Two forces `vec(F)_(1)=500N` due east and `vec(F)_(2)=250N` due north have their common initial point. `vec(F)_(2)-vec(F)_(1)` is

Options

A. `250sqrt(5)N,tan^(-1)(2) W of N`
B. `250 N, tan^(-1) (2) W` of N
C. Zero
D. `750N,tan^(-1)(3//4) N of W`

Explanation

To solve the problem of finding the vector difference \( \vec{F}_2 - \vec{F}_1 \), we will follow these steps:

### Step 1: Identify the Forces
We have two forces:
- \( \vec{F}_1 = 500 \, \text{N} \) due east
- \( \vec{F}_2 = 250 \, \text{N} \) due north

### Step 2: Represent the Forces as Vectors
We can represent these forces in vector form using Cartesian coordinates:
- \( \vec{F}_1 = 500 \hat{i} \) (where \( \hat{i} \) represents the east direction)
- \( \vec{F}_2 = 250 \hat{j} \) (where \( \hat{j} \) represents the north direction)

### Step 3: Calculate the Vector Difference
To find \( \vec{F}_2 - \vec{F}_1 \):
\[ \vec{F}_2 - \vec{F}_1 = 250 \hat{j} - 500 \hat{i} \]
This can be expressed as:
\[ \vec{F}_2 - \vec{F}_1 = -500 \hat{i} + 250 \hat{j} \]

### Step 4: Find the Magnitude of the Resultant Vector
To find the magnitude of the resultant vector \( \vec{R} = -500 \hat{i} + 250 \hat{j} \), we use the Pythagorean theorem:
\[ |\vec{R}| = \sqrt{(-500)^2 + (250)^2} \]
Calculating the squares:
\[ |\vec{R}| = \sqrt{250000 + 62500} = \sqrt{312500} \]
\[ |\vec{R}| = 250 \sqrt{5} \, \text{N} \]

### Step 5: Determine the Direction of the Resultant Vector
To find the angle \( \theta \) that \( \vec{R} \) makes with the negative x-axis (east direction), we can use the tangent function:
\[ \tan \theta = \frac{|\text{opposite}|}{|\text{adjacent}|} = \frac{250}{500} = \frac{1}{2} \]
Thus,
\[ \theta = \tan^{-1}(0.5) \]

### Step 6: Final Result
The vector \( \vec{F}_2 - \vec{F}_1 \) has a magnitude of \( 250 \sqrt{5} \, \text{N} \) and makes an angle \( \theta = \tan^{-1}(0.5) \) with the negative x-axis (east direction).