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 \(\vec{OA} \cdot \vec{OB}\) given the conditions about the points \(A\) and \(B\) on the curve \(y = x^2\).

### Step-by-Step Solution:

1. **Identify the Points**:
- Let point \(A\) be \((x_1, y_1)\) in the first quadrant, so \(y_1 = x_1^2\).
- Let point \(B\) be \((x_2, y_2)\) in the second quadrant, so \(y_2 = x_2^2\).

2. **Express Vectors**:
- The vector \(\vec{OA} = x_1 \hat{i} + y_1 \hat{j} = x_1 \hat{i} + x_1^2 \hat{j}\).
- The vector \(\vec{OB} = x_2 \hat{i} + y_2 \hat{j} = x_2 \hat{i} + x_2^2 \hat{j}\).

3. **Use Given Conditions**:
- From the problem, we have:
\[ \vec{OA} \cdot \hat{i} = 1 \implies x_1 = 1 \]
- Therefore, \(y_1 = x_1^2 = 1^2 = 1\). Thus, point \(A\) is \((1, 1)\).

4. **Find Point B**:
- The second condition is:
\[ \vec{OB} \cdot \hat{j} = 4 \implies y_2 = 4 \]
- Since \(y_2 = x_2^2\), we have:
\[ x_2^2 = 4 \implies x_2 = \pm 2 \]
- Since point \(B\) is in the second quadrant, we take \(x_2 = -2\). Thus, point \(B\) is \((-2, 4)\).

5. **Write the Vectors**:
- Now we can write:
\[ \vec{OA} = 1 \hat{i} + 1 \hat{j} \]
\[ \vec{OB} = -2 \hat{i} + 4 \hat{j} \]

6. **Calculate the Dot Product**:
- Now we calculate \(\vec{OA} \cdot \vec{OB}\):
\[ \vec{OA} \cdot \vec{OB} = (1 \hat{i} + 1 \hat{j}) \cdot (-2 \hat{i} + 4 \hat{j}) \]
- Using the properties of dot product:
\[ = (1)(-2) + (1)(4) = -2 + 4 = 2 \]

### Final Answer:
\[ \vec{OA} \cdot \vec{OB} = 2 \]