Find the angle between the vectors `hat(i)-hat(j)` and `hat(j)-hat(k)`.

Answer

To find the angle between the vectors \( \hat{i} - \hat{j} \) and \( \hat{j} - \hat{k} \), we can follow these steps:

### Step 1: Define the Vectors
Let:
- Vector \( \mathbf{A} = \hat{i} - \hat{j} \)
- Vector \( \mathbf{B} = \hat{j} - \hat{k} \)

### Step 2: Calculate the Dot Product
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = (a_1 \cdot b_1) + (a_2 \cdot b_2) + (a_3 \cdot b_3) \]
For our vectors:
- \( \mathbf{A} = (1, -1, 0) \)
- \( \mathbf{B} = (0, 1, -1) \)

Calculating the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = (1 \cdot 0) + (-1 \cdot 1) + (0 \cdot -1) = 0 - 1 + 0 = -1 \]

### Step 3: Calculate the Magnitude of Each Vector
The magnitude of a vector \( \mathbf{A} = (a_1, a_2, a_3) \) is given by:
\[ |\mathbf{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

**Magnitude of \( \mathbf{A} \)**:
\[ |\mathbf{A}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \]

**Magnitude of \( \mathbf{B} \)**:
\[ |\mathbf{B}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 4: Use the Cosine Formula
The cosine of the angle \( \theta \) between two vectors is given by:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

Substituting the values we found:
\[ \cos \theta = \frac{-1}{\sqrt{2} \cdot \sqrt{2}} = \frac{-1}{2} \]

### Step 5: Find the Angle
To find the angle \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(-\frac{1}{2}\right) \]

The angle whose cosine is \(-\frac{1}{2}\) is \(120^\circ\) or \(\frac{2\pi}{3}\) radians.

### Final Answer
The angle between the vectors \( \hat{i} - \hat{j} \) and \( \hat{j} - \hat{k} \) is:
\[ \theta = \frac{2\pi}{3} \text{ radians} \text{ or } 120^\circ \]