A class XII student appearing for a competitive examination was asked to attempt the following questions.
Let `vec(a),vec(b)` and `vec( c)` be theree non zero vectors. If `vec (a)=hat(i)-2hat(j)`, `vec(b)=2hat(i)+hat(j)+3hat(k)` then
The area of the parallelogram formed by `vec(a)` and `vec(b)` as diagonals is

Options

A. 70
B. 35
C. `(sqrt(70))/(2)`
D. `sqrt(70)`

Answer

To find the area of the parallelogram formed by the vectors \(\vec{a}\) and \(\vec{b}\) as diagonals, we can follow these steps:

### Step 1: Identify the vectors
Given:
\[ \vec{a} = \hat{i} - 2\hat{j} \]
\[ \vec{b} = 2\hat{i} + \hat{j} + 3\hat{k} \]

### Step 2: Calculate the cross product \(\vec{a} \times \vec{b}\)
The area of the parallelogram formed by two vectors \(\vec{u}\) and \(\vec{v}\) is given by the magnitude of the cross product of the vectors. We will first set up the determinant for the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 2 & 1 & 3 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -2 & 0 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -2 & 0 \\ 1 & 3 \end{vmatrix} = (-2)(3) - (0)(1) = -6\)
2. \(\begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix} = (1)(3) - (0)(2) = 3\)
3. \(\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 + 4 = 5\)

Putting it all together:
\[ \vec{a} \times \vec{b} = -6\hat{i} - 3\hat{j} + 5\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now we find the magnitude of \(\vec{a} \times \vec{b}\):
\[ |\vec{a} \times \vec{b}| = \sqrt{(-6)^2 + (-3)^2 + (5)^2} \]
Calculating the squares:
\[ = \sqrt{36 + 9 + 25} = \sqrt{70} \]

### Step 5: Calculate the area of the parallelogram
The area \(A\) of the parallelogram formed by the vectors \(\vec{a}\) and \(\vec{b}\) as diagonals is given by:
\[ A = \frac{1}{2} |\vec{a} \times \vec{b}| \]
Substituting the magnitude we found:
\[ A = \frac{1}{2} \sqrt{70} \]

### Final Answer
Thus, the area of the parallelogram is:
\[ \frac{\sqrt{70}}{2} \]