The vectors from origin to the points A and B are `vec(a) = 2 hat(i) - 3 hat(j) + 2 hat(k)` and `vec( b) = 2 hat(i) + 3 hat(j) + hat(k)`, respectively, then the area ( in sq. units ) of `Delta ABC` is

Options

A. `sqrt( 340)`
B. 5
C. `sqrt(229)`
D. `(1)/(2) sqrt( 229)`

Answer

To find the area of triangle ABC formed by the points A, B, and the origin O, we can use the formula for the area of a triangle given by vectors. The area \( \text{Area} \) of triangle ABC can be calculated using the formula:

\[ \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| \]

Where \( \vec{AB} \) and \( \vec{AC} \) are the vectors from A to B and from A to C, respectively. In this case, we can take point C as the origin O.

### Step 1: Define the vectors
The vectors from the origin to points A and B are given as:
\[ \vec{A} = 2\hat{i} - 3\hat{j} + 2\hat{k} \]
\[ \vec{B} = 2\hat{i} + 3\hat{j} + \hat{k} \]

### Step 2: Find the vector \( \vec{AB} \)
The vector \( \vec{AB} \) is calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} \]
Calculating this gives:
\[ \vec{AB} = (2\hat{i} + 3\hat{j} + \hat{k}) - (2\hat{i} - 3\hat{j} + 2\hat{k}) = 0\hat{i} + 6\hat{j} - 1\hat{k} = 6\hat{j} - \hat{k} \]

### Step 3: Find the vector \( \vec{AC} \)
The vector \( \vec{AC} \) is simply the vector \( \vec{A} \) since C is the origin:
\[ \vec{AC} = \vec{A} = 2\hat{i} - 3\hat{j} + 2\hat{k} \]

### Step 4: Calculate the cross product \( \vec{AB} \times \vec{AC} \)
We will calculate the cross product using the determinant of a matrix:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 6 & -1 \\ 2 & -3 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 6 & -1 \\ -3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & -1 \\ 2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 6 \\ 2 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ 6 \cdot 2 - (-1)(-3) = 12 - 3 = 9 \]
2. For \( -\hat{j} \):
\[ 0 \cdot 2 - (-1)(2) = 0 + 2 = 2 \]
3. For \( \hat{k} \):
\[ 0 \cdot (-3) - 6 \cdot 2 = 0 - 12 = -12 \]

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

### Step 5: Calculate the magnitude of the cross product
Now we find the magnitude:
\[ \| \vec{AB} \times \vec{AC} \| = \sqrt{9^2 + (-2)^2 + (-12)^2} = \sqrt{81 + 4 + 144} = \sqrt{229} \]

### Step 6: Calculate the area of triangle ABC
Finally, we can find the area:
\[ \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} \sqrt{229} \]

Thus, the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \sqrt{229} \text{ square units} \]

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