There are two vector `vec(A)=3hat(i)+hat(j)` and `vec(B)=hat(j)+2hat(k)`. For these two vectors-
(i) Find the component of `vec(A)` along `vec(B)` and perpendicular to `vec(B)` in vector form.
(ii) If `vec(A)` & `vec(B)` are the adjacent sides of parallelogram then find the magnitude of its area.
(iii) Find a unit vector which is perpendicular to both `vec(A)` & `vec(B)`.

Explanation

To solve the problem step by step, we will address each part of the question sequentially.

### Given Vectors:
- \( \vec{A} = 3\hat{i} + \hat{j} \)
- \( \vec{B} = \hat{j} + 2\hat{k} \)

### (i) Find the component of \( \vec{A} \) along \( \vec{B} \) and perpendicular to \( \vec{B} \).

1. **Calculate the dot product \( \vec{A} \cdot \vec{B} \)**:
\[ \vec{A} \cdot \vec{B} = (3\hat{i} + \hat{j}) \cdot (\hat{j} + 2\hat{k}) = 3(0) + 1(1) + 0(2) = 1 \]

2. **Calculate the magnitude of \( \vec{B} \)**:
\[ |\vec{B}| = \sqrt{(0)^2 + (1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

3. **Find the unit vector of \( \vec{B} \)**:
\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{\hat{j} + 2\hat{k}}{\sqrt{5}} \]

4. **Calculate the component of \( \vec{A} \) along \( \vec{B} \)**:
\[ \text{Component of } \vec{A} \text{ along } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} \hat{B} = \frac{1}{\sqrt{5}} \left(\frac{\hat{j} + 2\hat{k}}{\sqrt{5}}\right) = \frac{\hat{j} + 2\hat{k}}{5} \]

5. **Calculate the component of \( \vec{A} \) perpendicular to \( \vec{B} \)**:
\[ \text{Component of } \vec{A} \text{ perpendicular to } \vec{B} = \vec{A} - \text{Component along } \vec{B} \]
\[ = (3\hat{i} + \hat{j}) - \frac{\hat{j} + 2\hat{k}}{5} = 3\hat{i} + \left(1 - \frac{1}{5}\right)\hat{j} - \frac{2}{5}\hat{k} = 3\hat{i} + \frac{4}{5}\hat{j} - \frac{2}{5}\hat{k} \]

### (ii) Find the magnitude of the area of the parallelogram formed by \( \vec{A} \) and \( \vec{B} \).

1. **Calculate the cross product \( \vec{A} \times \vec{B} \)**:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 0 & 1 & 2 \end{vmatrix} \]
\[ = \hat{i}(1 \cdot 2 - 0 \cdot 1) - \hat{j}(3 \cdot 2 - 0 \cdot 0) + \hat{k}(3 \cdot 1 - 0 \cdot 1) \]
\[ = 2\hat{i} - 6\hat{j} + 3\hat{k} \]

2. **Calculate the magnitude of the cross product**:
\[ |\vec{A} \times \vec{B}| = \sqrt{(2)^2 + (-6)^2 + (3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]

### (iii) Find a unit vector which is perpendicular to both \( \vec{A} \) and \( \vec{B} \).

1. **Use the cross product calculated earlier**:
\[ \vec{A} \times \vec{B} = 2\hat{i} - 6\hat{j} + 3\hat{k} \]

2. **Calculate the magnitude again (already calculated)**:
\[ |\vec{A} \times \vec{B}| = 7 \]

3. **Find the unit vector**:
\[ \hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{2\hat{i} - 6\hat{j} + 3\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{3}{7}\hat{k} \]

### Summary of Results:
1. Component of \( \vec{A} \) along \( \vec{B} \): \( \frac{\hat{j} + 2\hat{k}}{5} \)
2. Component of \( \vec{A} \) perpendicular to \( \vec{B} \): \( 3\hat{i} + \frac{4}{5}\hat{j} - \frac{2}{5}\hat{k} \)
3. Area of the parallelogram: \( 7 \)
4. Unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \): \( \frac{2}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{3}{7}\hat{k} \)