Two point charges `q_(1) and q_(2)` are placed at (0, 0,0) and (1,2,2)m respectively. The repel each other with force of 3N. The force on `q_(2)` due to `q_(1) " is " F_(21) = (x hat(i) + y hat(j) + z hat(k))N`. Find the value of `x +y +z`

Answer

To solve the problem, we need to find the values of \( x \), \( y \), and \( z \) in the force vector \( \mathbf{F}_{21} = (x \hat{i} + y \hat{j} + z \hat{k}) \, \text{N} \) exerted on charge \( q_2 \) by charge \( q_1 \).

### Step-by-Step Solution:

1. **Identify the Positions of the Charges:**
- Charge \( q_1 \) is located at \( (0, 0, 0) \).
- Charge \( q_2 \) is located at \( (1, 2, 2) \).

2. **Calculate the Displacement Vector \( \mathbf{r} \):**
- The displacement vector \( \mathbf{r} \) from \( q_1 \) to \( q_2 \) is given by:
\[ \mathbf{r} = (1 - 0) \hat{i} + (2 - 0) \hat{j} + (2 - 0) \hat{k} = 1 \hat{i} + 2 \hat{j} + 2 \hat{k} \]

3. **Calculate the Magnitude of the Displacement Vector:**
- The magnitude \( |\mathbf{r}| \) is calculated as:
\[ |\mathbf{r}| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

4. **Determine the Unit Vector \( \hat{r} \):**
- The unit vector \( \hat{r} \) in the direction of \( \mathbf{r} \) is given by:
\[ \hat{r} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{1 \hat{i} + 2 \hat{j} + 2 \hat{k}}{3} = \frac{1}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \]

5. **Apply Coulomb's Law:**
- According to Coulomb's law, the force \( \mathbf{F}_{21} \) on charge \( q_2 \) due to charge \( q_1 \) can be expressed as:
\[ \mathbf{F}_{21} = k \frac{q_1 q_2}{|\mathbf{r}|^2} \hat{r} \]
- We know that the magnitude of this force is given as \( 3 \, \text{N} \).

6. **Express the Force Vector:**
- Since the magnitude of the force is \( 3 \, \text{N} \), we can write:
\[ \mathbf{F}_{21} = 3 \hat{r} = 3 \left( \frac{1}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \right) = 1 \hat{i} + 2 \hat{j} + 2 \hat{k} \]

7. **Identify the Components:**
- From the expression \( \mathbf{F}_{21} = x \hat{i} + y \hat{j} + z \hat{k} \), we can equate the components:
- \( x = 1 \)
- \( y = 2 \)
- \( z = 2 \)

8. **Calculate \( x + y + z \):**
- Finally, we sum the components:
\[ x + y + z = 1 + 2 + 2 = 5 \]

### Final Answer:
The value of \( x + y + z \) is \( 5 \).