Question

Se considera a , b doi vectori necoliniari.
a) Aratati ca vectorii u , v sunt coliniari unde :
- u=2a-3b , v=8a-12b
- u= 8/3a+10/3b , v= -3a-15/4b
b)Determinati numarul real ,,m'' astfel incat vectorii u=(m-1)a+2b si v=3a+mb sa fie coliniari.
c)Determinati numerele reale ,,m'' si ,,n'' astfel incat sa avem ma+nb=(n+1)a+(2-m)b

Answer 1

[tex]Daca\ avem\ doi \ vectori\ \vec{u}=x_1\cdot\vec{a}+y_1\cdot \vec{b}\ si\ \vec{b}=x_2\cdot \vec{a}+y_2\cdot \vec{b}\\ atunci\ conditia\ de\ perpendicularitate\ este:\boxed{\frac{x_1}{x_2}=\frac{y_1}{y_2}}\\ a)\vec{u}=2\cdot \vec{a}-3\cdot \vec{b}.\ \vec{v}=8\cdot \vec{a}-12\cdot \vec{b}\\ \frac{2}{8}=\frac{-3}{-12}\\ \frac{1}{4}=\frac{1}{4}(A)\Rightarrow Vectorii\ sunt\ coliniari.\\ \\ b)\vec{u}=\frac{8}{3}\cdot \vec{a}+\frac{10}{3}\cdot \vec{b},\ \vec{v}=-3\cdot \vec{a}-\frac{15}{4}\cdot \vec{b}\\ [/tex]
[tex]\frac{\frac{8}{3}}{-3}=\frac{\frac{10}{3}}{-\frac{15}{4}}\\ \frac{8}{3}\cdot (-\frac{1}{3})=\frac{10}{3}\cdot (-\frac{4}{15})\\ \frac{-8}{9}=\frac{-8}{9}(A)\Rightarrow Vectorii\ sunt\ coliniari.\\ \\ b)\vec{u}=(m-1)\cdot \vec{a}+2\cdot \vec{b}\ si\ \vec{v}=3\cdot \vec{a}+m\cdot \vec{b}\ sunt\ coliniari.\\ \frac{m-1}{3}=\frac{2}{m}\\ \\ m^2-m=6\\ m^2-m-6=0\\ \Delta=1+24=25\Rightarrow \sqrt{\Delta}=5\\\ m_1=\frac{1+5}{2}=3\\ m_2=\frac{1-5}{2}=-2\\ S:m\in \{3,-2\}\\ \\ [/tex]
[tex]c)m\cdot \vec{a}+n\cdot \vec{b}=(n+1)\cdot \vec{a}+(2-m)\cdot \vec{b}\\ Identificam\ coeficientii\ si\ vom\ obtine\ sistemul:\\ \left \{ {{n+1=m} \atop {2-m=n}} \right.\Leftrightarrow \left \{ {{n+1=m} \atop {2-n=m}} \right. \\ Adunam:3=2m\Rightarrow\boxed{m=\frac{3}{2}}\\ n+1=\frac{3}{2}\\ \boxed{n=\frac{1}{2}}[/tex]