Two point charge `q_(2) = 3xx10^(-6) C and q_(1) = 5xx10^(-6) C` are located at `(3,5,1) and (1,3,2)m`. Find `vec(F_(12)) and vec(F_(21))` using vector form of Coulomb's law. Also, find their magnitude.

Explanation

To solve this problem, we will use the vector form of Coulomb's law. Let's break down the steps:

### Step 1: Identify the given data
- Charge \( q_1 = 5 \times 10^{-6} \, \text{C} \)
- Charge \( q_2 = 3 \times 10^{-6} \, \text{C} \)
- Position of \( q_1 \): \( (1, 3, 2) \, \text{m} \)
- Position of \( q_2 \): \( (3, 5, 1) \, \text{m} \)

### Step 2: Determine the position vectors
- Position vector of \( q_1 \), \( \vec{r}_1 = 1\hat{i} + 3\hat{j} + 2\hat{k} \)
- Position vector of \( q_2 \), \( \vec{r}_2 = 3\hat{i} + 5\hat{j} + 1\hat{k} \)

### Step 3: Calculate the vector \( \vec{r}_{12} \) from \( q_1 \) to \( q_2 \)
\[ \vec{r}_{12} = \vec{r}_2 - \vec{r}_1 = (3\hat{i} + 5\hat{j} + 1\hat{k}) - (1\hat{i} + 3\hat{j} + 2\hat{k}) \]
\[ \vec{r}_{12} = (3 - 1)\hat{i} + (5 - 3)\hat{j} + (1 - 2)\hat{k} \]
\[ \vec{r}_{12} = 2\hat{i} + 2\hat{j} - 1\hat{k} \]

### Step 4: Calculate the magnitude of \( \vec{r}_{12} \)
\[ |\vec{r}_{12}| = \sqrt{(2)^2 + (2)^2 + (-1)^2} \]
\[ |\vec{r}_{12}| = \sqrt{4 + 4 + 1} \]
\[ |\vec{r}_{12}| = \sqrt{9} = 3 \, \text{m} \]

### Step 5: Calculate the unit vector \( \hat{r}_{12} \)
\[ \hat{r}_{12} = \frac{\vec{r}_{12}}{|\vec{r}_{12}|} = \frac{2\hat{i} + 2\hat{j} - 1\hat{k}}{3} \]
\[ \hat{r}_{12} = \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \]

### Step 6: Apply Coulomb's law to find \( \vec{F}_{12} \)
\[ \vec{F}_{12} = k_e \frac{q_1 q_2}{|\vec{r}_{12}|^2} \hat{r}_{12} \]
Where \( k_e = 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)
\[ \vec{F}_{12} = 9 \times 10^9 \frac{(5 \times 10^{-6})(3 \times 10^{-6})}{3^2} \left( \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \]
\[ \vec{F}_{12} = 9 \times 10^9 \frac{15 \times 10^{-12}}{9} \left( \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \]
\[ \vec{F}_{12} = 9 \times 10^9 \times 1.67 \times 10^{-12} \left( \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \]
\[ \vec{F}_{12} = 1.5 \times 10^{-2} \left( \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \]
\[ \vec{F}_{12} = 1 \times 10^{-2} \hat{i} + 1 \times 10^{-2} \hat{j} - 0.5 \times 10^{-2} \hat{k} \]

### Step 7: Calculate \( \vec{F}_{21} \)
By Newton's third law, \( \vec{F}_{21} = -\vec{F}_{12} \)
\[ \vec{F}_{21} = - (1 \times 10^{-2} \hat{i} + 1 \times 10^{-2} \hat{j} - 0.5 \times 10^{-2} \hat{k}) \]
\[ \vec{F}_{21} = -1 \times 10^{-2} \hat{i} - 1 \times 10^{-2} \hat{j} + 0.5 \times 10^{-2} \hat{k} \]

### Step 8: Calculate the magnitudes of \( \vec{F}_{12} \) and \( \vec{F}_{21} \)
\[ |\vec{F}_{12}| = |\vec{F}_{21}| = 1.5 \times 10^{-2} \, \text{N} \]

### Final Answer:
\[ \vec{F}_{12} = 1 \times 10^{-2} \hat{i} + 1 \times 10^{-2} \hat{j} - 0.5 \times 10^{-2} \hat{k} \, \text{N} \]
\[ \vec{F}_{21} = -1 \times 10^{-2} \hat{i} - 1 \times 10^{-2} \hat{j} + 0.5 \times 10^{-2} \hat{k} \, \text{N} \]
\[ |\vec{F}_{12}| = |\vec{F}_{21}| = 1.5 \times 10^{-2} \, \text{N} \]