The area of a triangle formed by vertices O,A and B where `vec(OA) = hat(i) + 2 hat(j) + 3 hat(k)` and `vec(OB) = - 3hat(i) - 2 hat(j) + hat(k)` is

Options

A. `3 sqrt(5)` sq. units
B. `5 sqrt(5)` sq. unit
C. `6 sqrt(5)` sq. units
D. 4 sq. units

Answer

To find the area of the triangle formed by the vertices O, A, and B, we will follow these steps:

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

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

We can represent this as a determinant:

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

### Step 3: Calculate the determinant
Now, we will calculate the determinant step by step:

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

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} = (2)(1) - (3)(-2) = 2 + 6 = 8\)

2. \(\begin{vmatrix} 1 & 3 \\ -3 & 1 \end{vmatrix} = (1)(1) - (3)(-3) = 1 + 9 = 10\)

3. \(\begin{vmatrix} 1 & 2 \\ -3 & -2 \end{vmatrix} = (1)(-2) - (2)(-3) = -2 + 6 = 4\)

Now substituting back into the equation:

\[ \vec{OA} \times \vec{OB} = 8\hat{i} - 10\hat{j} + 4\hat{k} \]

### Step 4: Find the magnitude of the cross product
The magnitude of the cross product is given by:

\[ |\vec{OA} \times \vec{OB}| = \sqrt{8^2 + (-10)^2 + 4^2} \]

Calculating each term:

\[ = \sqrt{64 + 100 + 16} = \sqrt{180} \]

### Step 5: Calculate the area of the triangle
The area \(A\) of the triangle is half the magnitude of the cross product:

\[ A = \frac{1}{2} |\vec{OA} \times \vec{OB}| = \frac{1}{2} \sqrt{180} = \frac{\sqrt{180}}{2} \]

### Step 6: Simplify the area
We can simplify \(\sqrt{180}\):

\[ \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \]

Thus, the area becomes:

\[ A = \frac{6\sqrt{5}}{2} = 3\sqrt{5} \]

### Final Answer
The area of the triangle formed by vertices O, A, and B is \(3\sqrt{5}\).

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