Angle between `hat(i)-hat(j)` and `hat(j)-hat(k)` is ____________.

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 Magnitudes of the Vectors
The magnitude of a vector \( \mathbf{V} = a\hat{i} + b\hat{j} + c\hat{k} \) is given by:
\[ |\mathbf{V}| = \sqrt{a^2 + b^2 + c^2} \]

For vector \( \mathbf{A} \):
\[ |\mathbf{A}| = |\hat{i} - \hat{j}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \]

For vector \( \mathbf{B} \):
\[ |\mathbf{B}| = |\hat{j} - \hat{k}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 3: Calculate the Dot Product of the Vectors
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = a_1a_2 + b_1b_2 + c_1c_2 \]

Calculating the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} - \hat{j}) \cdot (\hat{j} - \hat{k}) = (\hat{i} \cdot \hat{j}) + (-\hat{j} \cdot \hat{j}) + (-\hat{j} \cdot -\hat{k}) \]
Using the properties of dot products:
- \( \hat{i} \cdot \hat{j} = 0 \)
- \( \hat{j} \cdot \hat{j} = 1 \)
- \( \hat{j} \cdot \hat{k} = 0 \)

Thus,
\[ \mathbf{A} \cdot \mathbf{B} = 0 - 1 + 0 = -1 \]

### Step 4: Use the Dot Product to Find the Angle
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:
\[ \cos \theta = \frac{-1}{\sqrt{2} \cdot \sqrt{2}} = \frac{-1}{2} \]

### Step 5: Calculate the Angle
To find \( \theta \):
\[ \theta = \cos^{-1}\left(-\frac{1}{2}\right) \]
The angle whose cosine is \(-\frac{1}{2}\) is:
\[ \theta = 120^\circ \]

### Final Answer
The angle between \(\hat{i} - \hat{j}\) and \(\hat{j} - \hat{k}\) is \(120^\circ\).

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