The angle between the vectors `hat(i) - hat(j) ` and `hat(j) - hat(k)` is a) -π/3 b) 2π/3 c) π/6 d) 5π/6

Options

A. `- ( pi )/( 3)`
B. `( 2pi )/( 3)`
C. `( pi )/( 6)`
D. ` ( 5pi )/( 6)`

Answer

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

### Step 1: Define the Vectors
Let:
\[ \mathbf{a} = \hat{i} - \hat{j} \]
\[ \mathbf{b} = \hat{j} - \hat{k} \]

### Step 2: Calculate the Dot Product
The dot product \(\mathbf{a} \cdot \mathbf{b}\) can be calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = (\hat{i} - \hat{j}) \cdot (\hat{j} - \hat{k}) \]
Using the distributive property of the dot product:
\[ \mathbf{a} \cdot \mathbf{b} = \hat{i} \cdot \hat{j} - \hat{i} \cdot \hat{k} - \hat{j} \cdot \hat{j} + \hat{j} \cdot \hat{k} \]
Since \(\hat{i} \cdot \hat{j} = 0\), \(\hat{i} \cdot \hat{k} = 0\), \(\hat{j} \cdot \hat{j} = 1\), and \(\hat{j} \cdot \hat{k} = 0\), we have:
\[ \mathbf{a} \cdot \mathbf{b} = 0 - 0 - 1 + 0 = -1 \]

### Step 3: Calculate the Magnitudes of the Vectors
Now we calculate the magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\).

For \(\mathbf{a}\):
\[ |\mathbf{a}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

For \(\mathbf{b}\):
\[ |\mathbf{b}| = \sqrt{(0)^2 + (1)^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 4: Use the Cosine Formula
The cosine of the angle \(\theta\) between the 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:
\[ \cos \theta = \frac{-1}{\sqrt{2} \cdot \sqrt{2}} = \frac{-1}{2} \]

### Step 5: Find the Angle
Now we need to find the angle \(\theta\) such that:
\[ \cos \theta = -\frac{1}{2} \]
The angle that satisfies this equation is:
\[ \theta = \frac{2\pi}{3} \text{ (or } 120^\circ\text{)} \]

### Final Answer
Thus, the angle between the vectors \(\hat{i} - \hat{j}\) and \(\hat{j} - \hat{k}\) is:
\[ \theta = \frac{2\pi}{3} \]