The adjacent sides AB and AC of a triangle ABC are represented by the vectors `-2 hat(i) + 3 hat(j) + 2 hat(k) and -4hat(i) + 5 hat(j) + 2 hat(k)` respectively. The area of the triangle ABC is

Options

A. 6 square units
B. 5 square units
C. 4 square units
D. 3 square units

Explanation

To find the area of triangle ABC given the vectors for adjacent sides AB and AC, we can follow these steps:

### Step 1: Identify the vectors
The vectors representing the sides of the triangle are given as:
- Vector AB = \(-2 \hat{i} + 3 \hat{j} + 2 \hat{k}\)
- Vector AC = \(-4 \hat{i} + 5 \hat{j} + 2 \hat{k}\)

### Step 2: Calculate the cross product of the vectors
The area of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \]
To find the cross product \(\vec{AB} \times \vec{AC}\), we can use the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of the vectors AB and AC:

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 2 \\ -4 & 5 & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula, we can expand it as follows:

\[ \vec{AB} \times \vec{AC} = \hat{i} \begin{vmatrix} 3 & 2 \\ 5 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 2 \\ -4 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 3 \\ -4 & 5 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} 3 & 2 \\ 5 & 2 \end{vmatrix} = (3 \cdot 2) - (2 \cdot 5) = 6 - 10 = -4 \]
2. For \(\hat{j}\):
\[ \begin{vmatrix} -2 & 2 \\ -4 & 2 \end{vmatrix} = (-2 \cdot 2) - (2 \cdot -4) = -4 + 8 = 4 \]
3. For \(\hat{k}\):
\[ \begin{vmatrix} -2 & 3 \\ -4 & 5 \end{vmatrix} = (-2 \cdot 5) - (3 \cdot -4) = -10 + 12 = 2 \]

Putting it all together:
\[ \vec{AB} \times \vec{AC} = -4 \hat{i} - 4 \hat{j} + 2 \hat{k} \]

### Step 4: Calculate the magnitude of the cross product
Now, we find the magnitude of the vector:
\[ \left| \vec{AB} \times \vec{AC} \right| = \sqrt{(-4)^2 + (-4)^2 + (2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \]

### Step 5: Calculate the area of the triangle
Finally, we can find the area of triangle ABC:
\[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| = \frac{1}{2} \cdot 6 = 3 \]

Thus, the area of triangle ABC is \(3\) square units.

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