If `hati,hatj` and `hatk` represent unit vectors along the x,y and z-axes respectively, then the angle `theta` between the vectors `(hati+hatj+hatk)` and `(hati+hatj)` is equal to :

Options

A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for statement-I
B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for statement-I
C. Statement-I is true, Statement-II is false.
D. Statement-I is false and Statement-II is true.

Answer

To find the angle \( \theta \) between the vectors \( \hat{i} + \hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} \), we can use the formula for the cosine of the angle between two vectors:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]

where \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{B} = \hat{i} + \hat{j} \).

### Step 1: Calculate the dot product \( \vec{A} \cdot \vec{B} \)

The dot product of two vectors \( \vec{A} \) and \( \vec{B} \) is calculated as follows:

\[ \vec{A} \cdot \vec{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j}) \]

Expanding this, we have:

\[ \vec{A} \cdot \vec{B} = \hat{i} \cdot \hat{i} + \hat{i} \cdot \hat{j} + \hat{j} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{i} + \hat{k} \cdot \hat{j} \]

Since the dot product of different unit vectors is zero and the dot product of a unit vector with itself is one, we get:

\[ \vec{A} \cdot \vec{B} = 1 + 0 + 0 + 1 + 0 + 0 = 2 \]

### Step 2: Calculate the magnitudes of \( \vec{A} \) and \( \vec{B} \)

The magnitude of \( \vec{A} \) is calculated as follows:

\[ |\vec{A}| = |\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

The magnitude of \( \vec{B} \) is calculated as follows:

\[ |\vec{B}| = |\hat{i} + \hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

### Step 3: Substitute into the cosine formula

Now we can substitute the values we found into the cosine formula:

\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{2}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}} \]

### Step 4: Simplify the expression

We can simplify \( \frac{2}{\sqrt{6}} \):

\[ \cos \theta = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3} \]

### Step 5: Find the angle \( \theta \)

Finally, we find \( \theta \):

\[ \theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \]

### Summary of the Solution

The angle \( \theta \) between the vectors \( \hat{i} + \hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} \) is given by:

\[ \theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \]