Consider east as positive x-axis, north as positive y-axis. A girl walks `10 m` east first time than `10 m` in a direction `30^(@)` west of north for the second time and then third time in unknown direction and magnitude so as to return to her initial position. What is her third displacement in unit vector notation.?

Options

A. `-5hat(i)-5sqrt(3)hat(j)`
B. `5hat(i)-5sqrt(3)hat(j)`
C. `-5hat(i)+5sqrt(3)hat(j)`
D. She cannot return

Explanation

To solve the problem, we need to determine the third displacement vector of the girl after she has walked in two specified directions. Let's break down the steps:

### Step 1: Define the first displacement vector
The girl walks 10 meters east. In unit vector notation, this can be represented as:
\[ \vec{d_1} = 10 \hat{i} \]
where \(\hat{i}\) is the unit vector in the east direction (positive x-axis).

### Step 2: Define the second displacement vector
The girl then walks 10 meters in a direction that is 30 degrees west of north. To find the components of this vector, we need to break it down into its x (east-west) and y (north-south) components.

1. The angle with respect to the north (positive y-axis) is \(30^\circ\).
2. The x-component (west is negative x) is given by:
\[ d_{2x} = -10 \sin(30^\circ) = -10 \cdot \frac{1}{2} = -5 \]
3. The y-component (north is positive y) is given by:
\[ d_{2y} = 10 \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

Thus, the second displacement vector can be represented as:
\[ \vec{d_2} = -5 \hat{i} + 5\sqrt{3} \hat{j} \]

### Step 3: Find the resultant displacement after the first two movements
Now, we can find the total displacement after the first two movements:
\[ \vec{R} = \vec{d_1} + \vec{d_2} \]
Substituting the values:
\[ \vec{R} = (10 \hat{i}) + (-5 \hat{i} + 5\sqrt{3} \hat{j}) = (10 - 5) \hat{i} + 5\sqrt{3} \hat{j} = 5 \hat{i} + 5\sqrt{3} \hat{j} \]

### Step 4: Determine the third displacement vector
Since the girl returns to her initial position, the total displacement must equal zero:
\[ \vec{d_1} + \vec{d_2} + \vec{d_3} = 0 \]
This implies:
\[ \vec{d_3} = -\vec{R} = - (5 \hat{i} + 5\sqrt{3} \hat{j}) = -5 \hat{i} - 5\sqrt{3} \hat{j} \]

### Final Answer
The third displacement vector in unit vector notation is:
\[ \vec{d_3} = -5 \hat{i} - 5\sqrt{3} \hat{j} \]

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