The vectors from origin to the points A and B are `vec(a)=3hat(i)-6hat(j)+2hat(k) and vec(b)= 2hat(i) +hat(j)-2hat(k)` respectively. Find the area of :
(i) the triangle OAB
(ii) the parallelogram formed by `vec(OA) and vec(OB)` as adjacent sides.

Options

A. `(5)/(2)sqrt(17)`
B. `(2)/(5)sqrt(17)`
C. `(3)/(5)sqrt(17)`
D. `(5)/(3)sqrt(17)`

Answer

To solve the problem, we need to find the area of the triangle OAB and the area of the parallelogram formed by vectors OA and OB. Let's break it down step by step.

### Step 1: Define the Vectors
Given:
- \( \vec{A} = 3\hat{i} - 6\hat{j} + 2\hat{k} \)
- \( \vec{B} = 2\hat{i} + \hat{j} - 2\hat{k} \)

### Step 2: Find Vectors OA and OB
The vectors from the origin O to points A and B are:
- \( \vec{OA} = \vec{A} - \vec{O} = \vec{A} = 3\hat{i} - 6\hat{j} + 2\hat{k} \)
- \( \vec{OB} = \vec{B} - \vec{O} = \vec{B} = 2\hat{i} + \hat{j} - 2\hat{k} \)

### Step 3: Calculate the Cross Product \( \vec{OA} \times \vec{OB} \)
To find the area of the triangle, we need to calculate the cross product \( \vec{OA} \times \vec{OB} \).

\[ \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & 2 \\ 2 & 1 & -2 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -6 & 2 \\ 1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 2 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -6 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ (-6)(-2) - (2)(1) = 12 - 2 = 10 \]

2. For \( -\hat{j} \):
\[ (3)(-2) - (2)(2) = -6 - 4 = -10 \quad \Rightarrow \quad -(-10) = 10 \]

3. For \( \hat{k} \):
\[ (3)(1) - (-6)(2) = 3 + 12 = 15 \]

Putting it all together:
\[ \vec{OA} \times \vec{OB} = 10\hat{i} + 10\hat{j} + 15\hat{k} \]

### Step 4: Find the Magnitude of the Cross Product
Now, we calculate the magnitude of \( \vec{OA} \times \vec{OB} \):
\[ |\vec{OA} \times \vec{OB}| = \sqrt{(10)^2 + (10)^2 + (15)^2} = \sqrt{100 + 100 + 225} = \sqrt{425} \]

### Step 5: Calculate the Area of Triangle OAB
The area of triangle OAB is given by:
\[ \text{Area}_{\triangle OAB} = \frac{1}{2} |\vec{OA} \times \vec{OB}| = \frac{1}{2} \sqrt{425} = \frac{5\sqrt{17}}{2} \]

### Step 6: Calculate the Area of Parallelogram
The area of the parallelogram formed by vectors OA and OB is given by:
\[ \text{Area}_{\text{parallelogram}} = |\vec{OA} \times \vec{OB}| = \sqrt{425} = 5\sqrt{17} \]

### Final Answers
- Area of triangle OAB: \( \frac{5\sqrt{17}}{2} \)
- Area of the parallelogram: \( 5\sqrt{17} \)