Question

Терміново. 100 балів 1 завдання

Answer 1

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

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

Рисунок б

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

Рисунок в

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

Рисунок г

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