What is the area of the triangle OAB where O is the origin, `vec(OA)=3hat(i)-hat(j)+hat(k) and vec(OB)=2hat(i)+hat(j)-3hat(k)` ?

Options

A. `5sqrt(6)` square unit
B. `(5sqrt(6))/(2)` square unit
C. `sqrt(6)` square unit
D. `sqrt(30)` square unit

Explanation

To find the area of triangle OAB where O is the origin, and the vectors \(\vec{OA}\) and \(\vec{OB}\) are given, we can follow these steps:

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

### Step 2: Compute the cross product \(\vec{OA} \times \vec{OB}\)
To find the area of triangle OAB, we need to calculate the cross product \(\vec{OA} \times \vec{OB}\).

Using the determinant method:
\[ \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 2 & 1 & -3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 2 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} = (-1)(-3) - (1)(1) = 3 - 1 = 2\)
2. \(\begin{vmatrix} 3 & 1 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (1)(2) = -9 - 2 = -11\)
3. \(\begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = (3)(1) - (-1)(2) = 3 + 2 = 5\)

Putting it all together:
\[ \vec{OA} \times \vec{OB} = 2\hat{i} + 11\hat{j} + 5\hat{k} \]

### Step 3: Calculate the magnitude of the cross product
The magnitude of the vector \(\vec{OA} \times \vec{OB}\) is given by:
\[ |\vec{OA} \times \vec{OB}| = \sqrt{(2)^2 + (11)^2 + (5)^2} = \sqrt{4 + 121 + 25} = \sqrt{150} \]

### Step 4: Simplify the magnitude
We can simplify \(\sqrt{150}\):
\[ \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6} \]

### Step 5: Calculate the area of triangle OAB
The area \(A\) of triangle OAB is given by:
\[ A = \frac{1}{2} |\vec{OA} \times \vec{OB}| = \frac{1}{2} \times 5\sqrt{6} = \frac{5\sqrt{6}}{2} \]

Thus, the area of triangle OAB is:
\[ \boxed{\frac{5\sqrt{6}}{2}} \text{ square units} \]

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