Question

Геометрия.Задания во вложении.

Answer 1

$1)\; \; A(3;-2;4)\; ,\; B(-2;-3;0)\; ,\; \; C(0;0;0)\; ,\; \; D(3;4;5)\\\\\overline {AB}=(-2-3;-3+2;0-4)=(-5;-1;-4)\\\\\overline {CD}=(3;4;5)\\\\cos \alpha =\frac{\overline{AB}\cdot \overline {CD}}{|\overline {AB}|\cdot |\overline {CD}|}=\frac{-5\cdot 3-1\cdot 4-4\cdot 5}{\sqrt{5^2+1^2+4^2}\cdot \sqrt{3^2+4^2+5^2}}=\frac{-39}{\sqrt{42}\cdot \sqrt{50}}=-\frac{39}{10\sqrt{21}}\\\\ \alpha =\pi -arccos\frac{39}{10\sqrt{21}}$

$2)\; \; A(-3,7,4)\; ,\; B(1,1,1)\; ,\; \; C(2,-4,0)\\\\\left|\begin{array}{ccc}x-1&y-1&z-1\\-3-1&7-1&4-1\\2-1&-4-1&0-1\end{array}\right|=\left|\begin{array}{ccc}x-1&y-1&z-1\\-4&6&3\\1&-5&-1\end{array}\right|=\\\\=(x-1)\cdot (-6+15)-(y-1)\cdot (4-3)+(z-1)\cdot (20-6)=\\\\=9\cdot (x-1)-1\cdot (y-1)+14\cdot (z-20)=\\\\=9x-y+14z-288\; ;\\\\\pi :\; \; 9x-y+14z-288=0$