Для начала, найдем векторы BA, CD, AD, DS, BS, DA.
Вектор BA:
BA = (xA – xB, yA – yB) = (-2 – 3, 6 – 2) = (-5, 4)
Вектор CD:
CD = (xD – xC, yD – yC) = (7 – 4, 2 – 0) = (3, 2)
Вектор AD:
AD = (xD – xA, yD – yA) = (7 – (-2), 2 – 6) = (9, -4)
Вектор DS:
DS = (xS – xD, yS – yD) = (0 – 7, -2 – 2) = (-7, -4)
Вектор BS:
BS = (xS – xB, yS – yB) = (0 – (-2), -2 – 6) = (2, -8)
Вектор DA:
DA = (xA – xD, yA – yD) = (-2 – 7, 6 – 2) = (-9, 4)
Теперь, найдем скалярное произведение:
BA * CD = (-5 * 3) + (4 * 2) = -15 + 8 = -7
CD * AD = (3 * 9) + (2 * -4) = 27 – 8 = 19
AD * DS = (9 * -7) + (-4 * -4) = -63 + 16 = -47
DS * BS = (-7 * 2) + (-4 * -8) = -14 + 32 = 18
BS * DA = (2 * -9) + (-8 * 4) = -18 – 32 = -50
Таким образом, скалярное произведение векторов BA, CD, AD, DS, BS, DA соответственно равны -7, 19, -47, 18, -50.