What is the area of triangle formed by `vecA=2hati-3hatj+4hatk` and `vecB=hati-hatk` and their Resultant ?

Options

A. `sqrt13.5` units
B. 13.5 units
C. `sqrt38.7` units
D. 38.7 units

Explanation

To find the area of the triangle formed by the vectors \(\vec{A} = 2\hat{i} - 3\hat{j} + 4\hat{k}\) and \(\vec{B} = \hat{i} - \hat{k}\), and their resultant, we can follow these steps:

### Step 1: Calculate the Cross Product of \(\vec{A}\) and \(\vec{B}\)

The cross product \(\vec{A} \times \vec{B}\) can be calculated using the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of \(\vec{A}\) and \(\vec{B}\).

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 1 & 0 & -1 \end{vmatrix} \]

### Step 2: Expand the Determinant

Calculating the determinant, we have:

\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -3 & 4 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ 1 & 0 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ (-3)(-1) - (4)(0) = 3 \]

2. For \(-\hat{j}\):
\[ (2)(-1) - (4)(1) = -2 - 4 = -6 \quad \Rightarrow \quad +6\hat{j} \]

3. For \(\hat{k}\):
\[ (2)(0) - (-3)(1) = 0 + 3 = 3 \]

Thus, we have:

\[ \vec{A} \times \vec{B} = 3\hat{i} + 6\hat{j} + 3\hat{k} \]

### Step 3: Calculate the Magnitude of the Cross Product

Now, we find the magnitude of \(\vec{A} \times \vec{B}\):

\[ |\vec{A} \times \vec{B}| = \sqrt{3^2 + 6^2 + 3^2} = \sqrt{9 + 36 + 9} = \sqrt{54} \]

### Step 4: Calculate the Area of the Triangle

The area \(A\) of the triangle formed by the vectors is given by:

\[ A = \frac{1}{2} |\vec{A} \times \vec{B}| \]

Substituting the magnitude we found:

\[ A = \frac{1}{2} \sqrt{54} = \frac{\sqrt{54}}{2} \]

### Step 5: Simplify the Area

We can simplify \(\sqrt{54}\):

\[ \sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6} \]

Thus, the area becomes:

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

### Final Answer

The area of the triangle formed by the vectors \(\vec{A}\) and \(\vec{B}\) is:

\[ \frac{3\sqrt{6}}{2} \text{ square units} \]

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