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

Options

A. A) `5 sqrt6` sq units
B. B) `(5 sqrt6)/(2)` sq units
C. C) `sqrt6` sq units
D. D) `sqrt30` sq units

Answer

To find the area of triangle OAB where O is the origin, OA = \(3 \hat{i} - \hat{j} + \hat{k}\) and OB = \(2 \hat{i} + \hat{j} - 3 \hat{k}\), we can follow these steps:

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

### Step 2: Use the formula for the area of triangle in terms of vectors
The area \(A\) of triangle OAB can be calculated using the formula:
\[ A = \frac{1}{2} |\vec{OA} \times \vec{OB}| \]
where \(\times\) denotes the cross product.

### Step 3: Calculate the cross product \(\vec{OA} \times \vec{OB}\)
To find the cross product, we can use the determinant method:
\[ \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 2 & 1 & -3 \end{vmatrix} \]

### Step 4: Expand the determinant
Expanding the determinant:
\[ \vec{OA} \times \vec{OB} = \hat{i} \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 2 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ (-1)(-3) - (1)(1) = 3 - 1 = 2 \]
2. For \(-\hat{j}\):
\[ (3)(-3) - (1)(2) = -9 - 2 = -11 \quad \Rightarrow \quad +11 \hat{j} \]
3. For \(\hat{k}\):
\[ (3)(1) - (-1)(2) = 3 + 2 = 5 \]

Combining these results:
\[ \vec{OA} \times \vec{OB} = 2 \hat{i} + 11 \hat{j} + 5 \hat{k} \]

### Step 5: Calculate the magnitude of the cross product
Now, we find the magnitude:
\[ |\vec{OA} \times \vec{OB}| = \sqrt{(2)^2 + (11)^2 + (5)^2} = \sqrt{4 + 121 + 25} = \sqrt{150} \]

### Step 6: Simplify the magnitude
\[ \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6} \]

### Step 7: Calculate the area of triangle OAB
Substituting back into the area formula:
\[ A = \frac{1}{2} |\vec{OA} \times \vec{OB}| = \frac{1}{2} (5\sqrt{6}) = \frac{5\sqrt{6}}{2} \]

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

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