![\vec{a}_1=(2,1,4)\; ,\; \vec{a}_2=(-3,5,1)\; ,\; \vec{a}_3=(1,-4,-3)\; ,\; \vec{a}_4=(2,-5,-4)\\\\\\(\vec{a}_1,\vec{a}_2,\vec{a}_3)=\left|\begin{array}{ccc}2&1&4\\-3&5&1\\1&-4&-3\end{array}\right|=2\cdot (-15+4)-(9-1)+4\cdot (12-5)=-2\ne 0](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D_1%3D%282%2C1%2C4%29%5C%3B+%2C%5C%3B+%5Cvec%7Ba%7D_2%3D%28-3%2C5%2C1%29%5C%3B+%2C%5C%3B+%5Cvec%7Ba%7D_3%3D%281%2C-4%2C-3%29%5C%3B+%2C%5C%3B+%5Cvec%7Ba%7D_4%3D%282%2C-5%2C-4%29%5C%5C%5C%5C%5C%5C%28%5Cvec%7Ba%7D_1%2C%5Cvec%7Ba%7D_2%2C%5Cvec%7Ba%7D_3%29%3D%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D2%261%264%5C%5C-3%265%261%5C%5C1%26-4%26-3%5Cend%7Barray%7D%5Cright%7C%3D2%5Ccdot+%28-15%2B4%29-%289-1%29%2B4%5Ccdot+%2812-5%29%3D-2%5Cne+0)
Так как определитель не равен нулю, то векторы не компланарны (не лежат в одной плоскости), значит они образуют базис.
Если вектор
разложить по базису
, то можно записать:
![\vec{a}_4=\alpha \cdot \vec{a}_1+\beta \cdot \vec{a}_2+\gamma \cdot \vec{a}_3](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D_4%3D%5Calpha+%5Ccdot+%5Cvec%7Ba%7D_1%2B%5Cbeta+%5Ccdot+%5Cvec%7Ba%7D_2%2B%5Cgamma+%5Ccdot+%5Cvec%7Ba%7D_3)
Такой же линейной зависимостью будут связаны и координаты этих векторов. Это можно записать с помощью системы:
![\left\{\begin{array}{c}2\alpha -3\beta +\gamma =2\\\alpha +5\beta +4\gamma =-5\\4\alpha+\beta -3\gamma =-4\end{array}\right \; \; \left(\begin{array}{cccc}1&5&4&|-5\\2&-3&1&|\; \; \; 2\\4&1&-3&|-4\end{array}\right)\sim \\\\\\1str\cdot (-2)+2str\; \; ,\; \; 2str\cdot (-2)+3str\; \; ,\\\\\sim \left(\begin{array}{cccc}1&5&4&|-5\\0&-13&-7&|\; \; 12\\0&7&-5&|-8\end{array}\right)\sim\; \; 2str\cdot 7+3str\cdot 13\; \sim \left(\begin{array}{cccc}1&5&4&|-5\\0&-13&-7&|\; 12\\0&0&-114&|-20\end{array}\right)](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc%7D2%5Calpha+-3%5Cbeta+%2B%5Cgamma+%3D2%5C%5C%5Calpha+%2B5%5Cbeta+%2B4%5Cgamma+%3D-5%5C%5C4%5Calpha%2B%5Cbeta+-3%5Cgamma+%3D-4%5Cend%7Barray%7D%5Cright+%5C%3B+%5C%3B+%5Cleft%28%5Cbegin%7Barray%7D%7Bcccc%7D1%265%264%26%7C-5%5C%5C2%26-3%261%26%7C%5C%3B+%5C%3B+%5C%3B+2%5C%5C4%261%26-3%26%7C-4%5Cend%7Barray%7D%5Cright%29%5Csim+%5C%5C%5C%5C%5C%5C1str%5Ccdot+%28-2%29%2B2str%5C%3B+%5C%3B+%2C%5C%3B+%5C%3B+2str%5Ccdot+%28-2%29%2B3str%5C%3B+%5C%3B+%2C%5C%5C%5C%5C%5Csim+%5Cleft%28%5Cbegin%7Barray%7D%7Bcccc%7D1%265%264%26%7C-5%5C%5C0%26-13%26-7%26%7C%5C%3B+%5C%3B+12%5C%5C0%267%26-5%26%7C-8%5Cend%7Barray%7D%5Cright%29%5Csim%5C%3B+%5C%3B+2str%5Ccdot+7%2B3str%5Ccdot+13%5C%3B+%5Csim+%5Cleft%28%5Cbegin%7Barray%7D%7Bcccc%7D1%265%264%26%7C-5%5C%5C0%26-13%26-7%26%7C%5C%3B+12%5C%5C0%260%26-114%26%7C-20%5Cend%7Barray%7D%5Cright%29)
