Find a unit vector in the direction of `(vec(a)+vec(b))`, where : `vec(a)=2hat(i)+2hat(j)-5hat(k)` and `vec(b)=2hat(i)+hat(j)+3hat(k)`.

Answer

To find a unit vector in the direction of \(\vec{a} + \vec{b}\), where \(\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}\) and \(\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}\), we will follow these steps:

### Step 1: Add the vectors \(\vec{a}\) and \(\vec{b}\)
\[ \vec{a} + \vec{b} = (2\hat{i} + 2\hat{j} - 5\hat{k}) + (2\hat{i} + \hat{j} + 3\hat{k}) \]
Combine the like terms:
\[ = (2 + 2)\hat{i} + (2 + 1)\hat{j} + (-5 + 3)\hat{k} \]
\[ = 4\hat{i} + 3\hat{j} - 2\hat{k} \]

### Step 2: Calculate the magnitude of the resultant vector \(\vec{a} + \vec{b}\)
The magnitude (or modulus) 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 our resultant vector \(4\hat{i} + 3\hat{j} - 2\hat{k}\):
\[ |\vec{a} + \vec{b}| = \sqrt{4^2 + 3^2 + (-2)^2} \]
\[ = \sqrt{16 + 9 + 4} \]
\[ = \sqrt{29} \]

### 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:
\[ \hat{u} = \frac{4\hat{i} + 3\hat{j} - 2\hat{k}}{\sqrt{29}} \]
This can be expressed as:
\[ \hat{u} = \frac{4}{\sqrt{29}}\hat{i} + \frac{3}{\sqrt{29}}\hat{j} - \frac{2}{\sqrt{29}}\hat{k} \]

### Final Answer:
The unit vector in the direction of \(\vec{a} + \vec{b}\) is:
\[ \hat{u} = \frac{4}{\sqrt{29}}\hat{i} + \frac{3}{\sqrt{29}}\hat{j} - \frac{2}{\sqrt{29}}\hat{k} \]