Express the following vectors graphically: A = (2cm, 35°), B = (4cm, 120°), C = (3cm, 210°)

Given $\vec{AB} = (2 \text{cm}, 030^\circ)$:

$$x_{AB} = 2 \cos(30^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \text{ cm}$$ $$y_{AB} = 2 \sin(30^\circ) = 2 \times \frac{1}{2} = 1 \text{ cm}$$

So, $\vec{AB} = (\sqrt{3}, 1) \text{ cm} \approx (1.73, 1) \text{ cm}$.
To draw $\vec{AB}$: Start at the origin $(0,0)$. Draw an arrow 2 cm long at an angle of $30^\circ$ counter-clockwise from the positive x-axis.

Given $\vec{BC} = (6 \text{cm}, 270^\circ)$:

$$x_{BC} = 6 \cos(270^\circ) = 6 \times 0 = 0 \text{ cm}$$ $$y_{BC} = 6 \sin(270^\circ) = 6 \times (-1) = -6 \text{ cm}$$

So, $\vec{BC} = (0, -6) \text{ cm}$.
To draw $\vec{BC}$: Start at the origin $(0,0)$. Draw an arrow 6 cm long pointing directly downwards along the negative y-axis.

Given $\vec{CD} = (4 \text{cm}, 180^\circ)$:

$$x_{CD} = 4 \cos(180^\circ) = 4 \times (-1) = -4 \text{ cm}$$ $$y_{CD} = 4 \sin(180^\circ) = 4 \times 0 = 0 \text{ cm}$$

So, $\vec{CD} = (-4, 0) \text{ cm}$.
To draw $\vec{CD}$: Start at the origin $(0,0)$. Draw an arrow 4 cm long pointing directly left along the negative x-axis.

To express these vectors graphically as a sequence of displacements:

1. Draw a Cartesian coordinate system.
2. Place the tail of $\vec{AB}$ at the origin $(0,0)$. Draw an arrow from $(0,0)$ to $(\sqrt{3}, 1)$. Label the tail A and the head B.
3. Place the tail of $\vec{BC}$ at point B $(\sqrt{3}, 1)$. Draw an arrow from B to $(\sqrt{3}+0, 1-6) = (\sqrt{3}, -5)$. Label the head C.
4. Place the tail of $\vec{CD}$ at point C $(\sqrt{3}, -5)$. Draw an arrow from C to $(\sqrt{3}-4, -5+0) = (\sqrt{3}-4, -5)$. Label the head D.

The final positions are:

• Point A: $(0,0)$
• Point B: $(\sqrt{3}, 1)$
• Point C: $(\sqrt{3}, -5)$
• Point D: $(\sqrt{3}-4, -5)$

The graphical representation involves drawing these arrows on a coordinate plane according to their magnitudes and directions.