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

Options

A. `sqrt(13.5)` unit
B. 13.5 unit
C. `sqrt(38.7)` unit
D. 38.7 unit

Explanation

To find the area of the triangle formed by the vectors \(\vec{A} = 2\hat{i} - 3\hat{j} + 4\hat{k}\) and \(\vec{B} = \hat{i} - \hat{k}\), we can use the formula for the area of a triangle formed by two vectors:

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

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

To find the cross product, we can use the determinant of a matrix formed by the unit vectors and the components of the vectors \(\vec{A}\) 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: Calculate 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 \(\hat{i}\):
\[ \begin{vmatrix} -3 & 4 \\ 0 & -1 \end{vmatrix} = (-3)(-1) - (4)(0) = 3 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & 4 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (4)(1) = -2 - 4 = -6 \]

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

Putting it all together:

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

### 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

Finally, we can find the area of the triangle:

\[ \text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}| = \frac{1}{2} \sqrt{54} = \frac{\sqrt{54}}{2} \]

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

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

Thus, the area becomes:

\[ \text{Area} = \frac{3\sqrt{6}}{2} \]

### Final Answer

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

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