Решение:

Расставим оси на рисунке:

х - горизонтально вправо

у - горизонтально на нас

z - вертикально вверх

Тогда скалярное произведение векторов определим через проекции векторов.

Рисунок а

[tex]AB_y = -0.5a\sqrt{2} = -2\sqrt{2} ~; ~~~~~~~~AB_z = -0.5a\sqrt{2} = -2\sqrt{2} ~;\\CD_y = -0.5a\sqrt{2} = -2\sqrt{2} ~; ~~~~~~~~CD_z = 0.5a\sqrt{2} = 2\sqrt{2} ~;\\\\\vec{AB} \cdot \vec {CD} = AB_y\cdot CD_y + AB_z\cdot CD_z = -2\sqrt{2}\cdot(-2\sqrt{2}) + (-2\sqrt{2})\cdot 2\sqrt{2} = 0.[/tex]

Рисунок б

[tex]AB_y = -0.5a\sqrt{2} = -2\sqrt{2} ~; ~~~~~~~~AB_z = 0.5a\sqrt{2} = 2\sqrt{2} ~;\\CD_y = -0.5a\sqrt{2} = -2\sqrt{2} ~; ~~~~~~~~CD_z = 0.5a\sqrt{2} = 2\sqrt{2} ~;\\\\\vec{AB} \cdot \vec {CD} = AB_y\cdot CD_y + AB_z\cdot CD_z = -2\sqrt{2}\cdot(-2\sqrt{2}) + 2\sqrt{2}\cdot 2\sqrt{2} = 16.[/tex]

Рисунок в

[tex]AB_x = -2a\sqrt{2} = -2\sqrt{2}~;~~~~~~~~~~~CD_x = 0;\\ AB_y = 0~; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~CD_y = -0.5a\sqrt{2} = -2\sqrt{2} ~;\\ AB_z = 0.5a\sqrt{2} = 2\sqrt{2} ~~~~~~~~~~~~~~~~CD_z = 0.5a\sqrt{2} = 2\sqrt{2} ~;\\\\\vec{AB} \cdot \vec {CD} = AB_x \cdot CD_x + AB_y\cdot CD_y + AB_z\cdot CD_z =\\ = -2\sqrt{2}\cdot0+ 0\cdot (-2\sqrt{2}) + 2\sqrt{2} \cdot 2\sqrt{2} = 8.[/tex]

Рисунок г

[tex]AB_x = 2a\sqrt{2} = 2\sqrt{2}~;~~~~~~~~~~~CD_x = 0;\\ AB_y = 0~; ~~~~~~~~~~~~~~~~~~~~~~~~~~CD_y = -0.5a\sqrt{2} = -2\sqrt{2} ~;\\ AB_z =- 0.5a\sqrt{2} = -2\sqrt{2} ~~~~~~CD_z = 0.5a\sqrt{2} = 2\sqrt{2} ~;\\\\\vec{AB} \cdot \vec {CD} = AB_x \cdot CD_x + AB_y\cdot CD_y + AB_z\cdot CD_z =\\ = 2\sqrt{2}\cdot0+ 0\cdot (-2\sqrt{2}) - 2\sqrt{2} \cdot 2\sqrt{2} = -8.[/tex]