of \(\mathbf{b}\):
\[ |\mathbf{b}| = \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 4: Use the cosine formula to find \(\cos \theta\)
The cosine of the angle \(\theta\) between two vectors can be found using the formula:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]
Substituting the values we calculated:
\[ \cos \theta = \frac{0}{\sqrt{3} \cdot \sqrt{6}} = 0 \]

### Step 5: Find the angle \(\theta\)
The cosine of the angle is zero when:
\[ \theta = \frac{\pi}{2} \text{ (90 degrees)} \]

### Step 6: Calculate the sine of the angle
Using the identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Since \(\cos \theta = 0\):
\[ \sin^2 \theta = 1 \implies \sin \theta = 1 \]

### Final Answer
Thus, the sine of the angle between the vectors **a** and **b** is:
\[ \sin \theta = 1 \]