The vectors from origin to the points A and B are `vecA=3hati-6hatj+2hatk` and `vecB=2hati+hatj+2hatk` respectively. The area of triangle `OAB` be

Options

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

Explanation

To find the area of triangle OAB formed by the vectors \(\vec{A}\) and \(\vec{B}\), we can use the formula for the area of a triangle given by two vectors originating from the same point. The area \(A\) of triangle OAB can be calculated using the formula:

\[ A = \frac{1}{2} |\vec{A} \times \vec{B}| \]

### Step 1: Define the vectors
We are given:
\[ \vec{A} = 3\hat{i} - 6\hat{j} + 2\hat{k} \]
\[ \vec{B} = 2\hat{i} + \hat{j} + 2\hat{k} \]

### Step 2: Calculate the cross product \(\vec{A} \times \vec{B}\)
To find the cross product, we will use the determinant of a matrix formed by the unit vectors and the components of \(\vec{A}\) and \(\vec{B}\):

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

Calculating this determinant, we expand it as follows:

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

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -6 & 2 \\ 1 & 2 \end{vmatrix} = (-6)(2) - (2)(1) = -12 - 2 = -14\)
2. \(\begin{vmatrix} 3 & 2 \\ 2 & 2 \end{vmatrix} = (3)(2) - (2)(2) = 6 - 4 = 2\)
3. \(\begin{vmatrix} 3 & -6 \\ 2 & 1 \end{vmatrix} = (3)(1) - (-6)(2) = 3 + 12 = 15\)

Putting it all together:
\[ \vec{A} \times \vec{B} = -14\hat{i} - 2(-2)\hat{j} + 15\hat{k} = -14\hat{i} - 2\hat{j} + 15\hat{k} \]

### Step 3: Find the magnitude of \(\vec{A} \times \vec{B}\)
Now we calculate the magnitude of the cross product:
\[ |\vec{A} \times \vec{B}| = \sqrt{(-14)^2 + (-2)^2 + (15)^2} \]
Calculating each term:
\[ = \sqrt{196 + 4 + 225} = \sqrt{425} = 5\sqrt{17} \]

### Step 4: Calculate the area of triangle OAB
Now we can find the area:
\[ A = \frac{1}{2} |\vec{A} \times \vec{B}| = \frac{1}{2} (5\sqrt{17}) = \frac{5\sqrt{17}}{2} \]

Thus, the area of triangle OAB is:
\[ \boxed{\frac{5\sqrt{17}}{2}} \text{ square units} \]