The angle between vectors `(hati+hatj+hatk)` and `(hatj+hatk)` is :

Explanation

To find the angle between the vectors \(\vec{A} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{B} = \hat{j} + \hat{k}\), we will use the formula for the cosine of the angle between two vectors:

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

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

The dot product of two vectors \(\vec{A} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) and \(\vec{B} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\) is given by:

\[ \vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]

For our vectors:
- \(\vec{A} = 1\hat{i} + 1\hat{j} + 1\hat{k}\) (where \(a_1 = 1, a_2 = 1, a_3 = 1\))
- \(\vec{B} = 0\hat{i} + 1\hat{j} + 1\hat{k}\) (where \(b_1 = 0, b_2 = 1, b_3 = 1\))

Calculating the dot product:

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

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

The magnitude of a vector \(\vec{V} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) is given by:

\[ |\vec{V}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

Calculating the magnitude of \(\vec{A}\):

\[ |\vec{A}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

Calculating the magnitude of \(\vec{B}\):

\[ |\vec{B}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 3: Substitute into the cosine formula

Now we can substitute the dot product and the magnitudes 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: Calculate \(\theta\)

To find the angle \(\theta\), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{2}{\sqrt{6}}\right) \]

### Final Answer

Thus, the angle between the vectors \((\hat{i} + \hat{j} + \hat{k})\) and \((\hat{j} + \hat{k})\) is:

\[ \theta = \cos^{-1}\left(\frac{2}{\sqrt{6}}\right) \]