An insect starts from the origin in the argand plane and goes 4 km `(N 45^(@)E)` then it moves 3 km `(N 45^(@)W)` and then takes an angular movement of `(pi)/(3)` about origin in the anticlockwise direction. The final position of the insect is

Options

A. `(4-3i)e^((-5pi)/(6))`
B. `(4+3i)e^((-5pi)/(6))`
C. `(4-3i)e^((i3pi)/(4))`
D. `(4+3i)e^((-7pi)/(12))`

Answer

To solve the problem step by step, let's break down the movements of the insect in the Argand plane.

### Step 1: Determine the position after the first movement
The insect starts at the origin (0, 0) and moves 4 km in the direction of N 45° E.

In polar coordinates, this can be expressed as:
- The angle θ = 45° = π/4 radians
- The distance r = 4 km

Using the conversion from polar to Cartesian coordinates:
- x = r * cos(θ) = 4 * cos(π/4) = 4 * (√2/2) = 2√2
- y = r * sin(θ) = 4 * sin(π/4) = 4 * (√2/2) = 2√2

Thus, the position after the first movement (Point A) is:
\[ A = (2\sqrt{2}, 2\sqrt{2}) \]

### Step 2: Determine the position after the second movement
Next, the insect moves 3 km in the direction of N 45° W.

In polar coordinates:
- The angle θ = 180° - 45° = 135° = 3π/4 radians
- The distance r = 3 km

Using the conversion from polar to Cartesian coordinates:
- x = r * cos(θ) = 3 * cos(3π/4) = 3 * (-√2/2) = -3√2/2
- y = r * sin(θ) = 3 * sin(3π/4) = 3 * (√2/2) = 3√2/2

Now, we add this movement to the coordinates of Point A:
- New x-coordinate = 2√2 - 3√2/2 = (4√2/2 - 3√2/2) = (1√2/2) = √2/2
- New y-coordinate = 2√2 + 3√2/2 = (4√2/2 + 3√2/2) = (7√2/2)

Thus, the position after the second movement (Point B) is:
\[ B = \left(\frac{\sqrt{2}}{2}, \frac{7\sqrt{2}}{2}\right) \]

### Step 3: Apply the angular movement
The insect then takes an angular movement of \( \frac{\pi}{3} \) radians (60°) in the anticlockwise direction about the origin.

To rotate a point (x, y) by an angle θ about the origin, we use the rotation matrix:
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \]

Substituting \( \theta = \frac{\pi}{3} \):
- cos(π/3) = 1/2
- sin(π/3) = √3/2

Now substituting the coordinates of Point B:
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} \frac{\sqrt{2}}{2} \\ \frac{7\sqrt{2}}{2} \end{pmatrix} \]

Calculating the new coordinates:
\[ x' = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{7\sqrt{2}}{2} = \frac{\sqrt{2}}{4} - \frac{7\sqrt{6}}{4} = \frac{\sqrt{2} - 7\sqrt{6}}{4} \]
\[ y' = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{7\sqrt{2}}{2} = \frac{\sqrt{6}}{4} + \frac{7\sqrt{2}}{4} = \frac{\sqrt{6} + 7\sqrt{2}}{4} \]

Thus, the final position of the insect is:
\[ \text{Final Position} = \left(\frac{\sqrt{2} - 7\sqrt{6}}{4}, \frac{\sqrt{6} + 7\sqrt{2}}{4}\right) \]