If two points A and B lie on the curve `y=x^(2)` such that `vec(OA).hati=1 and vec(OB).hatj=4`, where O is origin and A and B lie in the `1^("st")` and `2^("nd")` quadrant respectively, then `vec(OA).vec(OB)` is equal to

Options

A. 0
B. 2
C. 4
D. 5

Answer

To solve the problem, we need to find the dot product of the vectors \( \vec{OA} \) and \( \vec{OB} \) given the conditions specified in the question.

### Step-by-Step Solution:

1. **Understanding the Points on the Curve:**
The points \( A \) and \( B \) lie on the curve \( y = x^2 \). Therefore, if \( A \) has coordinates \( (x_1, y_1) \), then \( y_1 = x_1^2 \). Similarly, for point \( B \) with coordinates \( (x_2, y_2) \), we have \( y_2 = x_2^2 \).

2. **Expressing Vectors:**
The position vector \( \vec{OA} \) can be expressed as:
\[ \vec{OA} = x_1 \hat{i} + y_1 \hat{j} = x_1 \hat{i} + x_1^2 \hat{j} \]
The position vector \( \vec{OB} \) can be expressed as:
\[ \vec{OB} = x_2 \hat{i} + y_2 \hat{j} = x_2 \hat{i} + x_2^2 \hat{j} \]

3. **Using Given Conditions:**
- The condition \( \vec{OA} \cdot \hat{i} = 1 \) implies:
\[ x_1 = 1 \]
- The condition \( \vec{OB} \cdot \hat{j} = 4 \) implies:
\[ x_2^2 = 4 \implies x_2 = \pm 2 \]
Since point \( B \) lies in the second quadrant, we take \( x_2 = -2 \).

4. **Finding Coordinates:**
Now we can find the coordinates of points \( A \) and \( B \):
- For point \( A \):
\[ A = (1, 1^2) = (1, 1) \]
- For point \( B \):
\[ B = (-2, (-2)^2) = (-2, 4) \]

5. **Calculating the Dot Product:**
Now we can find the dot product \( \vec{OA} \cdot \vec{OB} \):
\[ \vec{OA} = 1 \hat{i} + 1 \hat{j} \]
\[ \vec{OB} = -2 \hat{i} + 4 \hat{j} \]
The dot product is calculated as:
\[ \vec{OA} \cdot \vec{OB} = (1)(-2) + (1)(4) = -2 + 4 = 2 \]

### Final Answer:
Thus, \( \vec{OA} \cdot \vec{OB} = 2 \).