The vectors from origin to the points A and B are `vec(a)=2hat(i)-3hat(j)+2hat(k)` and `vec(b)=2hat(i)+3hat(j)+hat(k)` respectively then the area of triangle OAB is

Options

A. 340
B. `sqrt(25)`
C. `sqrt(229)`
D. `(1)/(2)sqrt(229)`

Answer

To find the area of triangle OAB formed by the vectors \(\vec{a}\) and \(\vec{b}\), we can follow these steps:

### Step 1: Identify the vectors
Given:
\[ \vec{a} = 2\hat{i} - 3\hat{j} + 2\hat{k} \]
\[ \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k} \]

### Step 2: Calculate the cross product \(\vec{a} \times \vec{b}\)
To find the area of triangle OAB, we need to calculate the cross product of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \]

### Step 3: Compute the determinant
Calculating the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -3 & 2 \\ 3 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ 2 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} -3 & 2 \\ 3 & 1 \end{vmatrix} = (-3)(1) - (2)(3) = -3 - 6 = -9 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} = (2)(1) - (2)(2) = 2 - 4 = -2 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & -3 \\ 2 & 3 \end{vmatrix} = (2)(3) - (-3)(2) = 6 + 6 = 12 \]

Putting it all together:
\[ \vec{a} \times \vec{b} = -9\hat{i} + 2\hat{j} + 12\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now, we find the magnitude of \(\vec{a} \times \vec{b}\):
\[ |\vec{a} \times \vec{b}| = \sqrt{(-9)^2 + 2^2 + 12^2} = \sqrt{81 + 4 + 144} = \sqrt{229} \]

### Step 5: Calculate the area of triangle OAB
The area \(A\) of triangle OAB is given by:
\[ A = \frac{1}{2} |\vec{a} \times \vec{b}| \]
Thus,
\[ A = \frac{1}{2} \sqrt{229} \]

### Final Answer
The area of triangle OAB is:
\[ \frac{1}{2} \sqrt{229} \]

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