Question

Determinați parametrul real m a.î. vectorii a=2√3 i + 2m j și b=m i + √3 j să formeze un unghi cu măsura π/4.
Mulțumesc anticipat!

Answer 1

[tex]Ne\ putem\ folosi\ de\ formul\ cosinusului:\\ Daca\ avem\ doi\ vectori\ \vec{a}=u_1\cdot\vec{i}+v_1\cdot\vec{j}\ si\ \vec{b}=u_2\cdot \vec{i}+v_2\cdot {j}\ atunci\\ cosinusul\ dintre\ cei\ doi\ vectori\ este\ egal\ cu:\\ \\ \boxed{cos\ \textless \ (\vec{a}, \vec{b})=\frac{u_1\cdot u_2+v_1\cdot v_2}{\sqrt{u_1^2+v_1^2}\cdot \sqrt{u_2^2+v_2^2}}}\\ Folosim\ aceasta\ formula:\\ \\ \cos \frac{\pi}{4}=\frac{2m\sqrt3+2m\sqrt3}{\sqrt{12+4m^2}\cdot \sqrt{m^2+3}}\\ \\ [/tex]
[tex]\frac{\sqrt2}{2}=\frac{4m\sqrt3}{\sqrt{4(m^2+3)}\cdot \sqrt{m^2+3}}\\ \\ \frac{\sqrt2}{2}=\frac{4m\sqrt3}{2\sqrt{m^2+3}\cdot \sqrt{m^2+3}}\\ \\ \frac{\sqrt2}{2}=\frac{4m\sqrt3}{2m^2+6}\\ 2\sqrt2m^2+6\sqrt2=8m\sqrt3\\ 2\sqrt2m^2-8m\sqrt3+6\sqrt2=0\\ \Delta=192-4\cdot2\sqrt2\cdot6\sqrt2\\ \Delta=192-96\\ \Delta=96\Rightarrow \sqrt{\Delta}}=4\sqrt6\\ \\ m_1=\frac{8\sqrt3+4\sqrt6}{4\sqrt2}=\frac{4(2\sqrt3+\sqrt6)}{4\sqrt2}=\frac{2\sqrt3+\sqrt6}{\sqrt2}=\frac{2\sqrt6+2\sqrt3}{2}=\sqrt6+\sqrt3\\ [/tex]
[tex]Analog:m_2=\sqrt6-\sqrt3\\ S:m\in\{\sqrt6-\sqrt3,\sqrt6+\sqrt3\}[/tex]