The area of the triangle formed by the adjacent sides with `vecA = 3hati + 2hatj - 4 hatk and vecB = -hati + 2hatj + hatk` is

Options

A. `(sqrt(165))/(2)` units
B. `(sqrt(137))/(2)` units
C. `sqrt(165)` units
D. `sqrt(137)` units

Explanation

To find the area of the triangle formed by the vectors \(\vec{A}\) and \(\vec{B}\), we can follow these steps:

### Step 1: Identify the vectors
We have:
\[ \vec{A} = 3\hat{i} + 2\hat{j} - 4\hat{k} \]
\[ \vec{B} = -\hat{i} + 2\hat{j} + \hat{k} \]

### Step 2: Calculate the cross product \(\vec{A} \times \vec{B}\)
The area of the triangle formed by the vectors is given by the formula:
\[ \text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}| \]
To find \(\vec{A} \times \vec{B}\), we will use the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of \(\vec{A}\) and \(\vec{B}\):

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -4 \\ -1 & 2 & 1 \end{vmatrix} \]

Calculating the determinant, we have:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 2 & -4 \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -4 \\ -1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ -1 & 2 \end{vmatrix} \]

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

Putting it all together:
\[ \vec{A} \times \vec{B} = 10\hat{i} - 1\hat{j} + 8\hat{k} = 10\hat{i} + 1\hat{j} + 8\hat{k} \]

### Step 3: Calculate the magnitude of the cross product
Now we find the magnitude:
\[ |\vec{A} \times \vec{B}| = \sqrt{10^2 + 1^2 + 8^2} = \sqrt{100 + 1 + 64} = \sqrt{165} \]

### Step 4: Calculate the area of the triangle
Now, substituting back into the area formula:
\[ \text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}| = \frac{1}{2} \sqrt{165} \]

### Final Answer
Thus, the area of the triangle formed by the vectors \(\vec{A}\) and \(\vec{B}\) is:
\[ \text{Area} = \frac{1}{2} \sqrt{165} \]

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