1. Stating the problem: We want to create 5 vectors such that their resultant vector lies in each of the four quadrants of the coordinate plane. 2. Understanding vectors and quadrants: Each vector can be represented as $\mathbf{v_i} = (x_i, y_i)$. The resultant vector $\mathbf{R}$ is the sum: $$\mathbf{R} = \sum_{i=1}^5 \mathbf{v_i} = \left(\sum_{i=1}^5 x_i, \sum_{i=1}^5 y_i \right).$$ To have $\mathbf{R}$ in a specific quadrant, we need: - Quadrant I: $x > 0$, $y > 0$ - Quadrant II: $x < 0$, $y > 0$ - Quadrant III: $x < 0$, $y < 0$ - Quadrant IV: $x > 0$, $y < 0$ 3. Constructing vectors: We will choose 5 vectors and calculate their resultant to lie in each quadrant. For example: - Quadrant I: Let vectors be $(1,2), (2,1), (1,1), (0.5,2), (2,0.5)$. Sum $x=1+2+1+0.5+2=6.5 > 0$, Sum $y=2+1+1+2+0.5=6.5 > 0$ - Quadrant II: Let vectors be $(-2,3), (-1,2), (-0.5,1), (-1.5,0.5), (0,1)$. Sum $x = -2-1-0.5-1.5+0 = -5 < 0$, Sum $y=3+2+1+0.5+1 = 7.5 > 0$ - Quadrant III: Let vectors be $(-1,-2), (-2,-1), (-1,-1), (-0.5,-2), (-2,-0.5)$. Sum $x = -1-2-1-0.5-2 = -6.5 < 0$, Sum $y = -2-1-1-2-0.5 = -6.5 < 0$ - Quadrant IV: Let vectors be $(2,-3), (1,-2), (0.5,-1), (1.5,-0.5), (0,-1)$. Sum $x = 2+1+0.5+1.5+0 = 5 > 0$, Sum $y = -3-2-1-0.5-1 = -7.5 < 0$ 4. Explanation: We designed each set so the sum of their x-components and y-components meets the criteria for the respective quadrant. 5. Final answer: The resultant vectors for the 4 quadrants are: - Quadrant I resultant: $(6.5, 6.5)$ - Quadrant II resultant: $(-5, 7.5)$ - Quadrant III resultant: $(-6.5, -6.5)$ - Quadrant IV resultant: $(5, -7.5)$ This demonstrates it is possible to have 5 vectors whose sum lies in each of the four quadrants.