The area of the triangle whose two sides are given by `2i-7j+k and 4j-3k` is

Options

A. 17
B. `17//2`
C. `17//4`
D. `(1)/(2)sqrt(389)`

Answer

To find the area of the triangle formed by the two vectors \( \vec{A} = 2\hat{i} - 7\hat{j} + \hat{k} \) and \( \vec{B} = 4\hat{j} - 3\hat{k} \), we can use the formula for the area of a triangle given by two vectors:

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

### Step 1: Identify the vectors
We have:
\[ \vec{A} = 2\hat{i} - 7\hat{j} + \hat{k} \]
\[ \vec{B} = 0\hat{i} + 4\hat{j} - 3\hat{k} \]

### Step 2: Set up the cross product
The cross product \( \vec{A} \times \vec{B} \) can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 1 \\ 0 & 4 & -3 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we expand it as follows:

\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -7 & 1 \\ 4 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 0 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -7 \\ 0 & 4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ (-7)(-3) - (1)(4) = 21 - 4 = 17 \]

2. For \( \hat{j} \):
\[ (2)(-3) - (1)(0) = -6 - 0 = -6 \]
(Remember to subtract this term, so it becomes \( +6\hat{j} \))

3. For \( \hat{k} \):
\[ (2)(4) - (-7)(0) = 8 - 0 = 8 \]

Putting it all together:
\[ \vec{A} \times \vec{B} = 17\hat{i} + 6\hat{j} + 8\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now we find the magnitude of \( \vec{A} \times \vec{B} \):
\[ \left| \vec{A} \times \vec{B} \right| = \sqrt{(17)^2 + (6)^2 + (8)^2} \]
Calculating each term:
\[ 17^2 = 289, \quad 6^2 = 36, \quad 8^2 = 64 \]
Adding these:
\[ \left| \vec{A} \times \vec{B} \right| = \sqrt{289 + 36 + 64} = \sqrt{389} \]

### Step 5: Calculate the area of the triangle
Finally, substituting back into the area formula:
\[ \text{Area} = \frac{1}{2} \sqrt{389} \]

### Final Answer
Thus, the area of the triangle is:
\[ \text{Area} = \frac{1}{2} \sqrt{389} \]

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