![\left\{\begin{array}{c}\alpha +5\beta +4\gamma =-5\\-13\beta -7\gamma =12\\-114\gamma=-20\end{array}\right \; \; \left\{\begin{array}{c}\alpha =-5-5\beta -4\gamma \\-13\beta =12+7\cdot \frac{10}{57}\\\gamma =\frac{10}{57}\end{array}\right \\\\\\\left\{\begin{array}{c}\alpha =-5+\frac{5\cdot 58}{57}-\frac{4\cdot 10}{57}\\\beta =-\frac{58}{57}\\\gamma =\frac{10}{57}\end{array}\right\; \; \; \left\{\begin{array}{ccc}\alpha =-\frac{35}{57}\\\beta =-\frac{58}{57}\\\gamma =\frac{10}{57}\end{array}\right](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc%7D%5Calpha+%2B5%5Cbeta+%2B4%5Cgamma+%3D-5%5C%5C-13%5Cbeta+-7%5Cgamma+%3D12%5C%5C-114%5Cgamma%3D-20%5Cend%7Barray%7D%5Cright+%5C%3B+%5C%3B+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc%7D%5Calpha+%3D-5-5%5Cbeta+-4%5Cgamma+%5C%5C-13%5Cbeta+%3D12%2B7%5Ccdot+%5Cfrac%7B10%7D%7B57%7D%5C%5C%5Cgamma+%3D%5Cfrac%7B10%7D%7B57%7D%5Cend%7Barray%7D%5Cright+%5C%5C%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc%7D%5Calpha+%3D-5%2B%5Cfrac%7B5%5Ccdot+58%7D%7B57%7D-%5Cfrac%7B4%5Ccdot+10%7D%7B57%7D%5C%5C%5Cbeta+%3D-%5Cfrac%7B58%7D%7B57%7D%5C%5C%5Cgamma+%3D%5Cfrac%7B10%7D%7B57%7D%5Cend%7Barray%7D%5Cright%5C%3B+%5C%3B+%5C%3B+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D%5Calpha+%3D-%5Cfrac%7B35%7D%7B57%7D%5C%5C%5Cbeta+%3D-%5Cfrac%7B58%7D%7B57%7D%5C%5C%5Cgamma+%3D%5Cfrac%7B10%7D%7B57%7D%5Cend%7Barray%7D%5Cright)
![\vec{a}_4=-\frac{35}{57}\, \vec{a}_1-\frac{58}{57}\, \vec{a}_2+\frac{10}{57}\, \vec{a}_3](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D_4%3D-%5Cfrac%7B35%7D%7B57%7D%5C%2C+%5Cvec%7Ba%7D_1-%5Cfrac%7B58%7D%7B57%7D%5C%2C+%5Cvec%7Ba%7D_2%2B%5Cfrac%7B10%7D%7B57%7D%5C%2C+%5Cvec%7Ba%7D_3)
(6*7)+(6*8)=130-40
42+48=90
90=90 - верно
7*(23-18)=42-8
7*5=34
35≠34- не верно
280+70=170+180
350=350 - верно
300*2=20*30
120:(6:2)=(120:6):2
120:3=20:2
40≠10 - не верно
(640-140)-100=640-(140-100)
500-100=640-40
400≠600 - не верно
(270+130):2=2*5=(270+130):(2*5)
400:2=10=400:10
200≠10≠40 - не верно
(24+36)*10:2=(24+36)*(10*2)
60*10:2=60*20
300≠1200 - не верно
8+3+6=17 см длина проволоки и он же периметр получившегося треугольника<span />
Скорость сближения = скорость первого + скорость второго.
Дальше проще.
<span>Расстояние делим на скорость сближения. </span>
Пусть у петра х дисков, тогда у кирилла 4х. составляю уравнение
4х-15=х
4х-х=15
3х=15
х=15:3
<span>х=5 </span>
<span>ответ: 5 дисков</span>