Find the angle between the vectors :
`vec(a)=hat(i)+hat(j)-hat(k) " and " vec(b)=hat(i)-hat(j)+hat(k)`

Answer

To find the angle between the vectors \(\vec{a} = \hat{i} + \hat{j} - \hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), we can use the formula for the cosine of the angle \(\theta\) 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 \(a_1 = 1\), \(a_2 = 1\), \(a_3 = -1\))
- \(\vec{b} = 1 \hat{i} - 1 \hat{j} + 1 \hat{k}\) (where \(b_1 = 1\), \(b_2 = -1\), \(b_3 = 1\))

Now, calculate the dot product:

\[ \vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1 \]

### Step 2: Calculate the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\)

The magnitude of a vector \(\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}\) is given by:

\[ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]

For \(\vec{a}\):

\[ |\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

For \(\vec{b}\):

\[ |\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 3: Substitute values into the cosine formula

Now we can substitute the values into the cosine formula:

\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3} \]

### Step 4: Find the angle \(\theta\)

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

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

This gives us the angle between the two vectors.

### Final Answer

The angle between the vectors \(\vec{a}\) and \(\vec{b}\) is:

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