Question

Пожалуйста помогите❤

Answer 1

$A_1(-2,2,4)\; ,\; \; A_2(-1,2,0)\; ,\; \; A_3(-1,4,2)\; ,\; \; A_4(3,-3,0)\\\\1)\; \; \overline {A_1A_2}=(1,0,-4)\; \; ,\; \; \overline {A_1A_3}=(1,2,-2)\\\\cos\varphi =\frac{\overline {A_1A_2}\, \cdot \, \overline {A_1A_3}}{|\overline {A_1A_2}|\, \cdot \, |\overline {A_1A_3}|}=\frac{1+0+8}{\sqrt{1+16}\cdot \sqrt{1+4+4}}=\frac{9}{\sqrt{17}\cdot 3}=\frac{3}{\sqrt{17}}\\\\\varphi =arccos\frac{3}{\sqrt{17}}\\\\\\\\2)\; \; S_{A_1A_2A_3}=\frac{1}{2}\cdot \Big |[\overline {A_1A_2}\times \overline {A_1A_3}]\Big |$

$.\; \; \; [\overline {A_1A_2}\times \overline {A_1A_3}]=\left|\begin{array}{ccc}i&j&k\\1&0&-4\\1&2&-2\end{array}\right|=8\vec{i}-2\vec{j}+2\vec{k}\\\\\\\Big |[\overline {A_1A_2}\times \overline {A_1A_3}]\Big |=\sqrt{8^2+2^2+2^2}=\sqrt{72}=6\sqrt2\\\\S=\frac{1}{2}\cdot 6\sqrt2=3\sqrt2$

$3)\; \; V=\frac{1}{6}\cdot \Big |(\overline {A_1A_2},\overline {A_1A_3},\overline {A_1A_4})\Big |\\\\\overline {A_1A_4}=(5,-5,-4)\\\\\\(\overline {A_1A_2},\overline {A_1A_3},\overline {A_1A_4})=\left|\begin{array}{ccc}1&0&-4\\1&2&-2\\5&-5&-4\end{array}\right|=1(-8-10)-4(-4+10)=-42\\\\\\V=\frac{1}{6}\cdot |-42|=\frac{42}{6}=7$