Answer
(a) $-104 \space m^2$
(b) $-148 \space m^2$
(c) $40.6 \space m^2$
Work Step by Step
As we have calculated in the exercise 1.27:
$$A_x = 0 , A_y = -8.00 \space m$$ $$B_x = 7.50 \space m, B_y = 13.0 \space m$$ $$C_x = -10.9 \space m, C_y = -5.07 \space m$$
(a)
$$\vec A \cdot \vec B = A_xB_x + A_yB_y = (0)(7.50 \space m) + (-8.00 \space m)(13.0 \space m)$$ $$\vec A \cdot \vec B = 0 - 104 \space m^2 ´= -104 \space m^2$$
(b)
$$\vec B \cdot \vec C = B_xC_x + B_yC_y = (7.50 \space m)(-10.9 \space m) + (13.0 \space m)(-5.07 \space m)$$ $$\vec B \cdot \vec C = -148 \space m^2$$
(c)
$$\vec A \cdot \vec C = A_xC_x + A_yC_y = (0)(-10.9 \space m) + (-8.00 \space m)(-5.07 \space m)$$ $$\vec B \cdot \vec C = 40.6 \space m^2$$
