If `vec(a)=2hat(i)-hat(j)+2hat(k)` and `vec(b)=6hat(i)+2hat(j)+3hat(k)`, find a unit vector parallel to `vec(a)+vec(b)`.

Answer

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

### Step 1: Find the vectors \(\vec{a}\) and \(\vec{b}\)
Given:
\[ \vec{a} = 2\hat{i} - \hat{j} + 2\hat{k} \]
\[ \vec{b} = 6\hat{i} + 2\hat{j} + 3\hat{k} \]

### Step 2: Calculate \(\vec{a} + \vec{b}\)
Add the two vectors component-wise:
\[ \vec{a} + \vec{b} = (2\hat{i} + 6\hat{i}) + (-\hat{j} + 2\hat{j}) + (2\hat{k} + 3\hat{k}) \]
\[ = (2 + 6)\hat{i} + (-1 + 2)\hat{j} + (2 + 3)\hat{k} \]
\[ = 8\hat{i} + 1\hat{j} + 5\hat{k} \]

### 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 \(\vec{a} + \vec{b} = 8\hat{i} + 1\hat{j} + 5\hat{k}\):
\[ |\vec{a} + \vec{b}| = \sqrt{8^2 + 1^2 + 5^2} \]
\[ = \sqrt{64 + 1 + 25} \]
\[ = \sqrt{90} \]
\[ = 3\sqrt{10} \]

### Step 4: Find the unit vector parallel to \(\vec{a} + \vec{b}\)
A unit vector in the direction of a vector \(\vec{v}\) is given by:
\[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} \]
Thus, the unit vector parallel to \(\vec{a} + \vec{b}\) is:
\[ \hat{u} = \frac{8\hat{i} + 1\hat{j} + 5\hat{k}}{3\sqrt{10}} \]

### Step 5: Write the final answer in standard form
We can express the unit vector as:
\[ \hat{u} = \frac{8}{3\sqrt{10}}\hat{i} + \frac{1}{3\sqrt{10}}\hat{j} + \frac{5}{3\sqrt{10}}\hat{k} \]

### Final Answer:
The unit vector parallel to \(\vec{a} + \vec{b}\) is:
\[ \hat{u} = \frac{8}{3\sqrt{10}}\hat{i} + \frac{1}{3\sqrt{10}}\hat{j} + \frac{5}{3\sqrt{10}}\hat{k} \]