Question

геометрия 2.65 помогите пожалуйста​

Answer 1

Ответ:

$1)\ \ \pi _1:\ 2x-5y-2z-4=0\ \ \ ,\ \ \ \pi _2:\ x+2y+z-6=0\\\\\vec{n}_1=(2;-5;-2)\ \ ,\ \ \vec{n}_2=(1;2;1)\\\\|\vec{n}_1|=\sqrt{4+25+4}=\sqrt{33}\ \ ,\ \ |\vec{n}_2|=\sqrt{1+4+1}=\sqrt6\\\\\vec{n}_1\cdot \vec{n}_2=2-10-2=-10\\\\cos\alpha =\dfrac{\vec{n}_1\cdot \vec{n}_2}{|\vec{n}_1|\cdot |\vec{n}_2|}=\dfrac{-10}{\sqrt{33}\cdot \sqrt6}=-\dfrac{10}{3\sqrt{22}}\ \ ,\ \ \alpha =arccos\dfrac{10}{3\sqrt{22}}$

$2)\ \ \pi _1:\ x+2y-2z-1=0\ \ \ ,\ \ \ \pi _2:\ 4x-y-2z+3=0\\\\\vec{n}_1=(1;2;-2)\ \ ,\ \ \vec{n}_2=(4;-1;-2)\\\\|\vec{n}_1|=\sqrt{1+4+4}=\sqrt{9}=3\ \ ,\ \ |\vec{n}_2|=\sqrt{16+1+4}=\sqrt{21}\\\\\vec{n}_1\cdot \vec{n}_2=4-2+4=6\\\\cos\alpha =\dfrac{\vec{n}_1\cdot \vec{n}_2}{|\vec{n}_1|\cdot |\vec{n}_2|}=\dfrac{6}{3\sqrt{21}}=\dfrac{2}{\sqrt{21}}\ \ ,\ \ \alpha =arccos\dfrac{2}{\sqrt{21}}$

$3)\ \ \vec{n}_1=(2;1;-2)\ \ ,\ \ \vec{n}_2=(1;3;-2)\\\\|\vec{n}_1|=\sqrt{9}=3\ \ ,\ \ |\vec{n}_2|=\sqrt{14}\\\\\vec{n}_1\cdot \vec{n}_2=2+3+4=9\\\\cos\alpha =\dfrac{\vec{n}_1\cdot \vec{n}_2}{|\vec{n}_1|\cdot |\vec{n}_2|}=\dfrac{9}{3\sqrt{14}}=\dfrac{3}{\sqrt{14}}\ \ ,\ \ \alpha =arccos\dfrac{3}{\sqrt{14}}$

$4)\ \ \vec{n}_1=(3;2;-1)\ \ ,\ \ \vec{n}_2=(2;3;-2)\\\\|\vec{n}_1|=\sqrt{14}\ \ ,\ \ |\vec{n}_2|=\sqrt{17}\\\\\vec{n}_1\cdot \vec{n}_2=6+6+2=14\\\\cos\alpha =\dfrac{\vec{n}_1\cdot \vec{n}_2}{|\vec{n}_1|\cdot |\vec{n}_2|}=\dfrac{14}{\sqrt{14}\cdot \sqrt{17}}=\dfrac{\sqrt{14}}{\sqrt{17}}\ \ ,\ \ \alpha =arccos\sqrt{\dfrac{14}{17}}$