If `vecA= 3hati+2hatj and vecB= 2hati+ 3hatj-hatk`, then find a unit vector along `(vecA-vecB)`.

Explanation

To find a unit vector along \( \vec{A} - \vec{B} \), we will follow these steps:

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

### Step 2: Calculate \( \vec{A} - \vec{B} \)
We need to subtract vector \( \vec{B} \) from vector \( \vec{A} \):
\[ \vec{A} - \vec{B} = (3\hat{i} + 2\hat{j}) - (2\hat{i} + 3\hat{j} - \hat{k}) \]
Distributing the negative sign:
\[ = 3\hat{i} + 2\hat{j} - 2\hat{i} - 3\hat{j} + \hat{k} \]
Now, combine the like terms:
\[ = (3 - 2)\hat{i} + (2 - 3)\hat{j} + 1\hat{k} \]
\[ = 1\hat{i} - 1\hat{j} + 1\hat{k} \]
Thus, we have:
\[ \vec{A} - \vec{B} = \hat{i} - \hat{j} + \hat{k} \]

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

### Step 4: Calculate the unit vector
The unit vector \( \hat{r} \) in the direction of \( \vec{R} \) is given by:
\[ \hat{r} = \frac{\vec{R}}{|\vec{R}|} \]
Substituting the values:
\[ \hat{r} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \]
This can be expressed as:
\[ \hat{r} = \frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \]

### Final Answer
Thus, the unit vector along \( \vec{A} - \vec{B} \) is:
\[ \hat{r} = \frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \]