If `vec(r )= x hat(i) + y hat(j) + z hat(k)`, then what is `vec(r ). (hat(i ) + hat(j) + hat(k))` equal to?

Options

A. x
B. `x+y`
C. `-(x+y+z)`
D. `(x+y+z)`

Answer

To solve the problem, we need to compute the dot product of the vector \(\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}\) with the vector \((\hat{i} + \hat{j} + \hat{k})\).

### Step-by-Step Solution:

1. **Write down the vectors**:
\[ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \]
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]

2. **Set up the dot product**:
We need to calculate:
\[ \vec{r} \cdot \vec{a} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) \]

3. **Apply the distributive property of the dot product**:
\[ \vec{r} \cdot \vec{a} = (x \hat{i}) \cdot (\hat{i}) + (x \hat{i}) \cdot (\hat{j}) + (x \hat{i}) \cdot (\hat{k}) + (y \hat{j}) \cdot (\hat{i}) + (y \hat{j}) \cdot (\hat{j}) + (y \hat{j}) \cdot (\hat{k}) + (z \hat{k}) \cdot (\hat{i}) + (z \hat{k}) \cdot (\hat{j}) + (z \hat{k}) \cdot (\hat{k}) \]

4. **Evaluate each dot product**:
- \((\hat{i}) \cdot (\hat{i}) = 1\)
- \((\hat{j}) \cdot (\hat{j}) = 1\)
- \((\hat{k}) \cdot (\hat{k}) = 1\)
- All other dot products between different unit vectors are zero:
\[ (\hat{i}) \cdot (\hat{j}) = 0, \quad (\hat{i}) \cdot (\hat{k}) = 0, \quad (\hat{j}) \cdot (\hat{k}) = 0 \]

5. **Combine the results**:
Thus, we have:
\[ \vec{r} \cdot \vec{a} = x \cdot 1 + y \cdot 1 + z \cdot 1 = x + y + z \]

6. **Final Answer**:
Therefore, the result of the dot product is:
\[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = x + y + z \]

### Conclusion:
The answer is \(x + y + z\).

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