`(hat(i) + hat(j) - hat(k)).(hat(i) - hat(j) + hat(k))`=_________

Options

A. `hat(i) - hat(j) - hat(k)`
B. `1`
C. -1
D. `-hat(j) + hat(k)`

Answer

To solve the expression \((\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k})\), we will use the properties of the dot product. Here are the steps:

### Step 1: Expand the Dot Product
We start by expanding the dot product using the distributive property:
\[ (\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = \hat{i} \cdot \hat{i} + \hat{i} \cdot (-\hat{j}) + \hat{i} \cdot \hat{k} + \hat{j} \cdot \hat{i} + \hat{j} \cdot (-\hat{j}) + \hat{j} \cdot \hat{k} - \hat{k} \cdot \hat{i} - \hat{k} \cdot (-\hat{j}) - \hat{k} \cdot \hat{k} \]

### Step 2: Calculate Each Dot Product
Now we calculate each of the dot products:
- \(\hat{i} \cdot \hat{i} = 1\) (since the dot product of a unit vector with itself is 1)
- \(\hat{i} \cdot (-\hat{j}) = 0\) (since \(\hat{i}\) and \(\hat{j}\) are perpendicular)
- \(\hat{i} \cdot \hat{k} = 0\) (since \(\hat{i}\) and \(\hat{k}\) are perpendicular)
- \(\hat{j} \cdot \hat{i} = 0\) (since \(\hat{j}\) and \(\hat{i}\) are perpendicular)
- \(\hat{j} \cdot (-\hat{j}) = -1\) (since the dot product of a unit vector with itself is 1, but we have a negative sign)
- \(\hat{j} \cdot \hat{k} = 0\) (since \(\hat{j}\) and \(\hat{k}\) are perpendicular)
- \(-\hat{k} \cdot \hat{i} = 0\) (since \(\hat{k}\) and \(\hat{i}\) are perpendicular)
- \(-\hat{k} \cdot (-\hat{j}) = 0\) (since \(\hat{k}\) and \(\hat{j}\) are perpendicular)
- \(-\hat{k} \cdot \hat{k} = -1\) (since the dot product of a unit vector with itself is 1, but we have a negative sign)

### Step 3: Combine the Results
Now we combine all the results:
\[ 1 + 0 + 0 + 0 - 1 + 0 + 0 + 0 - 1 = 1 - 1 - 1 = -1 \]

### Final Answer
Thus, the value of \((\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k})\) is:
\[ \boxed{-1} \]