If `veca = (-hati + hatj - hatk)` and `vecb = (2hati- 2hatj + 2hatk)` then find the unit vector in the direction of `(veca + vecb)`.

Answer

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

### Step 1: Find \(\vec{a} + \vec{b}\)

Given:
\[ \vec{a} = -\hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k} \]

Now, we add the two vectors:
\[ \vec{a} + \vec{b} = (-\hat{i} + \hat{j} - \hat{k}) + (2\hat{i} - 2\hat{j} + 2\hat{k}) \]

Combining the components:
- For \(\hat{i}\): \(-1 + 2 = 1\)
- For \(\hat{j}\): \(1 - 2 = -1\)
- For \(\hat{k}\): \(-1 + 2 = 1\)

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

### Step 2: 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 \(\vec{a} + \vec{b} = \hat{i} - \hat{j} + \hat{k}\):
- \(x = 1\), \(y = -1\), \(z = 1\)

Calculating the magnitude:
\[ |\vec{a} + \vec{b}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 3: Find the unit vector in the direction of \(\vec{a} + \vec{b}\)

The unit vector \(\hat{u}\) in the direction of a vector \(\vec{v}\) is given by:
\[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} \]

Thus, the unit vector in the direction of \(\vec{a} + \vec{b}\) is:
\[ \hat{u} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \]

### Final Answer

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

---