If ` vecA = 2 hati + hatj + hatk and vecB = hati + hatj + hatk` are two vectores, then the unit vector is

Options

A. perpendicular to `vecA is ((-hatj + hatk)/(sqrt2)).`
B. parallel to `vecA is ((2hati + hatj + hatk)/(sqrt6)).`
C. perpendicular to `vecA is ((-hatj + hatk)/( sqrt2)).`
D. parallel to `vecA is ((hati + hatj + hatk)/(sqrt3)).`

Answer

To solve the problem, we need to find the unit vectors of the given vectors \(\vec{A}\) and \(\vec{B}\). Let's break it down step by step.

### Step 1: Identify the vectors
The vectors are given as:
\[ \vec{A} = 2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{B} = \hat{i} + \hat{j} + \hat{k} \]

### Step 2: Calculate the magnitude of vector \(\vec{A}\)
The magnitude of a vector \(\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}\) is given by:
\[ |\vec{A}| = \sqrt{a^2 + b^2 + c^2} \]
For \(\vec{A}\):
- \(a = 2\), \(b = 1\), \(c = 1\)

Calculating the magnitude:
\[ |\vec{A}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 3: Find the unit vector of \(\vec{A}\)
The unit vector \(\hat{A}\) is given by:
\[ \hat{A} = \frac{\vec{A}}{|\vec{A}|} \]
Substituting the values:
\[ \hat{A} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} = \frac{2}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k} \]

### Step 4: Calculate the magnitude of vector \(\vec{B}\)
For \(\vec{B}\):
- \(a = 1\), \(b = 1\), \(c = 1\)

Calculating the magnitude:
\[ |\vec{B}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 5: Find the unit vector of \(\vec{B}\)
The unit vector \(\hat{B}\) is given by:
\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} \]
Substituting the values:
\[ \hat{B} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \]

### Summary of Results
- The unit vector of \(\vec{A}\) is:
\[ \hat{A} = \frac{2}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k} \]
- The unit vector of \(\vec{B}\) is:
\[ \hat{B} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \]