In a football game, a player wants to throw a ball to his one of teammate, who is running on the field. Taken thrower position as origin and receiver initial position at `2hat(i) + 3hat(j)`, where `hat(i)` & `hat(j)` are in the plane of field. In subsequent run receiver displacement is `5hat(i)` then `8hat(j)`, then `–2hat(i) + 4hat(j)` then `–6hat(j)`. How far is receiver from thrower? (all displacements are in meter)

Options

A. `sqrt(106)`
B. `sqrt(110)`
C. `sqrt(103)`
D. `sqrt(112)`

Explanation

To solve the problem, we need to determine the final position of the receiver after all the displacements and then calculate the distance from the thrower (origin) to the receiver's final position.

### Step-by-Step Solution:

1. **Identify the Initial Position of the Receiver**:
The initial position of the receiver is given as:
\[ \vec{R_0} = 2\hat{i} + 3\hat{j} \]

2. **List the Displacements**:
The receiver undergoes the following displacements:
- First displacement: \( \vec{D_1} = 5\hat{i} \)
- Second displacement: \( \vec{D_2} = 8\hat{j} \)
- Third displacement: \( \vec{D_3} = -2\hat{i} + 4\hat{j} \)
- Fourth displacement: \( \vec{D_4} = -6\hat{j} \)

3. **Calculate the Resultant Displacement**:
We will sum all the displacements:
\[ \vec{R} = \vec{R_0} + \vec{D_1} + \vec{D_2} + \vec{D_3} + \vec{D_4} \]
Substituting the vectors:
\[ \vec{R} = (2\hat{i} + 3\hat{j}) + (5\hat{i}) + (8\hat{j}) + (-2\hat{i} + 4\hat{j}) + (-6\hat{j}) \]

4. **Combine Like Terms**:
Combine the \( \hat{i} \) components:
\[ \text{Total } \hat{i} = 2 + 5 - 2 = 5 \]
Combine the \( \hat{j} \) components:
\[ \text{Total } \hat{j} = 3 + 8 + 4 - 6 = 9 \]

5. **Final Position of the Receiver**:
Therefore, the final position vector of the receiver is:
\[ \vec{R} = 5\hat{i} + 9\hat{j} \]

6. **Calculate the Distance from the Thrower**:
The distance from the thrower (origin) to the receiver is the magnitude of the position vector \( \vec{R} \):
\[ |\vec{R}| = \sqrt{(5)^2 + (9)^2} = \sqrt{25 + 81} = \sqrt{106} \]

### Final Answer:
The distance from the thrower to the receiver is:
\[ \sqrt{106} \text{ meters} \]

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