Find the unit vector in the direction of `vec(a)-vec(b)`, where :
`vec(a)=hat(i)+3hat(j)-hat(k), vec(b)=3hat(i)+2hat(j)+hat(k)`.

Answer

To find the unit vector in the direction of \(\vec{a} - \vec{b}\), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + 3\hat{j} - \hat{k} \]
\[ \vec{b} = 3\hat{i} + 2\hat{j} + \hat{k} \]

### Step 2: Calculate \(\vec{a} - \vec{b}\)
We subtract vector \(\vec{b}\) from vector \(\vec{a}\):
\[ \vec{a} - \vec{b} = (\hat{i} + 3\hat{j} - \hat{k}) - (3\hat{i} + 2\hat{j} + \hat{k}) \]
Calculating the components:
- For \(\hat{i}\): \(1 - 3 = -2\)
- For \(\hat{j}\): \(3 - 2 = 1\)
- For \(\hat{k}\): \(-1 - 1 = -2\)

Thus,
\[ \vec{a} - \vec{b} = -2\hat{i} + 1\hat{j} - 2\hat{k} \]

### Step 3: Find the magnitude of \(\vec{a} - \vec{b}\)
The magnitude of a vector \(\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}\) is given by:
\[ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \]
For \(\vec{a} - \vec{b} = -2\hat{i} + 1\hat{j} - 2\hat{k}\):
\[ |\vec{a} - \vec{b}| = \sqrt{(-2)^2 + (1)^2 + (-2)^2} \]
Calculating the squares:
\[ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 4: Calculate the unit vector
The unit vector in the direction of \(\vec{v}\) is given by:
\[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} \]
Thus, the unit vector in the direction of \(\vec{a} - \vec{b}\) is:
\[ \hat{u} = \frac{-2\hat{i} + 1\hat{j} - 2\hat{k}}{3} \]
This can be expressed as:
\[ \hat{u} = -\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} \]

### Final Answer
The unit vector in the direction of \(\vec{a} - \vec{b}\) is:
\[ \hat{u} = -\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} \]