Let ` vec(v_1) , vec(v_2) , vec(v_3), vec(v_4)` be unit vectors in the xy-plane, one each in the interior of the four quadrants. Which of the following statements is necessarily ture. ?

Options

A. `vec(v_1) + vec(v_2) + vec(v_3) + vec(v_4) =0`
B. There exist i, j with `1 le ilt j le 4 ` such ` vec(v_i) + vec(v_j)` is in the first quadrant
C. There exits i, j with ` 1 le ilt j le 4 ` such that ` vec(v_i)* vec(v_j) lt 0`
D. There exits i, j with ` 1 le ilt j le 4` such that ` vec(v_i)* vec(v_j) gt 0`

Answer

To solve the problem, we need to analyze the unit vectors \( \vec{v_1}, \vec{v_2}, \vec{v_3}, \vec{v_4} \) that lie in the interior of the four quadrants of the xy-plane. Each vector is a unit vector, meaning its magnitude is 1.

### Step-by-Step Solution:

1. **Understanding the Quadrants**:
- The xy-plane is divided into four quadrants:
- Quadrant I: \( x > 0, y > 0 \)
- Quadrant II: \( x < 0, y > 0 \)
- Quadrant III: \( x < 0, y < 0 \)
- Quadrant IV: \( x > 0, y < 0 \)

2. **Defining the Unit Vectors**:
- Let:
- \( \vec{v_1} = (a_1, b_1) \) in Quadrant I where \( a_1 > 0, b_1 > 0 \)
- \( \vec{v_2} = (a_2, b_2) \) in Quadrant II where \( a_2 < 0, b_2 > 0 \)
- \( \vec{v_3} = (a_3, b_3) \) in Quadrant III where \( a_3 < 0, b_3 < 0 \)
- \( \vec{v_4} = (a_4, b_4) \) in Quadrant IV where \( a_4 > 0, b_4 < 0 \)

3. **Magnitude of Unit Vectors**:
- Since all vectors are unit vectors, we have:
- \( \sqrt{a_1^2 + b_1^2} = 1 \)
- \( \sqrt{a_2^2 + b_2^2} = 1 \)
- \( \sqrt{a_3^2 + b_3^2} = 1 \)
- \( \sqrt{a_4^2 + b_4^2} = 1 \)

4. **Summing the Vectors**:
- We need to find the sum of these vectors:
\[ \vec{v_1} + \vec{v_2} + \vec{v_3} + \vec{v_4} = (a_1 + a_2 + a_3 + a_4, b_1 + b_2 + b_3 + b_4) \]

5. **Analyzing the Components**:
- The x-components:
- \( a_1 > 0 \) (from Quadrant I)
- \( a_2 < 0 \) (from Quadrant II)
- \( a_3 < 0 \) (from Quadrant III)
- \( a_4 > 0 \) (from Quadrant IV)
- The y-components:
- \( b_1 > 0 \) (from Quadrant I)
- \( b_2 > 0 \) (from Quadrant II)
- \( b_3 < 0 \) (from Quadrant III)
- \( b_4 < 0 \) (from Quadrant IV)

6. **Conclusion**:
- The positive x-component from Quadrants I and IV will be canceled out by the negative x-components from Quadrants II and III.
- Similarly, the positive y-components from Quadrants I and II will be canceled out by the negative y-components from Quadrants III and IV.
- Therefore, we conclude that:
\[ \vec{v_1} + \vec{v_2} + \vec{v_3} + \vec{v_4} = (0, 0) \]
- This means that the sum of the four vectors is necessarily equal to zero.

### Final Answer:
The statement that is necessarily true is:
\[ \vec{v_1} + \vec{v_2} + \vec{v_3} + \vec{v_4} = \vec{0} \]