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 systematically.

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

### Part (i): Find the component of \(\vec{A}\) along \(\vec{B}\) and perpendicular to \(\vec{B}\) in vector form.

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

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

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

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

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

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

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

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

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

#### Step 1: Use the result from the cross product.
The unit vector perpendicular to both \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{2\hat{i} - 6\hat{j} + 3\hat{k}}{7} \]
\[ = \frac{2}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{3}{7}\hat{k} \]

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