If `veca=hati+hatj, vecb=hatj+hatk, vec c hatk+hati`, a unit vector parallel to `veca+vecb+vecc`.

Options

A. `2hati+2hatj+2hatk`
B. `(hati+hatj+hatk)/(sqrt(3))`
C. `((hati+hatj+hatk))/(2sqrt(8))`
D. `(veca+vecb+vec c)/(sqrt(3))`

Answer

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

### Step 1: Write down the vectors
Given:
\[ \vec{a} = \hat{i} + \hat{j} \]
\[ \vec{b} = \hat{j} + \hat{k} \]
\[ \vec{c} = \hat{k} + \hat{i} \]

### Step 2: Calculate the sum of the vectors
We need to find \(\vec{a} + \vec{b} + \vec{c}\):
\[ \vec{a} + \vec{b} + \vec{c} = (\hat{i} + \hat{j}) + (\hat{j} + \hat{k}) + (\hat{k} + \hat{i}) \]
Combine like terms:
\[ = \hat{i} + \hat{j} + \hat{j} + \hat{k} + \hat{k} + \hat{i} = 2\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 3: Simplify the resultant vector
The resultant vector is:
\[ \vec{p} = 2\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 4: Find the magnitude of the resultant vector
The magnitude of \(\vec{p}\) is calculated as follows:
\[ |\vec{p}| = \sqrt{(2)^2 + (2)^2 + (2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

### Step 5: Find the unit vector in the direction of \(\vec{p}\)
The unit vector \(\hat{u}\) in the direction of \(\vec{p}\) is given by:
\[ \hat{u} = \frac{\vec{p}}{|\vec{p}|} = \frac{2\hat{i} + 2\hat{j} + 2\hat{k}}{2\sqrt{3}} \]
This simplifies to:
\[ \hat{u} = \frac{2}{2\sqrt{3}} \hat{i} + \frac{2}{2\sqrt{3}} \hat{j} + \frac{2}{2\sqrt{3}} \hat{k} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]

### Final Answer
Thus, the unit vector parallel to \(\vec{a} + \vec{b} + \vec{c}\) is:
\[ \hat{u} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \]

---