Question

геометрия помогите решить 2.61,2.62,2.64 просьба не знаете не пишите пожалуйста ​

Answer 1

Ответ:

$2.61.)\ 1.\ \ \vec{s}_1=(8;2;3)\ \ ,\ \ \vec{s}_2=(-1;2-3)\\\\\vec{s}_1\cdot \vec{s}_2=-8+4-9=-13\ \ ,\\\\|\vec{s}_1|=\sqrt{8^2+2^2+3^2}=\sqrt{77}\ \ \ ,\ \ \ |\vec{s}_2|=\sqrt{1^2+2^2+3^2}=\sqrt{14}\\\\cos\varphi =\dfrac{-13}{\sqrt{77}\cdot \sqrt{14}}=-\dfrac{13}{\sqrt{1078}}\\\\\varphi =arccos\dfrac{13}{\sqrt{1978}}$

$2.\ \ \vec{s}_1=(2;4;3)\ \ ,\ \ \vec{s}_2=(5;1;4)\\\\\vec{s}_1\cdot \vec{s}_2=10+4+12=26\ \ ,\\\\|\vec{s}_1|=\sqrt{2^2+4^2+3^2}=\sqrt{29}\ \ \ ,\ \ \ |\vec{s}_2|=\sqrt{5^2+1^2+4^2}=\sqrt{42}\\\\cos\varphi =\dfrac{26}{\sqrt{29}\cdot \sqrt{42}}=\dfrac{26}{\sqrt{1218}}\\\\\varphi =arccos\dfrac{26}{\sqrt{1218}}$

$2.62)\ \ 1.\ \ \vec{n}_1=(1;2;1)\ \ ,\ \ \vec{n}_2=(1;-1;-2)\\\\\vec{n}_1\cdot \vec{n}_2=1-2-2=-3\ \ ,\\\\|\vec{n}_1|=\sqrt{1^2+2^2+1^2}=\sqrt{6}\ \ \ ,\ \ \ |\vec{n}_2|=\sqrt{1^2+1^2+2^2}=\sqrt{6}\\\\cos\varphi =\dfrac{-3}{\sqrt{6}\cdot \sqrt{6}}=-\dfrac{3}{6}=-\dfrac{1}{2}\ \ \ ,\ \ \ \ \varphi =\dfrac{\pi }{3}$

$2.\ \ \vec{n}_1=(1;3;-2)\ \ ,\ \ \vec{n}_2=(2;3;-2)\\\\\vec{n}_1\cdot \vec{n}_2=2+9+4=15\ \ ,\\\\|\vec{n}_1|=\sqrt{1^2+3^2+2^2}=\sqrt{14}\ \ \ ,\ \ \ |\vec{n}_2|=\sqrt{2^2+3^2+2^2}=\sqrt{17}\\\\cos\varphi =\dfrac{15}{\sqrt{14}\cdot \sqrt{17}}=\dfrac{15}{\sqrt{238}}\ \ \ ,\ \ \ \ \varphi =arccos\dfrac{15}{\sqrt{238}}$

$2.64)\ \ 1.\ \ \vec{s}=(-1;2;5)\ \ ,\ \ \vec{n}=(1;2;3)\\\\\vec{s}\cdot \vec{n}=-1+4+15=18\ \ ,\\\\|\vec{s}|=\sqrt{1^2+2^2+5^2}=\sqrt{30}\ \ \ ,\ \ \ |\vec{n}|=\sqrt{1^2+2^2+3^2}=\sqrt{14}\\\\sin\varphi =\dfrac{|18|}{\sqrt{30}\cdot \sqrt{14}}=\dfrac{18}{\sqrt{420}}=\dfrac{9}{\sqrt{105}}\ \ \ ,\ \ \ \ \varphi =arcsin\dfrac{9}{\sqrt{105}}$

$2.\ \ \vec{s}=(-1;2;3)\ \ ,\ \ \vec{n}=(2;1;-2)\\\\\vec{s}\cdot \vec{n}=-2+2-6=-6\ \ ,\\\\|\vec{s}|=\sqrt{1^2+2^2+3^2}=\sqrt{14}\ \ \ ,\ \ \ |\vec{n}|=\sqrt{2^2+1^2+2^2}=\sqrt{9}=3\\\\sin\varphi =\dfrac{|-6|}{\sqrt{14}\cdot 3}=\dfrac{2}{\sqrt{14}}\ \ \ ,\ \ \ \ \varphi =arcsin\dfrac{2}{\sqrt{14}}$